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Unformatted text preview: 5E05(pp 314323) 1/17/06 3:37 PM Page 314 CHAPTER 5
To compute an area we approximate a region by rectangles
and let the number of rectangles become large. The precise
area is the limit of these sums of areas of rectangles. I ntegrals 5E05(pp 314323) 1/17/06 3:37 PM Page 315 In Chapter 2 we used the tangent and velocity problems to  Now is a good time to read (or reread)
A Preview of Calculus (see page 2). It discusses
the unifying ideas of calculus and helps put in
perspective where we have been and where we
are going. introduce the derivative, which is the central idea in differential calculus. In much the same way, this chapter starts
with the area and distance problems and uses them to formulate the idea of a deﬁnite integral, which is the basic
concept of integral calculus. We will see in Chapters 6 and 9 how to use the integral to solve problems concerning volumes, lengths of curves, population predictions, cardiac output, forces on a dam, work, consumer surplus, and baseball,
among many others.
There is a connection between integral calculus and differential calculus. The
Fundamental Theorem of Calculus relates the integral to the derivative, and we will
see in this chapter that it greatly simpliﬁes the solution of many problems.  5.1 Areas and Distances
In this section we discover that in trying to ﬁnd the area under a curve or the distance
traveled by a car, we end up with the same special type of limit. The Area Problem
We begin by attempting to solve the area problem: Find the area of the region S that lies
under the curve y f x from a to b. This means that S, illustrated in Figure 1, is bounded
0], the vertical lines x a and
by the graph of a continuous function f [where f x
x b, and the xaxis.
y y=ƒ
x=a
S
FIGURE 1 0 S=s (x, y)  a¯x¯b, 0¯y¯ƒ d a x=b
b x In trying to solve the area problem we have to ask ourselves: What is the meaning of
the word area ? This question is easy to answer for regions with straight sides. For a rectangle, the area is deﬁned as the product of the length and the width. The area of a triangle
is half the base times the height. The area of a polygon is found by dividing it into triangles (as in Figure 2) and adding the areas of the triangles. A™
w h
l FIGURE 2 A=lw A¡ A£
A¢ b
1
A= 2 bh A=A¡+A™+A£+A¢ 315 5E05(pp 314323) 316 ❙❙❙❙ 1/17/06 3:37 PM Page 316 CHAPTER 5 INTEGRALS However, it isn’t so easy to ﬁnd the area of a region with curved sides. We all have an
intuitive idea of what the area of a region is. But part of the area problem is to make this
intuitive idea precise by giving an exact deﬁnition of area.
Recall that in deﬁning a tangent we ﬁrst approximated the slope of the tangent line by
slopes of secant lines and then we took the limit of these approximations. We pursue a similar idea for areas. We ﬁrst approximate the region S by rectangles and then we take the
limit of the areas of these rectangles as we increase the number of rectangles. The following example illustrates the procedure.
x 2 from 0 to 1 EXAMPLE 1 Use rectangles to estimate the area under the parabola y
Try placing rectangles to estimate the area.
Resources / Module 6
/ What Is Area?
/ Estimating Area under a Parabola (the parabolic region S illustrated in Figure 3).
y
(1, 1) y=≈
S 0 x 1 FIGURE 3
SOLUTION We ﬁrst notice that the area of S must be somewhere between 0 and 1 because S
is contained in a square with side length 1, but we can certainly do better than that. Suppose we divide S into four strips S1, S2 , S3, and S4 by drawing the vertical lines x 1 ,
4
x 1 , and x 3 as in Figure 4(a).
2
4
y y
(1, 1) (1, 1) y=≈ S¢
S™ S£ S¡
0 1
4 1
2 FIGURE 4 x 1 3
4 0 1
4 1
2 (a) 3
4 1 x (b) We can approximate each strip by a rectangle whose base is the same as the strip and
whose height is the same as the right edge of the strip [see Figure 4(b)]. In other words,
the heights of these rectangles are the values of the function f x
x 2 at the right end
1
11
13
3
points of the subintervals [0, 4 ], [ 4 , 2 ], [ 2 , 4 ], and [ 4 , 1].
Each rectangle has width 1 and the heights are ( 1 )2, ( 1 )2, ( 3 )2, and 12. If we let R 4 be
4
4
2
4
the sum of the areas of these approximating rectangles, we get y
(1, 1) y=≈ R4 1
4 ( 1 )2
4 1
4 ( 1 )2
2 1
4 ( 3 )2
4 1
4 12 15
32 0.46875 From Figure 4(b) we see that the area A of S is less than R 4 , so
A
0 1
4 FIGURE 5 1
2 3
4 1 x 0.46875 Instead of using the rectangles in Figure 4(b) we could use the smaller rectangles in
Figure 5 whose heights are the values of f at the left endpoints of the subintervals. (The
leftmost rectangle has collapsed because its height is 0.) The sum of the areas of these 5E05(pp 314323) 1/17/06 3:37 PM Page 317 S ECTION 5.1 AREAS AND DISTANCES ❙❙❙❙ 317 approximating rectangles is
L4 1
4 02 1
4 ( 1 )2
4 1
4 ( 1 )2
2 1
4 ( 3 )2
4 7
32 0.21875 We see that the area of S is larger than L 4 , so we have lower and upper estimates for A:
0.21875 A 0.46875 We can repeat this procedure with a larger number of strips. Figure 6 shows what
happens when we divide the region S into eight strips of equal width. y y
(1, 1) (1, 1) y=≈ 0 F IGURE 6 Approximating S with eight rectangles 1
8 1 x 0 (a) Using left endpoints 1
8 1 x (b) Using right endpoints By computing the sum of the areas of the smaller rectangles L 8 and the sum of the
areas of the larger rectangles R 8 , we obtain better lower and upper estimates for A:
0.2734375
n Ln Rn 10
20
30
50
100
1000 0.2850000
0.3087500
0.3168519
0.3234000
0.3283500
0.3328335 0.3850000
0.3587500
0.3501852
0.3434000
0.3383500
0.3338335 A So one possible answer to the question is to say that the true area of S lies somewhere
between 0.2734375 and 0.3984375.
We could obtain better estimates by increasing the number of strips. The table at the
left shows the results of similar calculations (with a computer) using n rectangles whose
heights are found with left endpoints L n or right endpoints R n . In particular, we see
by using 50 strips that the area lies between 0.3234 and 0.3434. With 1000 strips we
narrow it down even more: A lies between 0.3328335 and 0.3338335. A good estimate is
obtained by averaging these numbers: A 0.3333335.
From the values in the table in Example 1, it looks as if R n is approaching
increases. We conﬁrm this in the next example. y
(1, 1) y=≈ lim R n 1
1
n FIGURE 7 1
3 as n EXAMPLE 2 For the region S in Example 1, show that the sum of the areas of the upper
approximating rectangles approaches 1 , that is,
3 nl 0 0.3984375 1
3 x SOLUTION R n is the sum of the areas of the n rectangles in Figure 7. Each rectangle has width 1 n and the heights are the values of the function f x
x 2 at the points
2
2
2
1 n, 2 n, 3 n, . . . , n n; that is, the heights are 1 n , 2 n , 3 n , . . . , n n 2. 5E05(pp 314323) 318 ❙❙❙❙ 1/17/06 3:37 PM Page 318 CHAPTER 5 INTEGRALS Thus
2 1
n 1
n 1
n Rn 1
n 12
1
n2 12
1
n3 2 2
n
22 22 1
n 2 3
n 32 1
n n
n 2 n2 32 n2 Here we need the formula for the sum of the squares of the ﬁrst n positive integers: The ideas in Examples 1 and 2 are
explored in Module 5.1/5.2/8.7 for a
variety of functions. 12 1 22 32 nn n2 1 2n
6 1 Perhaps you have seen this formula before. It is proved in Example 5 in Appendix E.
Putting Formula 1 into our expression for R n , we get
Rn 1
n3 nn 1 2n
6 1 n 1 2n
6n 2 1 Thus, we have
 Here we are computing the limit of the
sequence R n . Sequences were discussed in
A Preview of Calculus and will be studied in
detail in Chapter 12. Their limits are calculated
in the same way as limits at inﬁnity (Section 4.4).
In particular, we know that
1
0
lim
nl
n lim R n nl n lim 1 2n
6n 2 nl lim 1
6 lim 1
6 nl nl
1
6 n 1 1 2n n
1
n 1 12 1
n 2 1
n 1
3 It can be shown that the lower approximating sums also approach 1 , that is,
3
lim L n nl 1
3 From Figures 8 and 9 it appears that, as n increases, both L n and R n become better and better approximations to the area of S. Therefore, we deﬁne the area A to be the limit of the
y y n=10 R¡¸=0.385 0 F IGURE 8 y n=50 R∞¸=0.3434 n=30 R£¸Å0.3502 1 x 0 1 x 0 1 x 5E05(pp 314323) 1/17/06 3:37 PM Page 319 S ECTION 5.1 AREAS AND DISTANCES y y 0 1 FIGURE 9
The area is the number that is smaller
than all upper sums and larger than
all lower sums x 319 y n=50 L∞¸=0.3234 n=30 L£¸Å0.3169 n=10 L¡¸=0.285 ❙❙❙❙ 0 1 x 0 1 x sums of the areas of the approximating rectangles, that is,
A lim R n 1
3 lim L n nl nl Let’s apply the idea of Examples 1 and 2 to the more general region S of Figure 1. We
start by subdividing S into n strips S1, S2 , . . . , Sn of equal width as in Figure 10. The width
of the interval a, b is b a, so the width of each of the n strips is
x b a
n These strips divide the interval [a, b] into n subintervals
x0, x1 ,
where x 0 a and x n
x1 x 1, x 2 , x2, x3 , ..., x n 1, x n b. The right endpoints of the subintervals are
a x, x2 a 2 x, x3 a 3 x, ... y y=ƒ S¡ 0 a S™ ⁄ S£ ¤ Si ‹ . . . xi1 Sn xi . . . xn1 b x FIGURE 10 Let’s approximate the i th strip Si by a rectangle with width x and height f x i , which
is the value of f at the right endpoint (see Figure 11). Then the area of the i th rectangle
y Îx f(x i) 0 F IGURE 11 a ⁄ ¤ ‹ xi1 xi b x 5E05(pp 314323) 320 ❙❙❙❙ 1/17/06 3:37 PM Page 320 CHAPTER 5 INTEGRALS y is f x i x. What we think of intuitively as the area of S is approximated by the sum of the
areas of these rectangles, which is
Rn 0 a ⁄ bx (a) n=2 f x1 x f x2 x f xn x Figure 12 shows this approximation for n 2, 4, 8, and 12. Notice that this approximation appears to become better and better as the number of strips increases, that is, as
n l . Therefore, we deﬁne the area A of the region S in the following way. y 2 Definition The area A of the region S that lies under the graph of the continuous function f is the limit of the sum of the areas of approximating rectangles: A
0 a ⁄ ¤ ‹ b lim R n lim f x 1 nl x nl f x2 x f xn x x (b) n=4
y It can be proved that the limit in Deﬁnition 2 always exists, since we are assuming that
f is continuous. It can also be shown that we get the same value if we use left endpoints:
3 0 a bx (c) n=8
y A lim L n a x nl f x1 x f xn 1 x In fact, instead of using left endpoints or right endpoints, we could take the height of the
i th rectangle to be the value of f at any number x* in the i th subinterval x i 1, x i . We call
i
**
*
the numbers x 1 , x 2 , . . . , x n the sample points. Figure 13 shows approximating rectangles when the sample points are not chosen to be endpoints. So a more general expression
for the area of S is
A 4
0 lim f x 0 nl *
lim f x 1 *
f x2 x nl *
f xn x x y bx Îx (d) n=12
FIGURE 12
f(x*)
i 0 a
x*
¡ ⁄ ¤ ‹ *
x™ xi1 *
x£ xi b xn1 x*
i x *
xn FIGURE 13
This tells us to
end with i=n.
This tells us
to add.
This tells us to
start with i=m. n μ f(xi) Îx
i=m We often use sigma notation to write sums with many terms more compactly. For
instance,
n f xi
i1 x f x1 x f x2 x f xn x 5E05(pp 314323) 1/17/06 3:37 PM Page 321 S ECTION 5.1 AREAS AND DISTANCES ❙❙❙❙ 321 So the expressions for area in Equations 2, 3, and 4 can be written as follows:
 If you need practice with sigma notation,
look at the examples and try some of the exercises in Appendix E. n A lim f xi nl x i1
n A lim f xi nl x 1 i1
n A f x*
i lim nl x i1 We could also rewrite Formula 1 in the following way:
n nn i2 1 2n
6 i1 1 EXAMPLE 3 Let A be the area of the region that lies under the graph of f x
cos x
between x 0 and x b, where 0 b
2.
(a) Using right endpoints, ﬁnd an expression for A as a limit. Do not evaluate the limit.
(b) Estimate the area for the case b
2 by taking the sample points to be midpoints
and using four subintervals.
SOLUTION (a) Since a 0, the width of a subinterval is
b x f x1 x ib n, and x n f x2 x cos x 1 x b
n b
n cos nb n. The sum of the areas of f xn cos x 2 cos b
n n So x 1 b n, x 2 2b n, x 3 3b n, x i
the approximating rectangles is
Rn 0 x x
cos x n 2b
n b
n cos 2b
n cos x
nb
n b
n According to Deﬁnition 2, the area is
A lim R n nl lim nl b
n cos b
n cos 3b
n cos nb
n Using sigma notation we could write
A lim nl b
n n cos
i1 ib
n It is very difﬁcult to evaluate this limit directly by hand, but with the aid of a computer
algebra system it isn’t hard (see Exercise 25). In Section 5.3 we will be able to ﬁnd A
more easily using a different method. 5E05(pp 314323) 322 ❙❙❙❙ 1/17/06 3:37 PM Page 322 CHAPTER 5 INTEGRALS y (b) With n 4 and b
0, 8 ,
8, 4 ,
vals are y=cos x 1 x*
1
0 π
8 FIGURE 14 π
4 3π
8 π
2 x 2 we have x
4, 3 8 , and 3
3
16 x*
2 16 24
8, so the subintervals are
2 . The midpoints of these subinter 8, 5
16 x*
3 7
16 x*
4 and the sum of the areas of the four approximating rectangles (see Figure 14) is
4 f x*
i x 16 M4 x i1 f cos 8 16 cos f3 16
cos 8
16 cos x 3
16 f5 3
16 cos 5
16 x cos 8 16 5
16 cos f7 16
cos 8 7
16 x
7
16 8 1.006 So an estimate for the area is
A 1.006 The Distance Problem
Now let’s consider the distance problem: Find the distance traveled by an object during a
certain time period if the velocity of the object is known at all times. (In a sense this is the
inverse problem of the velocity problem that we discussed in Section 2.1.) If the velocity
remains constant, then the distance problem is easy to solve by means of the formula
distance velocity time But if the velocity varies, it’s not so easy to ﬁnd the distance traveled. We investigate the
problem in the following example.
EXAMPLE 4 Suppose the odometer on our car is broken and we want to estimate the distance driven over a 30second time interval. We take speedometer readings every ﬁve
seconds and record them in the following table: Time (s)
Velocity (mi h) 0 5 10 15 20 25 30 17 21 24 29 32 31 28 In order to have the time and the velocity in consistent units, let’s convert the velocity
readings to feet per second (1 mi h 5280 3600 ft s):
Time (s)
Velocity (ft s) 0 5 10 15 20 25 30 25 31 35 43 47 46 41 During the ﬁrst ﬁve seconds the velocity doesn’t change very much, so we can estimate
the distance traveled during that time by assuming that the velocity is constant. If we
take the velocity during that time interval to be the initial velocity (25 ft s), then we 5E05(pp 314323) 1/17/06 3:37 PM Page 323 S ECTION 5.1 AREAS AND DISTANCES ❙❙❙❙ 323 obtain the approximate distance traveled during the ﬁrst ﬁve seconds:
25 ft s 5s 125 ft Similarly, during the second time interval the velocity is approximately constant and we
take it to be the velocity when t 5 s. So our estimate for the distance traveled from
t 5 s to t 10 s is
31 ft s 5 s 155 ft
If we add similar estimates for the other time intervals, we obtain an estimate for the
total distance traveled:
25 5 31 5 35 5 43 5 47 5 46 5 1135 ft We could just as well have used the velocity at the end of each time period instead of
the velocity at the beginning as our assumed constant velocity. Then our estimate
becomes
31 5 35 5 43 5 47 5 46 5 41 5 1215 ft If we had wanted a more accurate estimate, we could have taken velocity readings every
two seconds, or even every second.
√
40 20 0 FIGURE 15 10 20 30 t Perhaps the calculations in Example 4 remind you of the sums we used earlier to estimate areas. The similarity is explained when we sketch a graph of the velocity function of
the car in Figure 15 and draw rectangles whose heights are the initial velocities for each
time interval. The area of the ﬁrst rectangle is 25 5 125, which is also our estimate
for the distance traveled in the ﬁrst ﬁve seconds. In fact, the area of each rectangle can be
interpreted as a distance because the height represents velocity and the width represents
time. The sum of the areas of the rectangles in Figure 15 is L 6 1135, which is our initial estimate for the total distance traveled.
In general, suppose an object moves with velocity v f t , where a t b and
ft
0 (so the object always moves in the positive direction). We take velocity readings
at times t0
a , t1, t2 , . . . , tn
b so that the velocity is approximately constant on each
subinterval. If these times are equally spaced, then the time between consecutive readings
b a n. During the ﬁrst time interval the velocity is approximately f t0 and so
is t
the distance traveled is approximately f t0 t. Similarly, the distance traveled during the
second time interval is about f t1 t and the total distance traveled during the time interval a, b is approximately
n f t0 t f t1 t f tn 1 t f ti t 1 i1 If we use the velocity at right endpoints instead of left endpoints, our estimate for the total
distance becomes
n f t1 t f t2 t f tn t f ti t i1 The more frequently we measure the velocity, the more accurate our estimates become, so
it seems plausible that the exact distance d traveled is the limit of such expressions:
n 5 d nl n f ti lim 1 t i1 We will see in Section 5.4 that this is indeed true. lim nl f ti
i1 t 5E05(pp 324333) 324 ❙❙❙❙ 1/17/06 3:38 PM Page 324 CHAPTER 5 INTEGRALS Because Equation 5 has the same form as our expressions for area in Equations 2 and
3, it follows that the distance traveled is equal to the area under the graph of the velocity
function. In Chapters 6 and 9 we will see that other quantities of interest in the natural and
social sciences—such as the work done by a variable force or the cardiac output of the
heart—can also be interpreted as the area under a curve. So when we compute areas in this
chapter, bear in mind that they can be interpreted in a variety of practical ways.  5.1 Exercises 1. (a) By reading values from the given graph of f , use ﬁve rect right endpoints. Sketch the graph and the rectangles. Is
your estimate an underestimate or an overestimate?
(b) Repeat part (a) using left endpoints. angles to ﬁnd a lower estimate and an upper estimate for
the area under the given graph of f from x 0 to x 10.
In each case sketch the rectangles that you use.
(b) Find new estimates using 10 rectangles in each case. 1 x 2 from
x
1 to x 2 using three rectangles and right endpoints. Then improve your estimate by using six rectangles.
Sketch the curve and the approximating rectangles.
(b) Repeat part (a) using left endpoints.
(c) Repeat part (a) using midpoints.
(d) From your sketches in parts (a)–(c), which appears to be
the best estimate? 5. (a) Estimate the area under the graph of f x y 5 y=ƒ 1 1 x 2 , 2 x 2.
(b) Estimate the area under the graph of f using four approximating rectangles and taking the sample points to be
(i) right endpoints
(ii) midpoints
In each case sketch the curve and the rectangles.
(c) Improve your estimates in part (b) by using eight
rectangles. ; 6. (a) Graph the function f x
0 10 x 5 2. (a) Use six rectangles to ﬁnd estimates of each type for the area under the given graph of f from x 0 to x 12.
(i) L 6 (sample points are left endpoints)
(ii) R 6 (sample points are right endpoints)
(iii) M6 (sample points are midpoints)
(b) Is L 6 an underestimate or overestimate of the true area?
(c) Is R 6 an underestimate or overestimate of the true area?
(d) Which of the numbers L 6, R 6, or M6 gives the best
estimate? Explain. 7–8  With a programmable calculator (or a computer), it is possible to evaluate the expressions for the sums of areas of approximating rectangles, even for large values of n, using looping. (On a
TI use the Is command or a ForEndFor loop, on a Casio use Isz,
on an HP or in BASIC use a FORNEXT loop.) Compute the sum
of the areas of approximating rectangles using equal subintervals
and right endpoints for n 10, 30, and 50. Then guess the value of
the exact area. y
8 7. The region under y sin x from 0 to 8. The region under y 1 x 2 from 1 to 2 y=ƒ
4
■ CAS 0 4 8 3. (a) Estimate the area under the graph of f x 25 x 2 from
5 using ﬁve approximating rectangles and 4. (a) Estimate the area under the graph of f x x 0 to x ■ ■ ■ ■ ■ ■ ■ ■ ■ CAS ■ 9. Some computer algebra systems have commands that will draw approximating rectangles and evaluate the sums of their areas,
at least if x* is a left or right endpoint. (For instance, in Maple
i
use leftbox, rightbox, leftsum, and rightsum.)
(a) If f x
s x, 1 x 4, ﬁnd the left and right sums for
n 10, 30, and 50.
(b) Illustrate by graphing the rectangles in part (a).
(c) Show that the exact area under f lies between 4.6 and 4.7. 12 x 1 x from
x 1 to x 5 using four approximating rectangles and
right endpoints. Sketch the graph and the rectangles. Is
your estimate an underestimate or an overestimate?
(b) Repeat part (a) using left endpoints. ■ 10. (a) If f x 2, use the commands dissin sin x , 0 x
cussed in Exercise 9 to ﬁnd the left and right sums for
n 10, 30, and 50. 5E05(pp 324333) 1/17/06 3:38 PM Page 325 S ECTION 5.1 AREAS AND DISTANCES (b) Illustrate by graphing the rectangles in part (a).
(c) Show that the exact area under f lies between 0.87
and 0.91. ❙❙❙❙ 325 Use these data to estimate the height above the Earth’s surface
of the space shuttle Endeavour, 62 seconds after liftoff.
15. The velocity graph of a braking car is shown. Use it to estimate the distance traveled by the car while the brakes are applied. 11. The speed of a runner increased steadily during the ﬁrst three seconds of a race. Her speed at halfsecond intervals is given in
the table. Find lower and upper estimates for the distance that
she traveled during these three seconds. √
(ft /s)
60
40 t (s) 0 0.5 1.0 1.5 2.0 2.5 3.0 v (ft s) 0 6.2 10.8 14.9 18.1 19.4 20.2 20 12. Speedometer readings for a motorcycle at 12second intervals 0 are given in the table.
(a) Estimate the distance traveled by the motorcycle during this
time period using the velocities at the beginning of the time
intervals.
(b) Give another estimate using the velocities at the end of the
time periods.
(c) Are your estimates in parts (a) and (b) upper and lower
estimates? Explain. 4 2 6 t
(seconds) 16. The velocity graph of a car accelerating from rest to a speed of 120 km h over a period of 30 seconds is shown. Estimate the
distance traveled during this period.
√
(km / h)
80 t (s) 0 12 24 36 48 60 v (ft s) 30 28 25 22 24 27 40 0 13. Oil leaked from a tank at a rate of r t liters per hour. The rate decreased as time passed and values of the rate at 2hour time
intervals are shown in the table. Find lower and upper estimates
for the total amount of oil that leaked out.
th
r t (L h) 17–19  Use Deﬁnition 2 to ﬁnd an expression for the area under
the graph of f as a limit. Do not evaluate the limit. 0 2 4 6 8 10 17. f x 4
sx, 1 8.7 7.6 6.8 6.2 5.7 5.3 18. f x 1 x 4, 2 19. f x x cos x, 0 14. When we estimate distances from velocity data, it is sometimes necessary to use times t0 , t1, t2 , t3 , . . . that are not equally
spaced. We can still estimate distances using the time periods
ti ti ti 1. For example, on May 7, 1992, the space shuttle
Endeavour was launched on mission STS49, the purpose of
which was to install a new perigee kick motor in an Intelsat
communications satellite. The table, provided by NASA, gives
the velocity data for the shuttle between liftoff and the jettisoning of the solid rocket boosters. ■ ■ ■ n 20. lim
nl i1 0
185
319
447
742
1325
1445
4151 x 5 x
■ 2
■ ■ ■ ■ ■ ■ ■ 2i
n 5 i1 4n tan 10 i
4n Velocity (ft s) 0
10
15
20
32
59
62
125 2
n n ■ Launch
Begin roll maneuver
End roll maneuver
Throttle to 89%
Throttle to 67%
Throttle to 104%
Maximum dynamic pressure
Solid rocket booster separation ■ 16  Determine a region whose area is equal to the given limit.
Do not evaluate the limit. nl Time (s) x 20–21 21. lim
Event t
30
(seconds) 20 10 ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 22. (a) Use Deﬁnition 2 to ﬁnd an expression for the area under the curve y x 3 from 0 to 1 as a limit.
(b) The following formula for the sum of the cubes of the ﬁrst
n integers is proved in Appendix E. Use it to evaluate the
limit in part (a).
13 23 33 n3 nn 1
2 2 ■ 5E05(pp 324333) 326 CAS ❙❙❙❙ 1/17/06 3:38 PM Page 326 CHAPTER 5 INTEGRALS x 5 from 0 to 2 as a 23. (a) Express the area under the curve y CAS 25. Find the exact area under the cosine curve y cos x from
x 0 to x b, where 0 b
2. (Use a computer algebra
system both to evaluate the sum and compute the limit in
Example 3.) In particular, what is the area if b
2? Compare your answer with the estimate obtained in Example 3(b). limit.
(b) Use a computer algebra system to ﬁnd the sum in your
expression from part (a).
(c) Evaluate the limit in part (a).
CAS x4 24. (a) Express the area under the curve y 5x 2 x from 2
to 7 as a limit.
(b) Use a computer algebra system to evaluate the sum in
part (a).
(c) Use a computer algebra system to ﬁnd the exact area by
evaluating the limit of the expression in part (b).  5.2 26. (a) Let A n be the area of a polygon with n equal sides inscribed in a circle with radius r. By dividing the polygon into n
congruent triangles with central angle 2 n, show that
A n 1 nr 2 sin 2 n .
2
r 2. [Hint: Use Equation 3.5.2.] (b) Show that lim n l A n The Definite Integral
We saw in Section 5.1 that a limit of the form
n 1 f x*
i lim nl *
lim f x 1 x *
f x2 x nl i1 *
f xn x x arises when we compute an area. We also saw that it arises when we try to ﬁnd the distance traveled by an object. It turns out that this same type of limit occurs in a wide variety of situations even when f is not necessarily a positive function. In Chapters 6 and 9 we
will see that limits of the form (1) also arise in ﬁnding lengths of curves, volumes of solids,
centers of mass, force due to water pressure, and work, as well as other quantities. We
therefore give this type of limit a special name and notation.
2 Definition of a Definite Integral If f is a continuous function deﬁned for x b, we divide the interval a, b into n subintervals of equal width
x
b a n. We let x 0
a , x 1, x 2 , . . . , x n ( b) be the endpoints of these
**
subintervals and we let x 1 , x 2 , . . . , x * be any sample points in these subintervals,
n
so x * lies in the i th subinterval x i 1, x i . Then the deﬁnite integral of f from a
i
to b is
a y b a n f x dx f x*
i lim nl x i1 Because we have assumed that f is continuous, it can be proved that the limit in
Deﬁnition 2 always exists and gives the same value no matter how we choose the sample
points x *. (See Note 4 for a precise deﬁnition of this type of limit.) If we take the sample
i
points to be right endpoints, then x *
x i and the deﬁnition of an integral becomes
i
3 y b a n f x dx lim nl f xi
i1 x 5E05(pp 324333) 1/17/06 3:38 PM Page 327 S ECTION 5.2 THE DEFINITE INTEGRAL If we choose the sample points to be left endpoints, then x *
i
becomes y b a xi 1 ❙❙❙❙ 327 and the deﬁnition n f x dx lim nl f xi 1 x i1 Alternatively, we could choose x * to be the midpoint of the subinterval or any other numi
ber between x i 1 and x i .
Although most of the functions that we encounter are continuous, the limit in Deﬁnition 2 also exists if f has a ﬁnite number of removable or jump discontinuities (but not
inﬁnite discontinuities). (See Section 2.5.) So we can also deﬁne the deﬁnite integral for
such functions.
The symbol x was introduced by Leibniz and is called an integral sign. It is
an elongated S and was chosen because an integral is a limit of sums. In the notation
xab f x d x, f x is called the integrand and a and b are called the limits of integration;
a is the lower limit and b is the upper limit. The symbol dx has no ofﬁcial meaning
by itself; xab f x d x is all one symbol. The procedure of calculating an integral is called
integration.
NOTE 1 ■ The deﬁnite integral xab f x d x is a number; it does not depend on x. In fact, we
could use any letter in place of x without changing the value of the integral:
NOTE 2 ■ y b a NOTE 3 ■ y f x dx b a y f t dt b a f r dr The sum
n f x*
i x i1  Bernhard Riemann received his Ph.D. under
the direction of the legendary Gauss at the
University of Göttingen and remained there to
teach. Gauss, who was not in the habit of praising other mathematicians, spoke of Riemann’s
“creative, active, truly mathematical mind and
gloriously fertile originality.” The deﬁnition (2) of
an integral that we use is due to Riemann. He
also made major contributions to the theory of
functions of a complex variable, mathematical
physics, number theory, and the foundations of
geometry. Riemann’s broad concept of space and
geometry turned out to be the right setting, 50
years later, for Einstein’s general relativity theory.
Riemann’s health was poor throughout his life,
and he died of tuberculosis at the age of 39. that occurs in Deﬁnition 2 is called a Riemann sum after the German mathematician
Bernhard Riemann (1826–1866). We know that if f happens to be positive, then the
Riemann sum can be interpreted as a sum of areas of approximating rectangles (see Figure 1). By comparing Deﬁnition 2 with the deﬁnition of area in Section 5.1, we see that
the deﬁnite integral xab f x d x can be interpreted as the area under the curve y f x from
a to b. (See Figure 2.)
y y Îx 0 a x*
i y=ƒ b x 0 a b x FIGURE 1 FIGURE 2 If ƒ˘0, the Riemann sum μ f(x*) Îx
i
is the sum of areas of rectangles. If ƒ˘0, the integral ja ƒ dx is the
area under the curve y=ƒ from a to b. b 5E05(pp 324333) 328 ❙❙❙❙ 1/17/06 3:38 PM Page 328 CHAPTER 5 INTEGRALS y y=ƒ
+ + 0a b _ x If f takes on both positive and negative values, as in Figure 3, then the Riemann sum is
the sum of the areas of the rectangles that lie above the xaxis and the negatives of the areas
of the rectangles that lie below the xaxis (the areas of the gold rectangles minus the areas
of the blue rectangles). When we take the limit of such Riemann sums, we get the situation illustrated in Figure 4. A deﬁnite integral can be interpreted as a net area, that is, a
difference of areas: FIGURE 3 μ f(x*) Î x is an approximation to
i
the net area
y + +
_ bx b a f x dx In the spirit of the precise deﬁnition of the limit of a function in Section 2.4,
we can write the precise meaning of the limit that deﬁnes the integral in Deﬁnition 2 as
follows:
0 there is an integer N such that FIGURE 4
b a A2 ■ For every number j A1 where A 1 is the area of the region above the xaxis and below the graph of f , and A 2 is the
area of the region below the xaxis and above the graph of f .
NOTE 4 y=ƒ
0a y y ƒ dx is the net area b a n f x*
i f x dx x i1 N and for every choice of x* in x i 1, x i .
i for every integer n This means that a deﬁnite integral can be approximated to within any desired degree of
accuracy by a Riemann sum.
Although we have deﬁned xab f x d x by dividing a, b into subintervals of
equal width, there are situations in which it is advantageous to work with subintervals
of unequal width. For instance, in Exercise 14 in Section 5.1 NASA provided velocity data
at times that were not equally spaced, but we were still able to estimate the distance traveled. And there are methods for numerical integration that take advantage of unequal
subintervals.
If the subinterval widths are x 1, x 2 , . . . , x n , we have to ensure that all these widths
approach 0 in the limiting process. This happens if the largest width, max x i , approaches
0. So in this case the deﬁnition of a deﬁnite integral becomes
NOTE 5 ■ y b a n f x dx EXAMPLE 1 Express lim max xi l 0 i 1 f x*
i xi n x3
i lim nl as an integral on the interval 0, x i sin x i x i1 . SOLUTION Comparing the given limit with the limit in Deﬁnition 2, we see that they will
be identical if we choose x3 fx
x sin x x*
i and xi (So the sample points are right endpoints and the given limit is of the form of Equation 3.) We are given that a 0 and b
. Therefore, by Deﬁnition 2 or Equation 3,
we have
n x3
i lim nl i1 x i sin x i x y 0 x3 x sin x d x 5E05(pp 324333) 1/17/06 3:38 PM Page 329 S ECTION 5.2 THE DEFINITE INTEGRAL ❙❙❙❙ 329 Later, when we apply the deﬁnite integral to physical situations, it will be important to
recognize limits of sums as integrals, as we did in Example 1. When Leibniz chose the
notation for an integral, he chose the ingredients as reminders of the limiting process. In
general, when we write
n f x*
i lim nl y x b f x dx a i1 by x, x * by x, and x by dx.
i we replace lim Evaluating Integrals
When we use the deﬁnition to evaluate a deﬁnite integral, we need to know how to work
with sums. The following three equations give formulas for sums of powers of positive
integers. Equation 4 may be familiar to you from a course in algebra. Equations 5 and 6
were discussed in Section 5.1 and are proved in Appendix E.
n nn i 4
i1
n nn i2 5 1
2
1 2n
6 i1
n nn i3 6 1 2 1
2 i1 The remaining formulas are simple rules for working with sigma notation:
n c 7
 Formulas 7–10 are proved by writing out
each side in expanded form. The left side of
Equation 8 is
ca 1
The right side is
c a1 ca 2 an n n ca i 8 c ai i1 ca n a2 nc i1 i1 n n ai 9 bi i1 These are equal by the distributive property. The
other formulas are discussed in Appendix E. bi i1 n i1 n ai 10 n ai bi n ai i1 bi i1 i1 EXAMPLE 2 x3 (a) Evaluate the Riemann sum for f x
endpoints and a 0, b 3, and n 6.
3 (b) Evaluate y x 3
0 Try more problems like this one.
Resources / Module 6
/ What Is Area?
/ Problems and Tests 6 x taking the sample points to be right 6 x d x. SOLUTION (a) With n 6 the interval width is
x b a
n 3 0
6 1
2 5E05(pp 324333) 330 ❙❙❙❙ 1/17/06 3:38 PM Page 330 CHAPTER 5 INTEGRALS and the right endpoints are x 1 0.5, x 2
x 6 3.0. So the Riemann sum is 1.0, x 3 1.5, x 4 2.0, x 5 2.5, and 6 R6 y f xi x i1 f 0.5
5 y=˛6x 1
2 x 2.875 f 1.0 f 1.5 x 5.625 5 x f 2.0 4 0.625 x f 2.5 x f 3.0 x 9 3.9375
0 x 3 Notice that f is not a positive function and so the Riemann sum does not represent a sum
of areas of rectangles. But it does represent the sum of the areas of the gold rectangles
(above the xaxis) minus the sum of the areas of the blue rectangles (below the xaxis) in
Figure 5.
(b) With n subintervals we have FIGURE 5 b x a 3
n n Thus x 0 0, x 1 3 n, x 2 6 n, x 3 9 n, and, in general, x i
using right endpoints, we can use Equation 3: y 3 n x3 6x dx n lim f xi nl 3
n nl lim nl lim nl 5 0 81
4 A™ 81
4 27 nl 3 FIGURE 6
3 81
n4 (˛6x) dx=A¡A™=_6.75 x (Equation 8 with c i1 i (Equations 10 and 8) nn
2 27
4 n
i1
2 1 1
n 3 n) 18
i
n
54
n2 i3 3
n 3i
n 6 n 1 i1 3 27 3
i
n3 i1 lim y=˛6x
A¡ 0 n 3i
n f nl 3i
n i1 81
n4 nl j 3
n n lim y lim i1 lim 0 x 3i n. Since we are 54 n n 1
n2
2 2 27 1 1
n 6.75 This integral can’t be interpreted as an area because f takes on both positive and negative values. But it can be interpreted as the difference of areas A 1 A 2 , where A 1 and A 2
are shown in Figure 6.
Figure 7 illustrates the calculation by showing the positive and negative terms in the
right Riemann sum R n for n 40. The values in the table show the Riemann sums
approaching the exact value of the integral, 6.75, as n l . 5E05(pp 324333) 1/17/06 3:38 PM Page 331 S ECTION 5.2 THE DEFINITE INTEGRAL y n 5 0 x 3 331 Rn 40
100
500
1000
5000 y=˛6x ❙❙❙❙ 6.3998
6.6130
6.7229
6.7365
6.7473 FIGURE 7 R¢¸Å_6.3998 A much simpler method for evaluating the integral in Example 2 will be given in
Section 5.3.
 Because f x
x 4 is positive, the integral
in Example 3 represents the area shown in
Figure 8. EXAMPLE 3 (a) Set up an expression for x25 x 4 dx as a limit of sums.
(b) Use a computer algebra system to evaluate the expression. y SOLUTION x 4, a (a) Here we have f x 2, b 5, and y=x$ So x0
0 b x 300 2 5 2, x 1 2 3 n, x 2 2 a 3
n n 6 n, x 3 2 9 n, and xi 2 3i
n f xi x lim x FIGURE 8 From Equation 3, we get y 5 n x 4 dx n lim nl i1 lim 2 3
n nl n 2
i1 nl 3i
n f2
i1 3i
n 3
n 4 (b) If we ask a computer algebra system to evaluate the sum and simplify, we obtain
n 3i
n 2
i1 4 2062n 4 3045n 3 1170n 2
10n 3 27 Now we ask the CAS to evaluate the limit: y 5 2 x 4 dx lim nl 3
n n 2
i1 3 2062
10 3i
n 3093
5 4 lim nl 3 2062n 4 3045n 3 1170n 2
10n 4 27 618.6 We will learn a much easier method for the evaluation of integrals in the next section. 5E05(pp 324333) 332 ❙❙❙❙ 1/17/06 3:38 PM Page 332 CHAPTER 5 INTEGRALS EXAMPLE 4 Evaluate the following integrals by interpreting each in terms of areas. (a) y 1 0 s1 x 2 dx (b) y 3 0 x 1 dx SOLUTION
y (a) Since f x
s1 x 2 0, we can interpret this integral as the area under the curve
2 from 0 to 1. But, since y 2
y s1 x
1 x 2, we get x 2 y 2 1, which shows
that the graph of f is the quartercircle with radius 1 in Figure 9. Therefore y= œ„„„„„
1≈
or
≈+¥=1 1 1 y s1
0 0 1 x 1
4 x 2 dx 1 2 4 (In Section 8.3 we will be able to prove that the area of a circle of radius r is r 2.)
(b) The graph of y x 1 is the line with slope 1 shown in Figure 10. We compute the
integral as the difference of the areas of the two triangles: FIGURE 9 y 3 0 x 1 dx A1 1
2 A2 22 y 1
2 11 1.5 (3, 2) y=x1
A¡
0 A™ 1 3 x _1 FIGURE 10 The Midpoint Rule
We often choose the sample point x * to be the right endpoint of the i th subinterval because
i
it is convenient for computing the limit. But if the purpose is to ﬁnd an approximation to
an integral, it is usually better to choose x * to be the midpoint of the interval, which we
i
denote by x i . Any Riemann sum is an approximation to an integral, but if we use midpoints
we get the following approximation.
Module 5.1/5.2/8.7 shows how the
Midpoint Rule estimates improve as n
increases. Midpoint Rule y b a n f x dx f xi x x f x1 f xn i1 where x and xi b a
n 1
2 xi 1 xi EXAMPLE 5 Use the Midpoint Rule with n midpoint of x i 1, x i 5 to approximate y 2 1 1
dx.
x 5E05(pp 324333) 1/17/06 3:38 PM Page 333 S ECTION 5.2 THE DEFINITE INTEGRAL ❙❙❙❙ 333 SOLUTION The endpoints of the ﬁve subintervals are 1, 1.2, 1.4, 1.6, 1.8, and 2.0,
so the midpoints are 1.1, 1.3, 1.5, 1.7, and 1.9. The width of the subintervals is
x
2 1 5 1 , so the Midpoint Rule gives
5 y 1
y= x y 2 1 1
dx
x x f 1.1
1
5 1
1.1 f 1.3
1
1.3 f 1.5
1
1.5 f 1.7 1
1.7 f 1.9 1
1.9 0.691908
0 1 2 x FIGURE 11 Since f x
1 x 0 for 1 x 2, the integral represents an area, and the approximation given by the Midpoint Rule is the sum of the areas of the rectangles shown in
Figure 11.
At the moment we don’t know how accurate the approximation in Example 5 is, but in
Section 8.7 we will learn a method for estimating the error involved in using the Midpoint
Rule. At that time we will discuss other methods for approximating deﬁnite integrals.
If we apply the Midpoint Rule to the integral in Example 2, we get the picture in Figure 12. The approximation M40
6.7563 is much closer to the true value 6.75 than the
right endpoint approximation, R 40
6.3998, shown in Figure 7.
y 5 y=˛6x 0 3 x FIGURE 12 M¢¸Å_6.7563 Properties of the Definite Integral
When we deﬁned the deﬁnite integral xab f x dx, we implicitly assumed that a b. But the
deﬁnition as a limit of Riemann sums makes sense even if a b. Notice that if we reverse
a and b, then x changes from b a n to a b n. Therefore y a b If a b, then x f x dx y b a f x dx 0 and so y a a f x dx 0 We now develop some basic properties of integrals that will help us to evaluate integrals
in a simple manner. We assume that f and t are continuous functions. 5E05(pp 334343) 334 ❙❙❙❙ 1/17/06 3:39 PM Page 334 CHAPTER 5 INTEGRALS Properties of the Integral
1.
2. area=c(ba)
0 a b b y f+g g b a
b a
b a c dx cb fx a, t x dx where c is any constant y b a b c y f x dx, c f x dx t x dx y b a b t x dx a where c is any constant a fx y f x dx y f x dx b t x dx a Property 1 says that the integral of a constant function f x
c is the constant times
the length of the interval. If c 0 and a b, this is to be expected because c b a is
the area of the shaded rectangle in Figure 13.
Property 2 says that the integral of a sum is the sum of the integrals. For positive functions it says that the area under f t is the area under f plus the area under t. Figure 14
helps us understand why this is true: In view of how graphical addition works, the corresponding vertical line segments have equal height.
In general, Property 2 follows from Equation 3 and the fact that the limit of a sum is the
sum of the limits: c dx=c(ba) a y b a x F IG URE 13 j y 4. y=c c y 3. y y y f b a n fx t x dx lim nl f xi t xi x i1
n 0 bx a lim nl x t xi i1 n lim b nl [ ƒ+©] dx=
a j b a b ƒ dx+ja © dx  Property 3 seems intuitively reasonable
because we know that multiplying a function by
a positive number c stretches or shrinks its graph
vertically by a factor of c. So it stretches or
shrinks each approximating rectangle by a factor
c and therefore it has the effect of multiplying
the area by c. y b a x i1 n FIGURE 14 j n f xi
f xi x lim nl i1 y f x dx b a t xi x i1 t x dx Property 3 can be proved in a similar manner and says that the integral of a constant
times a function is the constant times the integral of the function. In other words, a constant (but only a constant) can be taken in front of an integral sign. Property 4 is proved by
writing f t f
t and using Properties 2 and 3 with c
1.
EXAMPLE 6 Use the properties of integrals to evaluate y 1 4 0 3x 2 dx. SOLUTION Using Properties 2 and 3 of integrals, we have y 1 0 4 3x 2 d x y 1 0 1 3x 2 dx 4 dx y 4 dx 41 0 y We know from Property 1 that y 1 0 0 1 0 4 4 dx 1 3 y x 2 dx
0 5E05(pp 334343) 1/17/06 3:39 PM Page 335 S ECTION 5.2 THE DEFINITE INTEGRAL
1 1
3 and we found in Example 2 in Section 5.1 that y x 2 dx
0 y 1 3x 2 d x 4 0 y 1 0 3 335 . So 1 3 y x 2 dx 4 dx 4 ❙❙❙❙ 0 1
3 5 The next property tells us how to combine integrals of the same function over adjacent
intervals: y y=ƒ y 5. 0 a c b x FIGURE 15 c a y f x dx b c f x dx y b a f x dx This is not easy to prove in general, but for the case where f x
0 and a c b
Property 5 can be seen from the geometric interpretation in Figure 15: The area under
y f x from a to c plus the area from c to b is equal to the total area from a to b.
EXAMPLE 7 If it is known that x010 f 17 and x08 f x dx x dx 12, ﬁnd x810 f x dx. SOLUTION By Property 5, we have y 8 0 y so 10 8 f x dx y f x dx 10 0 y 10 8 f x dx y f x dx 8 0 y 10 f x dx 0 f x dx 17 12 5 Notice that Properties 1–5 are true whether a b, a b, or a b. The following
properties, in which we compare sizes of functions and sizes of integrals, are true only
if a b.
Comparison Properties of the Integral
6. If f x 0 for a 7. If f x t x for a 8. If m fx b b, then y f x dx x a M for a
mb y b b, then y f x dx x a 0. y b a t x dx. b, then x
a y b a f x dx Mb a M y=ƒ
m
0 a FIGURE 16 b x If f x
0, then xab f x d x represents the area under the graph of f , so the geometric
interpretation of Property 6 is simply that areas are positive. But the property can be
proved from the deﬁnition of an integral (Exercise 64). Property 7 says that a bigger function has a bigger integral. It follows from Properties 6 and 4 because f t 0.
Property 8 is illustrated by Figure 16 for the case where f x
0. If f is continuous we
could take m and M to be the absolute minimum and maximum values of f on the inter 5E05(pp 334343) 336 ❙❙❙ 1/17/06 3:39 PM Page 336 CHAPTER 5 INTEGRALS val a, b . In this case Property 8 says that the area under the graph of f is greater than the
area of the rectangle with height m and less than the area of the rectangle with height M .
Proof of Property 8 Since m M , Property 7 gives fx y b a m dx y b a y f x dx b a M dx Using Property 1 to evaluate the integrals on the left and right sides, we obtain
mb y a b a f x dx Mb a Property 8 is useful when all we want is a rough estimate of the size of an integral without going to the bother of using the Midpoint Rule.
EXAMPLE 8 Use Property 8 to estimate y 4 1 s x d x. S OLUTION Since f x s x is an increasing function, its absolute minimum on 1, 4 is
m f1
1 and its absolute maximum on 1, 4 is M f 4
s4 2. Thus, Property 8 gives
y y=œ„
x 14 2 1 y 4 1 sx dx 24 sx dx 1 6 or
1 3
0 1 4 4 1 x The result of Example 8 is illustrated in Figure 17. The area under y s x from 1 to 4
is greater than the area of the lower rectangle and less than the area of the large rectangle. FIGURE 17  5.2 y Exercises 2 x 2, 0 x 2, with
four subintervals, taking the sample points to be right endpoints. Explain, with the aid of a diagram, what the Riemann
sum represents. 1. Evaluate the Riemann sum for f x 2. If f x 3x 7, 0 x 3, evaluate the Riemann sum with
n 6, taking the sample points to be left endpoints. What does
the Riemann sum represent? Illustrate with a diagram. 3. If f x sx 2, 1 x 6, ﬁnd the Riemann sum with
n 5 correct to six decimal places, taking the sample points to
be midpoints. What does the Riemann sum represent? Illustrate
with a diagram. 4. (a) Find the Riemann sum for f x x 2 sin 2 x, 0 x 3,
with six terms, taking the sample points to be right endpoints. (Give your answer correct to six decimal places.) Explain what the Riemann sum represents with the aid of a
sketch.
(b) Repeat part (a) with midpoints as the sample points.
8
5. The graph of a function f is given. Estimate x0 f x d x using four subintervals with (a) right endpoints, (b) left endpoints,
and (c) midpoints.
y f
1
0 1 x 5E05(pp 334343) 1/17/06 3:40 PM Page 337 ❙❙❙❙ S ECTION 5.2 THE DEFINITE INTEGRAL 6. The graph of t is shown. Estimate x 3 3 t x dx with six sub 15. Use a calculator or computer to make a table of values of right Riemann sums R n for the integral x0 sin x d x with n 5, 10,
50, and 100. What value do these numbers appear to be
approaching? intervals using (a) right endpoints, (b) left endpoints, and
(c) midpoints.
y 16. Use a calculator or computer to make a table of values of g left and right Riemann sums L n and R n for the integral
x02 s1 x 4 d x with n 5, 10, 50, and 100. Between what two
numbers must the value of the integral lie? Can you make a
similar statement for the integral x 2 1 s1 x 4 d x ? Explain. 1 0 337 x 1 17–20 Express the limit as a deﬁnite integral on the given  interval.
n 17. lim
nl 7. A table of values of an increasing function f is shown. Use the
25
0 table to ﬁnd lower and upper estimates for x f x d x. x i sin x i x, 0, i1
n xi 18. lim
nl 1 i1 x, xi 1, 5 n x 0 5 10 15 20 25 nl fx 42 37 25 6 15 s 2 x*
i 19. lim x*
i 2 x, 1, 8] i1 36 n 20. lim 8. The table gives the values of a function obtained from an nl 6
0 experiment. Use them to estimate x f x d x using three equal
subintervals with (a) right endpoints, (b) left endpoints, and
(c) midpoints. If the function is known to be a decreasing function, can you say whether your estimates are less than or
greater than the exact value of the integral?
x 0 1 2 3 4 fx 9.3 9.0 8.3 6.5 5 2.3 11.
■ CAS y 10 2 y 1 0 1 dx, sin x 2 dx,
■ ■ n
n ■ 10. 4 12. 5
■ ■ y 0 y 5 1 ■ sec x 3 dx, n y
y 25. y ■ ■ 5 1 1 sin x 2 d x 0, 2 ■ ■ ■ ■ 2 2 1 y 24. x 2 dx 2 0 22. y 4 1
5 0 x2 ■ 1
dx,
1
■ n
■ ■ 1 2x ■ ■ 5 dx 2x 3 dx x 3 dx
■ ■ ■ ■ ■ ■ ■ ■ 26. (a) Find an approximation to the integral x x
x ■ ■ ■ ■ ■ b2 b 27. Prove that y x dx a2
2 a b3 b 28. Prove that y x 2 dx . a3
3 a . 29–30  Express the integral as a limit of Riemann sums. Do not
evaluate the limit. 0.315 Deduce that the approximation using the Midpoint Rule with
n 5 in Exercise 11 is accurate to two decimal places. 2 x
3x d x using
a Riemann sum with right endpoints and n 8.
(b) Draw a diagram like Figure 3 to illustrate the approximation
in part (a).
(c) Use Equation 3 to evaluate x04 x 2 3x d x.
(d) Interpret the integral in part (c) as a difference of areas and
illustrate with a diagram like Figure 4. 4 tions for Exercise 7 in Section 5.1), compute the left and right
sin x 2 on the interval
Riemann sums for the function f x
0, 1 with n 100. Explain why these estimates show that
0 ■ 3x dx 14. With a programmable calculator or computer (see the instruc y x, 4
0 and graphs the corresponding rectangles (use middlesum and
middlebox commands in Maple), check the answer to Exercise 11 and illustrate with a graph. Then repeat with n 10
and n 20. 0.306 5 6 13. If you have a CAS that evaluates midpoint approximations 1 6 x*
i  Use the form of the deﬁnition of the integral given in
Equation 3 to evaluate the integral. ■ sx 3 2 21–25 23. 10.5 9–12  Use the Midpoint Rule with the given value of n to
approximate the integral. Round the answer to four decimal places. 9. ■ 21. 6 7.6 ■ 3 x*
i 4
i1 29.
■ y x 6 1 2
■ x5
■ 30. dx
■ ■ ■ ■ y 2 0 x 2 sin x dx
■ ■ ■ ■ ■ 5E05(pp 334343) 338 CAS ❙❙❙❙ 1/17/06 3:40 PM Page 338 CHAPTER 5 INTEGRALS 31–32  Express the integral as a limit of sums. Then evaluate,
using a computer algebra system to ﬁnd both the sum and the limit.
31. y ■ ■ y 32. sin 5x dx 0 ■ ■ ■ ■ 10 2 ■ evaluate x25 1 integrals to evaluate x14 2 x 2 ■ ■ ■ (c) 2 0 y 7 5 (b) f x dx (d) f x dx y 5 f x dx 0 y 9 b
47. Write as a single integral in the form xa f x d x : f x dx 0 y y=ƒ 2
2 2 4 6 x 9
0 x 8 2f x y f x dx 5
48. If x1 f x dx 49. If x09 f x d x
0 2 cos x dx 1
(from Exercise 25 in Section 5.1), together with the properties
of integrals, to evaluate x0 2 2 cos x 5x dx. y
2 1 dx. 46. Use the result of Exercise 27 and the fact that x0 it in terms of areas. y 3x ■ 33. The graph of f is shown. Evaluate each integral by interpreting (a) 3 x 4 dx. 45. Use the results of Exercises 27 and 28 and the properties of x 6 dx ■ 44. Use the properties of integrals and the result of Example 3 to 5 2 12 and x45 f x dx 1 y f x dx 2 f x dx 3.6, ﬁnd x14 f x dx. 37 and x09 t x d x
3 t x dx. 16, ﬁnd 5
50. Find x0 f x dx if 3 for x
x for x fx 3
3 34. The graph of t consists of two straight lines and a semicircle. Use it to evaluate each integral.
(a) y 2 0 (b) t x dx y 6 2 (c) t x dx y 7 0 51–54  Use the properties of integrals to verify the inequality
without evaluating the integrals. t x dx 51. y=© 2 y 4 y 52. y
4 0
2 s5 1 7x 4 54. 35.  31
2
0 y( x y 36. 1 dx 39.
■ 0 y (1 s9 3 y 2
1
■ 0 y ■ 2 sx x 2 dx s1 6 2 1 sin 2 x dx 2 s2 sin x dx ■ ■ 1 dx 3
■ ■ ■ ■ ■ ■ ■ ■ Evaluate the integral by interpreting it in terms of areas.
55–60 2
2 x 2 ) dx y 38. y 40. x dx
■ ■ ■ 9 41. Given that y s x d x
4 ■ 38
3 3
1
10 0 ■ 55. 3 y 59. y 1
dx
x 2 y Use Property 8 to estimate the value of the integral. 1 56. y 58. y 60. y 2 0 sx 3 1 dx 2x dx x ■  x 2 dx s4 57. 37. 1 y 6 ■ 35–40 1 4 y x dx y 53. 2
0 sin 3 x dx 3
4 tan x dx 2 x3 0 3x 3 dx 5 dx
■ ■ ■ 4 ■ 1
1 ■ , what is y st dt ? x 4 dx s1 ■ ■ ■ ■ ■ 3 ■ 4
4 ■ sin2x dx
■ ■ ■ ■ 9 61–62  Use properties of integrals, together with Exercises 27
and 28, to prove the inequality. 1 42. Evaluate y x 2 cos x dx.
1 1
43. In Example 2 in Section 5.1 we showed that x0 x 2 dx Use this fact and the properties of integrals to evaluate
x01 5 6 x 2 dx. 1
3 . 61.
■ y 3 sx 4 1
■ 1 dx
■ ■ 26
3
■ 62.
■ ■ 2 y 0
■ 2 x sin x dx
■ ■ 8
■ ■ 5E05(pp 334343) 1/17/06 3:40 PM Page 339 D ISCOVERY PROJECT AREA FUNCTIONS 63. Prove Property 3 of integrals. 67–68 i1 65. If f is continuous on a, b , show that y b a [Hint: fx y f x dx fx 68. lim b nl f x dx a ■ f x .] 2 0 f x sin 2 x dx 1
n i1 ■ xi y 2 0 [Hint: Consider f x n 1
in 1 ■ 69. Find x12 x 66. Use the result of Exercise 65 to show that y i4
n5 67. lim
nl 1 ■ x 4.] 2
■ ■ ■ ■ ■ ■ ■ 2 dx. Hint: Choose x* to be the geometric mean of
i
and x i (that is, x*
sx i 1 x i ) and use the identity
i
1
mm 1 f x dx 1
m 1
m 1 DISCOVERY PROJECT
Area Functions
1. (a) Draw the line y 2 t 1 and use geometry to ﬁnd the area under this line, above the
taxis, and between the vertical lines t 1 and t 3.
(b) If x 1, let A x be the area of the region that lies under the line y 2 t 1 between
t 1 and t x. Sketch this region and use geometry to ﬁnd an expression for A x .
(c) Differentiate the area function A x . What do you notice? 2. (a) If x 339 Express the limit as a deﬁnite integral.  n 64. Prove Property 6 of integrals. ❙❙❙❙ 1, let y Ax x
1 1 t 2 dt A x represents the area of a region. Sketch that region.
(b) Use the result of Exercise 28 in Section 5.2 to ﬁnd an expression for A x .
(c) Find A x . What do you notice?
(d) If x
1 and h is a small positive number, then A x h
A x represents the area
of a region. Describe and sketch the region.
(e) Draw a rectangle that approximates the region in part (d). By comparing the areas of
these two regions, show that
Ax h
h Ax 1 x2 (f) Use part (e) to give an intuitive explanation for the result of part (c). ; 3. (a) Draw the graph of the function f x cos x 2 in the viewing rectangle 0, 2 by 1.25, 1.25 .
(b) If we deﬁne a new function t by
tx y x 0 cos t 2 dt then t x is the area under the graph of f from 0 to x [until f x becomes negative, at
which point t x becomes a difference of areas]. Use part (a) to determine the value of
x at which t x starts to decrease. [Unlike the integral in Problem 2, it is impossible to
evaluate the integral deﬁning t to obtain an explicit expression for t x .]
(c) Use the integration command on your calculator or computer to estimate t(0.2), t(0.4),
t(0.6), . . . , t(1.8), t(2). Then use these values to sketch a graph of t.
(d) Use your graph of t from part (c) to sketch the graph of t using the interpretation of
t x as the slope of a tangent line. How does the graph of t compare with the graph
of f ? ■ 5E05(pp 334343) 340 ❙❙❙❙ 1/17/06 3:40 PM Page 340 CHAPTER 5 INTEGRALS 4. Suppose f is a continuous function on the interval a, b and we deﬁne a new function t by the equation y tx x a f t dt Based on your results in Problems 1–3, conjecture an expression for t x .  5.3 The Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus is appropriately named because it establishes a
connection between the two branches of calculus: differential calculus and integral calculus. Differential calculus arose from the tangent problem, whereas integral calculus arose
from a seemingly unrelated problem, the area problem. Newton’s teacher at Cambridge,
Isaac Barrow (1630–1677), discovered that these two problems are actually closely
related. In fact, he realized that differentiation and integration are inverse processes. The
Fundamental Theorem of Calculus gives the precise inverse relationship between the
derivative and the integral. It was Newton and Leibniz who exploited this relationship and
used it to develop calculus into a systematic mathematical method. In particular, they saw
that the Fundamental Theorem enabled them to compute areas and integrals very easily
without having to compute them as limits of sums as we did in Sections 5.1 and 5.2.
The ﬁrst part of the Fundamental Theorem deals with functions deﬁned by an equation
of the form Investigate the area function interactively.
Resources / Module 6
/ Areas and Derivatives
/ Area as a Function
y y tx 1 x a f t dt y=f(t)
area=© 0 a x b FIGURE 1 t where f is a continuous function on a, b and x varies between a and b. Observe that t
depends only on x, which appears as the variable upper limit in the integral. If x is a ﬁxed
number, then the integral xax f t dt is a deﬁnite number. If we then let x vary, the number
xax f t dt also varies and deﬁnes a function of x denoted by t x .
If f happens to be a positive function, then t x can be interpreted as the area under the
graph of f from a to x, where x can vary from a to b. (Think of t as the “area so far” function; see Figure 1.) x0x f EXAMPLE 1 If f is the function whose graph is shown in Figure 2 and t x t dt,
ﬁnd the values of t 0 , t 1 , t 2 , t 3 , t 4 , and t 5 . Then sketch a rough graph of t. x00 f SOLUTION First we notice that t 0 y
2 t dt 0. From Figure 3 we see that t 1 is the area of a triangle: y=f(t)
1
0 1 2 4 y t1 t 1 0 1
2 f t dt 12 1 f t dt 1 To ﬁnd t 2 we add to t 1 the area of a rectangle: FIGURE 2 t2 y 2 0 f t dt y 1 0 f t dt y 2 1 12 3 5E05(pp 334343) 1/17/06 3:40 PM Page 341 SECTION 5.3 THE FUNDAMENTAL THEOREM OF CALCULUS ❙❙❙❙ 341 We estimate that the area under f from 2 to 3 is about 1.3, so
t3 y t2 3 2 f t dt 3 1.3 4.3 y
2 y
2 y
2 y
2 y
2 1 1 1 1 1 0 1 t 0 1 g(1)=1 t 2 0 g(2)=3 1 2 3 t 0 1 2 4 t 0 1 2 4 t g(3)Å4.3 FIGURE 3 g(4)Å3 For t g(5)Å1.7 3, f t is negative and so we start subtracting areas: y t4 g t3 y t5 4 t4 y 4 3 f t dt 4.3 f t dt 1.3 3 3.0 3
2 5 4 1.3 1.7 1
0 1 2 3 4 We use these values to sketch the graph of t in Figure 4. Notice that, because f t is
positive for t 3, we keep adding area for t 3 and so t is increasing up to x 3,
where it attains a maximum value. For x 3, t decreases because f t is negative. 5x FIGURE 4
x ©=j f(t) dt If we take f t a t and a 0, then, using Exercise 27 in Section 5.2, we have y tx h
ƒ
a FIGURE 5 x 0 t dt x2
2 Notice that t x
x, that is, t
f . In other words, if t is deﬁned as the integral of f by
Equation 1, then t turns out to be an antiderivative of f , at least in this case. And if we
sketch the derivative of the function t shown in Figure 4 by estimating slopes of tangents,
f in Example 1 too.
we get a graph like that of f in Figure 2. So we suspect that t
To see why this might be generally true we consider any continuous function f with
fx
0. Then t x
xax f t dt can be interpreted as the area under the graph of f from
a to x, as in Figure 1.
In order to compute t x from the deﬁnition of derivative we ﬁrst observe that,
t x is obtained by subtracting areas, so it is the area under the
for h 0, t x h
graph of f from x to x h (the gold area in Figure 5). For small h you can see from the
ﬁgure that this area is approximately equal to the area of the rectangle with height f x and
width h : y 0 x x+h b tx t so h tx tx h
h tx hf x
fx 5E05(pp 334343) 342 ❙❙❙❙ 1/17/06 3:41 PM Page 342 CHAPTER 5 INTEGRALS Intuitively, we therefore expect that
tx tx lim h
h hl0 tx fx The fact that this is true, even when f is not necessarily positive, is the ﬁrst part of the
Fundamental Theorem of Calculus.
The Fundamental Theorem of Calculus, Part 1 If f is continuous on a, b , then the func We abbreviate the name of this theorem as
FTC1. In words, it says that the derivative of a
deﬁnite integral with respect to its upper limit is
the integrand evaluated at the upper limit. tion t deﬁned by y tx x f t dt a a x b is continuous on a, b and differentiable on a, b , and t x
Proof If x and x tx h are in a, b , then h tx y xh x a y xh x 2
y m y x f t dt a xh x y f t dt x a f t dt (by Property 5) f t dt 0,
tx h
h tx 1
h y xh x f t dt M mh
that is, F IGURE 6 y For now let us assume that h 0. Since f is continuous on x, x h , the Extreme
Value Theorem says that there are numbers u and v in x, x h such that f u
m and
fv
M , where m and M are the absolute minimum and maximum values of f on
x, x h . (See Figure 6.)
By Property 8 of integrals, we have y=ƒ 0 f t dt
f t dt a y and so, for h f x. xu √=x+h y f uh y xh x x Since h xh x f t dt Mh f t dt f vh 0, we can divide this inequality by h :
fu 1
h y xh x f t dt fv Now we use Equation 2 to replace the middle part of this inequality:
3 fu tx h
h tx fv Inequality 3 can be proved in a similar manner for the case h 0. (See Exercise 55.) 5E05(pp 334343) 1/17/06 3:41 PM Page 343 SECTION 5.3 THE FUNDAMENTAL THEOREM OF CALCULUS Module 5.3 provides visual evidence
for FTC1. ❙❙❙❙ Now we let h l 0. Then u l x and v l x, since u and v lie between x and x
Therefore
lim f u lim f u lim f v h. fx lim f v 343 fx hl0 u lx and
hl0 v lx because f is continuous at x. We conclude, from (3) and the Squeeze Theorem, that
tx 4 lim tx h
h hl0 tx fx If x a or b, then Equation 4 can be interpreted as a onesided limit. Then Theorem 3.2.4 (modiﬁed for onesided limits) shows that t is continuous on a, b .
Using Leibniz notation for derivatives, we can write FTC1 as
d
dx 5 y x a f t dt fx when f is continuous. Roughly speaking, Equation 5 says that if we ﬁrst integrate f and
then differentiate the result, we get back to the original function f . y EXAMPLE 2 Find the derivative of the function t x
SOLUTION Since f t s1 x 0 s1 t 2 dt. t 2 is continuous, Part 1 of the Fundamental Theorem of Calculus gives
tx s1 EXAMPLE 3 Although a formula of the form t x x2 xax f t dt may seem like a strange way
of deﬁning a function, books on physics, chemistry, and statistics are full of such functions. For instance, the Fresnel function
Sx y x 0 sin t 2 2 dt is named after the French physicist Augustin Fresnel (1788–1827), who is famous for
his works in optics. This function ﬁrst appeared in Fresnel’s theory of the diffraction of
light waves, but more recently it has been applied to the design of highways.
Part 1 of the Fundamental Theorem tells us how to differentiate the Fresnel function:
Sx sin x2 2 This means that we can apply all the methods of differential calculus to analyze S (see
Exercise 49). 5E05(pp 344353) 344 ❙❙❙❙ 1/17/06 3:44 PM Page 344 CHAPTER 5 INTEGRALS Figure 7 shows the graphs of f x
sin x 2 2 and the Fresnel function
x
Sx
x0 f t dt. A computer was used to graph S by computing the value of this integral
for many values of x. It does indeed look as if S x is the area under the graph of f from
0 to x [until x 1.4 when S x becomes a difference of areas]. Figure 8 shows a larger
part of the graph of S.
y y
1 f 0 0.5 S
x 1 x 1 FIGURE 7 FIGURE 8 x The Fresnel function S(x)=j0 sin (πt@/2) dt ƒ=sin (π≈/2)
x S(x)=j sin (πt@/2) dt
0 If we now start with the graph of S in Figure 7 and think about what its derivative
should look like, it seems reasonable that S x
f x . [For instance, S is increasing
when f x
0 and decreasing when f x
0.] So this gives a visual conﬁrmation of
Part 1 of the Fundamental Theorem of Calculus.
EXAMPLE 4 Find d
dx y x4 1 sec t dt. SOLUTION Here we have to be careful to use the Chain Rule in conjunction with FTC1. Let u x 4. Then
d
dx y x4 1 sec t dt d
dx y u 1 sec t dt d
du y sec u du
dx u 1 sec t dt sec x 4 du
dx (by the Chain Rule) (by FTC1) 4x 3 In Section 5.2 we computed integrals from the deﬁnition as a limit of Riemann sums
and we saw that this procedure is sometimes long and difﬁcult. The second part of the
Fundamental Theorem of Calculus, which follows easily from the ﬁrst part, provides us
with a much simpler method for the evaluation of integrals.
The Fundamental Theorem of Calculus, Part 2 If f is continuous on a, b , then
 We abbreviate this theorem as FTC2. y b a f x dx Fb Fa where F is any antiderivative of f , that is, a function such that F f. 5E05(pp 344353) 1/17/06 3:44 PM Page 345 S ECTION 5.3 THE FUNDAMENTAL THEOREM OF CALCULUS Proof Let t x ❙❙❙❙ 345 xax f t dt. We know from Part 1 that t x
f x ; that is, t is an antiderivative of f . If F is any other antiderivative of f on a, b , then we know from
Corollary 4.2.7 that F and t differ by a constant:
Fx 6 tx C for a x b. But both F and t are continuous on a, b and so, by taking limits of both
sides of Equation 6 (as x l a and x l b ), we see that it also holds when x a and
x b.
If we put x a in the formula for t x , we get y ta
So, using Equation 6 with x
Fb b and x
Fa a a f t dt 0 a, we have tb C tb ta ta C y tb b a f t dt Part 2 of the Fundamental Theorem states that if we know an antiderivative F of f , then
we can evaluate xab f x d x simply by subtracting the values of F at the endpoints of the
interval a, b . It’s very surprising that xab f x d x, which was deﬁned by a complicated procedure involving all of the values of f x for a x b, can be found by knowing the values of F x at only two points, a and b.
Although the theorem may be surprising at ﬁrst glance, it becomes plausible if we interpret it in physical terms. If v t is the velocity of an object and s t is its position at time t,
s t , so s is an antiderivative of v. In Section 5.1 we considered an object that
then v t
always moves in the positive direction and made the guess that the area under the velocity
curve is equal to the distance traveled. In symbols: y b a v t dt sb sa That is exactly what FTC2 says in this context.
EXAMPLE 5 Evaluate the integral y 1
2 x 3 dx. x 3 is continuous on 2, 1 and we know from Sec14
tion 4.10 that an antiderivative is F x
4 x , so Part 2 of the Fundamental Theorem
gives
SOLUTION The function f x y 1
2 x 3 dx F1 F 2 1
4 1 4 1
4 2 4 15
4 Notice that FTC2 says we can use any antiderivative F of f. So we may as well use
14
14
7 or 1 x 4 C.
the simplest one, namely F x
4 x , instead of 4 x
4
We often use the notation
Fx b
a Fb Fa 5E05(pp 344353) 346 ❙❙❙❙ 1/17/06 3:44 PM Page 346 CHAPTER 5 INTEGRALS So the equation of FTC2 can be written as y b f x dx a Fx
b
a Other common notations are F x b where a b
a and F x x 2 is F x
using Part 2 of the Fundamental Theorem:
SOLUTION An antiderivative of f x y A 1 0 1 x3
3 2 x dx f .
x 2 from 0 to 1. EXAMPLE 6 Find the area under the parabola y  In applying the Fundamental Theorem we
use a particular antiderivative F of f . It is not
necessary to use the most general antiderivative. F 0 1
3 x 3. The required area A is found 13
3 03
3 1
3 If you compare the calculation in Example 6 with the one in Example 2 in Section 5.1,
you will see that the Fundamental Theorem gives a much shorter method. y y=cos x 1 EXAMPLE 7 Find the area under the cosine curve from 0 to b, where 0
SOLUTION Since an antiderivative of f x cos x is F x b 2. sin x, we have area=1
0 π
2 x A y b 0 cos x dx In particular, taking b
from 0 to 2 is sin
2 FIGURE 9 sin x b
0 sin b sin 0 sin b 2, we have proved that the area under the cosine curve
1. (See Figure 9.) When the French mathematician Gilles de Roberval ﬁrst found the area under the sine
and cosine curves in 1635, this was a very challenging problem that required a great deal
of ingenuity. If we didn’t have the beneﬁt of the Fundamental Theorem, we would have to
compute a difﬁcult limit of sums using obscure trigonometric identities (or a computer
algebra system as in Exercise 25 in Section 5.1). It was even more difﬁcult for Roberval
because the apparatus of limits had not been invented in 1635. But in the 1660s and 1670s,
when the Fundamental Theorem was discovered by Barrow and exploited by Newton and
Leibniz, such problems became very easy, as you can see from Example 7.
EXAMPLE 8 What is wrong with the following calculation?  y 3
1 1
dx
x2 1 x 1 3 1 1
3 1 4
3 SOLUTION To start, we notice that this calculation must be wrong because the answer is
negative but f x
1 x 2 0 and Property 6 of integrals says that xab f x d x 0 when
f 0. The Fundamental Theorem of Calculus applies to continuous functions. It can’t
be applied here because f x
1 x 2 is not continuous on 1, 3 . In fact, f has an inﬁnite discontinuity at x 0, so y 3
1 1
dx
x2 does not exist 5E05(pp 344353) 1/17/06 3:44 PM Page 347 SECTION 5.3 THE FUNDAMENTAL THEOREM OF CALCULUS ❙❙❙❙ 347 D ifferentiation and Integration as Inverse Processes
We end this section by bringing together the two parts of the Fundamental Theorem.
The Fundamental Theorem of Calculus Suppose f is continuous on a, b .
1. If t x
2. y b a y x a f x dx f t dt, then t x
Fb f x. F a , where F is any antiderivative of f , that is, F f. We noted that Part 1 can be rewritten as
d
dx y x a f t dt fx which says that if f is integrated and then the result is differentiated, we arrive back at the
f x , Part 2 can be rewritten as
original function f . Since F x y b a F x dx Fb Fa This version says that if we take a function F , ﬁrst differentiate it, and then integrate the
F a . Taken
result, we arrive back at the original function F , but in the form F b
together, the two parts of the Fundamental Theorem of Calculus say that differentiation
and integration are inverse processes. Each undoes what the other does.
The Fundamental Theorem of Calculus is unquestionably the most important theorem
in calculus and, indeed, it ranks as one of the great accomplishments of the human mind.
Before it was discovered, from the time of Eudoxus and Archimedes to the time of Galileo
and Fermat, problems of ﬁnding areas, volumes, and lengths of curves were so difﬁcult
that only a genius could meet the challenge. But now, armed with the systematic method
that Newton and Leibniz fashioned out of the Fundamental Theorem, we will see in the
chapters to come that these challenging problems are accessible to all of us.  5.3 Exercises 1. Explain exactly what is meant by the statement that “differenti ation and integration are inverse processes.”
2. Let t x x0x f t dt, where f is the function whose graph is shown.
(a) Evaluate t x for x 0, 1, 2, 3, 4, 5, and 6.
(b) Estimate t 7 .
(c) Where does t have a maximum value? Where does it have a
minimum value? (d) Sketch a rough graph of t.
y
3
2
1
0 1 4 6 t 5E05(pp 344353) 348 ❙❙❙❙ 1/17/06 3:45 PM Page 348 CHAPTER 5 INTEGRALS x0x f 3. Let t x t dt, where f is the function whose graph 12. F x
13. h x is shown.
(a) Evaluate t 0 , t 1 , t 2 , t 3 , and t 6 .
(b) On what interval is t increasing?
(c) Where does t have a maximum value?
(d) Sketch a rough graph of t. y
y 15. y y 17. y y t 1 ■ y 21. y 23. y 25. y 27. y 29. y 31. y 33. y 35. y 36. 3 1
0 y t 1 Sketch the area represented by t x . Then ﬁnd t x in two
ways: (a) by using Part 1 of the Fundamental Theorem and (b) by
evaluating the integral using Part 2 and then differentiating.
 ■ ■ x 1 t 2 dt ■ 6. t x ■ ■ ■ ■ ■ x y (1 9. t y
11. F x y x 0 y y 2 y 2 x 2 s1 ■ ■ y ■ 10. t u t sin t dt x 1 y ■ t 4 5 dt 2 y x 2 cos t 2 dt ■ ■ t sin t dt sin3t dt ■ ■ 22. y x 4 5 dx 24. y 3
dt
t4 26. y 28. y 30. y 32. y (3 34. y 8 4x 2 1 0 2 1 5 2 3 dx 2
dx
x3 5 x5 dx x2 0 4 0 2 sec 2 t dt csc 2 d 2 f x dx 0 ■ x x2 5 y x4
x5 where f x f x dx
■ ■ ■ 2 ■ ■ ■ 4 6 dx 1 0 ■ ■ sx dx 3
2 2 4 1 x 5 dx cos d
1
dx
sx 1 x sx ) dx 0 6 0 if 0
if 1 ■ y 2 dy 3y 83 1 csc cot d x
x x
if
sin x if 0 where f x ■ 1
2
x 0 x
■ ■ ■ ■ ; 37–40  Use a graph to give a rough estimate of the area of the
region that lies beneath the given curve. Then ﬁnd the exact area. 37. y 1 u 3 cos t 2 dt ■ x 5 dx 3
sx, 0 x 27 38. y x 4, 1 x 6 39. y 2 Hint: y cos t 2 dt
x 8. t x 2t dt st ) dt 0 7–18  Use Part 1 of the Fundamental Theorem of Calculus to
ﬁnd the derivative of the function. 7. t x 0 1 x2 20. 1 f t dt, where f is the function whose graph is f y y cos x 1 r 3 dr s1  Use Part 2 of the Fundamental Theorem of Calculus to
evaluate the integral, or explain why it does not exist. y 5. t x y 18. y du x2 0 19–36 shown.
(a) Evaluate t 3 and t 3 .
(b) Estimate t 2 , t 1 , and t 0 .
(c) On what interval is t increasing?
(d) Where does t have a maximum value?
(e) Sketch a rough graph of t.
(f) Use the graph in part (e) to sketch the graph of t x .
Compare with the graph of f . 5–6 u2 1 1 3x y 14. h x
16. y u3 1 ■ 19.
3 sin4t dt cos t
dt
t sx 3 ■ 0 x 1x f 1 4. Let t x tan d 2 y x 10 x sin x, 0 x 40. y sec2x, 0 x dx ■ ■ ■ ■ 3
■ ■ ■ ■ ■ ■ ■ ■ 5E05(pp 344353) 1/17/06 3:45 PM Page 349 S ECTION 5.3 THE FUNDAMENTAL THEOREM OF CALCULUS 41– 42  Evaluate the integral and interpret it as a difference of
areas. Illustrate with a sketch. 41. y 2 ■ x 3 dx 1
■ y 42. ■ ■ ■ ■ 5 2
4 ■ y
■ ■ y 43. t x 3x 2x u2
u2 1
du
1 3x Hint: y f u du y 2x y 44. t x s2 tan x y f u du 3x 0 f u du  51. 1 x2 0 2x t 4 x0x f Let t x y
3 dt 2 y 45. y
■ x3 sx ■ 47. If F x ■ ■ y x 1 y 46. y st sin t dt
■ ■ ■ cos x ■ y f t dt, where f t 5x t2 u4 s1
u 1 cos u 2 du ■ 1 t dt, where f is the function whose graph
is shown.
(a) At what values of x do the local maximum and minimum
values of t occur?
(b) Where does t attain its absolute maximum value?
(c) On what intervals is t concave downward?
(d) Sketch the graph of t. Find the derivative of the function.  sin t
dt
t ■ 51–52
43–46 x 0 ■ 349 (d) Does this function have horizontal asymptotes?
(e) Solve the following equation correct to one decimal place: sin x dx ■ ❙❙❙❙ ■ f 1
■ 0
_1 ■ 2 4 6 t 8 _2 du, ﬁnd F 2 .
52. 48. Find the interval on which the curve y y 1
t x 0 1 y f
0.4 t2 dt
0.2 is concave upward.
0 49. The Fresnel function S was deﬁned in Example 3 and graphed CAS in Figures 7 and 8.
(a) At what values of x does this function have local maximum
values?
(b) On what intervals is the function concave upward?
(c) Use a graph to solve the following equation correct to two
decimal places: y 0 CAS sin 2 t 2 dt 0.2 50. The sine integral function ■ ■ ■ 53. lim
nl 54. lim y x 0 sin t
dt
t is important in electrical engineering. [The integrand
ft
sin t t is not deﬁned when t 0, but we know that its
1 and this makes f
limit is 1 when t l 0. So we deﬁne f 0
a continuous function everywhere.]
(a) Draw the graph of Si.
(b) At what values of x does this function have local maximum
values?
(c) Find the coordinates of the ﬁrst inﬂection point to the right
of the origin. 5 7 t 9 ■ ■ ■ ■ ■ ■ ■ ■ ■ 53–54  Evaluate the limit by ﬁrst recognizing the sum as a Riemann sum for a function deﬁned on 0, 1 . nl Si x 3 _0.2 n
x 1 ■ ■ i1 i3
n4 1
n 1
n
■ 2
n
■ 3
n ■ ■ n
n
■ 55. Justify (3) for the case h ■ ■ ■ ■ ■ 0. 56. If f is continuous and t and h are differentiable functions, ﬁnd a formula for
d
dx
57. (a) Show that 1 (b) Show that 1 y hx tx f t dt s1 x 3 1
x01 s1 x 3 dx x 3 for x
1.25. 0. 5E05(pp 344353) 350 ❙❙❙❙ 1/17/06 3:45 PM Page 350 CHAPTER 5 INTEGRALS 62. A hightech company purchases a new computing system 58. Let 0
x
2
0 fx and if
if
x if
if y tx x
0
1
x x 0
x
x
2 whose initial value is V . The system will depreciate at the rate
f f t and will accumulate maintenance costs at the rate
t t t , where t is the time measured in months. The company
wants to determine the optimal time to replace the system.
(a) Let 1
2 0 59. Find a function f and a number a such that 6
for all x y a and
2, h 1
2,
13, and h is continu ■ ■ 1
t 63–68 63. A y t 0 y
y f s ds What does C represent and why would the company want
to minimize C?
(c) Show that C has a minimum value at the numbers t T
where C T
fT. 30 30 t 0 ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ The following exercises are intended only for those who have
already covered Chapter 7. 65. Ct  5.4 Vt 2
12,900 t Determine the length of time T for the total depreciation
x0t f s ds to equal the initial value V .
Dt
(c) Determine the absolute minimum of C on 0, T .
(d) Sketch the graphs of C and f t in the same coordinate
system, and verify the result in part (a) in this case. 61. A manufacturing company owns a major piece of equipment that depreciates at the (continuous) rate f f t , where t is the
time measured in months since its last overhaul. Because a
ﬁxed cost A is incurred each time the machine is overhauled,
the company wants to determine the optimal time T (in
months) between overhauls.
(a) Explain why x0t f s ds represents the loss in value of the
machine over the period of time t since the last overhaul.
(b) Let C C t be given by t s ds V
t if 0
450
if t 0. h1
3, h 2
6, h 2
5, h 2
ous everywhere. Evaluate x12 h u du. fs tt 2 sx 60. Suppose h is a function such that h 1 t 0 V
15
0 ft ft
dt
t2 y Show that the critical numbers of C occur at the numbers t
where C t
ft
tt.
(b) Suppose that (a) Find an expression for t x similar to the one for f x .
(b) Sketch the graphs of f and t.
(c) Where is f differentiable? Where is t differentiable? x 1
t Ct f t dt  1
dx
2x 9 1 ■ y 1
1 64. s1
eu ■ 1 66. dt t2 du
■ ■ ■ ■ y
y 68. 6 s3 2 12 67. Evaluate the integral. y ■ 1 0 10 x dx
4 1 t2 0
2 1
u2 4
u 1
■ 3
■ dt
du
■ ■ ■ Indefinite Integrals and the Net Change Theorem
We saw in Section 5.3 that the second part of the Fundamental Theorem of Calculus provides a very powerful method for evaluating the deﬁnite integral of a function, assuming
that we can ﬁnd an antiderivative of the function. In this section we introduce a notation
for antiderivatives, review the formulas for antiderivatives, and use them to evaluate deﬁnite integrals. We also reformulate FTC2 in a way that makes it easier to apply to science
and engineering problems. 5E05(pp 344353) 1/17/06 3:46 PM Page 351 SECTION 5.4 INDEFINITE INTEGRALS AND THE NET CHANGE THEOREM ❙❙❙❙ 351 I ndefinite Integrals
Both parts of the Fundamental Theorem establish connections between antiderivatives and
deﬁnite integrals. Part 1 says that if f is continuous, then xax f t dt is an antiderivative of f.
F a , where F is an antiPart 2 says that xab f x dx can be found by evaluating F b
derivative of f.
We need a convenient notation for antiderivatives that makes them easy to work with.
Because of the relation given by the Fundamental Theorem between antiderivatives and
integrals, the notation x f x d x is traditionally used for an antiderivative of f and is called
an indeﬁnite integral. Thus yf x dx Fx Fx means fx For example, we can write yx 2 x3
3 dx C x3
3 d
dx because x2 C So we can regard an indeﬁnite integral as representing an entire family of functions (one
antiderivative for each value of the constant C ).
 You should distinguish carefully between deﬁnite and indeﬁnite integrals. A deﬁnite
integral xab f x d x is a number, whereas an indeﬁnite integral x f x d x is a function (or
family of functions). The connection between them is given by Part 2 of the Fundamental
Theorem. If f is continuous on a, b , then y b a yf f x dx x dx b
a The effectiveness of the Fundamental Theorem depends on having a supply of antiderivatives of functions. We therefore restate the Table of Antidifferentiation Formulas
from Section 4.10, together with a few others, in the notation of indeﬁnite integrals. Any
formula can be veriﬁed by differentiating the function on the right side and obtaining the
integrand. For instance 1 y cf C d
tan x
dx C y tan x because fx t x dx yf yx 2 y sec x dx n xn 1
n1 sec2x Table of Indefinite Integrals x dx y k dx kx y sin x dx
2 y sec x dx c y f x dx
C
cos x
tan x y sec x tan x dx y cos x dx C 2 y csc x dx C
sec x dx C sin x
cot x y csc x cot x dx C ytx x dx
n 1 C
C
csc x C dx 5E05(pp 344353) 352 ❙❙❙❙ 1/17/06 3:46 PM Page 352 CHAPTER 5 INTEGRALS Recall from Theorem 4.10.1 that the most general antiderivative on a given interval is
obtained by adding a constant to a particular antiderivative. We adopt the convention that
when a formula for a general indeﬁnite integral is given, it is valid only on an interval. Thus, we write
1
1
C
y x 2 dx
x
with the understanding that it is valid on the interval 0,
or on the interval
, 0 . This
is true despite the fact that the general antiderivative of the function f x
1 x 2, x 0, is
1
x
1
x Fx if x 0 C2 if x 0 EXAMPLE 1 Find the general indeﬁnite integral  The indeﬁnite integral in Example 1 is
graphed in Figure 1 for several values of C.
The value of C is the yintercept. y 4 10 x 4 2 sec 2x d x SOLUTION Using our convention and Table 1, we have 10 x 4 y
_1.5 C1 2 sec2x d x 10 y x 4 dx 1.5 10
_4 FIGURE 1 2 y sec2x dx x5
5 2x 5 2 tan x
2 tan x C
C You should check this answer by differentiating it.
EXAMPLE 2 Evaluate cos
d.
sin2 y SOLUTION This indeﬁnite integral isn’t immediately apparent in Table 1, so we use trigonometric identities to rewrite the function before integrating: y cos
d
sin2 y 1
sin y csc
EXAMPLE 3 Evaluate y 3 0 x3 cos
sin
cot d d
csc C 6 x d x. SOLUTION Using FTC2 and Table 1, we have y 3 0 x 3 6x dx x4
4 ( 1
4 81
4 3 x2
6
2
3 4 27 0 3 32 )
0 0 (1
4 04
6.75 Compare this calculation with Example 2(b) in Section 5.2. 3 02 ) 5E05(pp 344353) 1/17/06 3:46 PM Page 353 SECTION 5.4 INDEFINITE INTEGRALS AND THE NET CHANGE THEOREM  Figure 2 shows the graph of the integrand in
Example 4. We know from Section 5.2 that the
value of the integral can be interpreted as the
sum of the areas labeled with a plus sign minus
the areas labeled with a minus sign. E XAMPLE 4 Find y 12 SOLUTION The Fundamental Theorem gives y y 12 x 0 12 x2
2 12 sin x dx y=x12 sin x 1
2 + 12 cos x
0 12 2 12 cos 12 0 72 +
_ _ 12 cos 12 60 10 353 12 sin x dx. x 0 ❙❙❙❙ cos 0 12 12 cos 12 12 x This is the exact value of the integral. If a decimal approximation is desired, we can use
a calculator to approximate cos 12. Doing so, we get y FIGURE 2 12 x 0 EXAMPLE 5 Evaluate y 9 2t 2 12 sin x dx t 2 st
t2 1 1 70.1262 dt. SOLUTION First we need to write the integrand in a simpler form by carrying out the division: y 9 2t 2 1 t 2 st
t2 1 y dt 9 1 18 t 3
2 18 9 2t 2 1
t 2 32
3 t 1
1
9 32 1
9 dt
9 1 1
2
3 9 2 t t3 2 2t [2 t1 2 2 (2
2
3 1 2
3 1
32 1 9 1 32 1
1 ) 4
9 Applications
Part 2 of the Fundamental Theorem says that if f is continuous on a, b , then y b a f x dx Fb where F is any antiderivative of f. This means that F
ten as y b a F x dx Fb Fa
f , so the equation can be rewrit Fa We know that F x represents the rate of change of y F x with respect to x and
Fb
F a is the change in y when x changes from a to b. [Note that y could, for instance,
increase, then decrease, then increase again. Although y might change in both directions,
Fb
F a represents the net change in y.] So we can reformulate FTC2 in words as
follows. 5E05(pp 354363) 354 ❙❙❙❙ 1/17/06 3:47 PM Page 354 CHAPTER 5 INTEGRALS The Net Change Theorem The integral of a rate of change is the net change: y b a F x dx Fb Fa This principle can be applied to all of the rates of change in the natural and social sciences that we discussed in Section 3.4. Here are a few instances of this idea:
■ If V t is the volume of water in a reservoir at time t, then its derivative V t is
the rate at which water ﬂows into the reservoir at time t. So y t2 t1 V t dt V t2 V t1 is the change in the amount of water in the reservoir between time t1 and time t2 .
■ If C t is the concentration of the product of a chemical reaction at time t, then
the rate of reaction is the derivative d C dt. So y dC
dt
dt t2 t1 C t2 C t1 is the change in the concentration of C from time t1 to time t2 .
■ If the mass of a rod measured from the left end to a point x is m x , then the
linear density is x
m x . So y b x dx a mb ma is the mass of the segment of the rod that lies between x
■ a and x b. If the rate of growth of a population is dn dt, then y t2 t1 dn
dt
dt n t2 n t1 is the net change in population during the time period from t1 to t2 . (The population increases when births happen and decreases when deaths occur. The net
change takes into account both births and deaths.)
■ If C x is the cost of producing x units of a commodity, then the marginal cost is
the derivative C x . So y x2 x1 C x dx C x2 C x1 is the increase in cost when production is increased from x1 units to x2 units.
■ If an object moves along a straight line with position function s t , then its
velocity is v t
s t , so
2 y t2 t1 v t dt s t2 s t1 is the net change of position, or displacement, of the particle during the time
period from t1 to t2 . In Section 5.1 we guessed that this was true for the case
where the object moves in the positive direction, but now we have proved that it
is always true. 5E05(pp 354363) 1/17/06 3:47 PM Page 355 S ECTION 5.4 INDEFINITE INTEGRALS AND THE NET CHANGE THEOREM ■ ❙❙❙❙ If we want to calculate the distance traveled during the time interval, we have to
consider the intervals when v t
0 (the particle moves to the right) and also the
intervals when v t
0 (the particle moves to the left). In both cases the distance
is computed by integrating v t , the speed. Therefore y 3 t2 t1 vt dt total distance traveled Figure 3 shows how both displacement and distance traveled can be interpreted
in terms of areas under a velocity curve.
√ √(t) t™ displacement=j √(t) dt=A¡A™+A£
t¡ A¡ t™ A£
0 t¡ t t™ A™ distance=j  √(t) dt=A¡+A™+A£
t¡ F IGURE 3
■ The acceleration of the object is a t y t2 t1 v t , so a t dt v t2 v t1 is the change in velocity from time t1 to time t2 .
Resources / Module 7
/ Physics and Engineering
/ Start of Spy Tracks EXAMPLE 6 A particle moves along a line so that its velocity at time t is
vt
t 2 t 6 (measured in meters per second). (a) Find the displacement of the particle during the time period 1
(b) Find the distance traveled during this time period. t 4. SOLUTION (a) By Equation 2, the displacement is
s4 y s1 4 1 y v t dt t3
3 4 1 t2 t 4 t2
2 6 dt
9
2 6t
1 This means that the particle moved 4.5 m toward the left.
(b) Note that v t
t2 t 6
t 3 t 2 and so v t
0 on the interval 1, 3
and v t
0 on 3, 4 . Thus, from Equation 3, the distance traveled is
 To integrate the absolute value of v t , we
use Property 5 of integrals from Section 5.2 to
split the integral into two parts, one where
vt
0 and one where v t
0. y 4 1 vt dt y 3 1 y 3 1 v t dt t2
t3
3 61
6 y 4 3 v t dt t 6 dt t2
2 6t 10.17 m 3 1 y 4 3 t2
t3
3 t 6 dt t2
2 6t 4 3 355 5E05(pp 354363) 356 ❙❙❙❙ 1/17/06 3:48 PM Page 356 CHAPTER 5 INTEGRALS EXAMPLE 7 Figure 4 shows the power consumption in the city of San Francisco for a day
in September (P is measured in megawatts; t is measured in hours starting at midnight).
Estimate the energy used on that day.
P
800
600
400
200 0 FIGURE 4 3 6 9 12 15 18 t 21 Pacific Gas & Electric SOLUTION Power is the rate of change of energy: P t
Theorem, y 24 0 y P t dt 24 0 E t dt E t . So, by the Net Change
E 24 E0 is the total amount of energy used that day. We approximate the value of the integral
using the Midpoint Rule with 12 subintervals and t 2: y 24 P t dt P1 P3 440 400 420 620 790 840 840 0 P5 P 21 P 23 810 690 670 t 550 2 850 15,840
The energy used was approximately 15,840 megawatthours.
How did we know what units to use for energy in Example 7? The integral x024 P t dt is
deﬁned as the limit of sums of terms of the form P t * t. Now P t * is measured in
i
i
megawatts and t is measured in hours, so their product is measured in megawatthours.
The same is true of the limit. In general, the unit of measurement for xab f x dx is the product of the unit for f x and the unit for x.  A note on units  5.4
1– 4 1.
2.  Exercises Verify by differentiation that the formula is correct. y sx x
2 1 dx sx 2 1 4. yx x sin x cos x ■ 5–14 C ■  5.
3. y sa 1
2 x2 3 dx x
a 2 sa 2 sx 2 a sx 2 a 2
a 2x dx 2 C C
■ y x cos x dx 1
2 x2 C yx 7. y ■ ■ ■ ■ ■ ■ ■ Find the general indeﬁnite integral.
34 x3 6. dx
6x 1 dx y sx dx 8. yx1 3 2 x 4 dx ■ ■ 5E05(pp 354363) 1/17/06 3:48 PM Page 357 S ECTION 5.4 INDEFINITE INTEGRALS AND THE NET CHANGE THEOREM 9. y 1 10. t 2 dt t2 y u2 11. y (2 s x ) 2 dx 12. y sin 13. y1 sin x
dx
sin2x 14. y 357 43. The area of the region that lies to the right of the yaxis and to 1
u2 1 ❙❙❙❙ du the left of the parabola x 2y y 2 (the shaded region in the
ﬁgure) is given by the integral x02 2y y 2 d y. (Turn your head
clockwise and think of the region as lying below the curve
x 2y y 2 from y 0 to y 2.) Find the area of the region. sin 2 x
dx
sin x 3 cos d y
2 ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ x=2y¥ ■ ; 15–16  Find the general indeﬁnite integral. Illustrate by graphing
several members of the family on the same screen. 15. y x sx dx 16. y cos x 0 2 sin x d x x 1
■ ■ ■ 17– 40 17. y 19. y 21. y 23. y 25. 27. 6x 0
0 2 6y 2 y 1 2
dy
y3 4y 3 2 26. sx ) d x 3 28. 4 5
dx
x 4 y 4 sin 0 64 1 1
1 39. y (x 4 y 2 ■ ■ ■ ■ ■ 2 2v 9 5y7 y y
y 1 dv 0 dx x 46. The current in a wire is deﬁned as the derivative of the charge: It xab I t sec
sin Q t . (See Example 3 in Section 3.4.) What does
dt represent? 47. If oil leaks from a tank at a rate of r t gallons per minute at tan d
sin
sec2 3 1 45. If w t is the rate of growth of a child in pounds per year, what
does x510 w t dt represent? 2 time t, what does x0120 r t dt represent?
tan2 48. A honeybee population starts with 100 bees and increases d at a rate of n t bees per week. What does 100
represent? x015 n t dt x 2 3 dx 1 49. In Section 4.8 we deﬁned the marginal revenue function R x 8 1 as the derivative of the revenue function R x , where x is the
5000
number of units sold. What does x1000 R x d x represent? x1
dx
3
sx 2 3 0
■ 2 50. If f x is the slope of a trail at a distance of x miles from the sin x d x
■ start of the trail, what does x35 f x d x represent? ■ ■ ■ y x x 2 x 4. Then use this information to estimate the
area of the region that lies under the curve and above the xaxis.
2x 3x 4 51. If x is measured in meters and f x is measured in newtons, what are the units for x0100 f x d x ? ; 41. Use a graph to estimate the xintercepts of the curve ; 42. Repeat Exercise 41 for the curve y y=1
y=$ „
œx 3x 2
dx
sx
3 1 y dy 1
x x 4 0 dx u 2 du 5 3v 0 y 4
y 1, and the curve y sx . Find the area of this region by
writing x as a function of y and integrating with respect to y
(as in Exercise 43). s 2 t dt 2 y ■ u3 y3 y 40.
■ 9 4x 3 1 4 1 38. 2 x ) dx 1 u5 2 1 y 36. 5
sx 4 ) dx 5 0 ■ ■ 2x 1 3 sx
dx
sx y (sx y 32. d 0 0 34. 3 cos 1 0 30. cos2
d
cos2 1 4 37. ■ y 3 1 24. 1 y 20. 14 d y y t dt 0 33. 18. 5 dx y 0 ■ 44. The boundaries of the shaded region are the yaxis, the line 22. y 35. 4x y x (sx 31. ■ 1 2 du 1 y ■ st 1 4 1 29. ■ 3u 2 y 2 5y 4 3 ■ Evaluate the integral.  2 ■ 6 2x . 52. If the units for x are feet and the units for a x are pounds per foot, what are the units for da d x? What units does x28 a x d x
have? 5E05(pp 354363) 358 ❙❙❙❙ 1/17/06 3:48 PM Page 358 CHAPTER 5 INTEGRALS 62. Water ﬂows in and out of a storage tank. A graph of the rate of 53–54  The velocity function (in meters per second) is given
for a particle moving along a line. Find (a) the displacement and
(b) the distance traveled by the particle during the given time
interval. 53. v t 3t 5, 54. v t 2 2t ■ t ■ ■ 0 t ■ 3 1 8, t change r t of the volume of water in the tank, in liters per day,
is shown. If the amount of water in the tank at time t 0 is
25,000 L, use the Midpoint Rule to estimate the amount of
water four days later. ■ r
2000 6 ■ ■ ■ ■ ■ ■ ■ The acceleration function (in m s2 ) and the initial velocity are given for a particle moving along a line. Find (a) the
velocity at time t and (b) the distance traveled during the given
time interval.
55–56 55. a t t 56. a t 2t ■ 1000  ■ 4, 5, v0 0 3, v 0 ■ ■ t 4, 0
■ ■ 0 t 2 4t 3 _1000 10 ■ 1 3
■ ■ ■ ■ ■ 57. The linear density of a rod of length 4 m is given by x
9 2 s x measured in kilograms per meter, where
x is measured in meters from one end of the rod. Find the total
mass of the rod.
58. Water ﬂows from the bottom of a storage tank at a rate of rt
200 4 t liters per minute, where 0 t 50. Find the
amount of water that ﬂows from the tank during the ﬁrst
10 minutes.
59. The velocity of a car was read from its speedometer at 10second intervals and recorded in the table. Use the Midpoint
Rule to estimate the distance traveled by the car.
t (s) v (mi h) t (s) v (mi h) 0
10
20
30
40
50 0
38
52
58
55
51 60
70
80
90
100 56
53
50
47
45 63. Economists use a cumulative distribution called a Lorenz curve to describe the distribution of income between households in a
given country. Typically, a Lorenz curve is deﬁned on 0, 1
with endpoints 0, 0 and 1, 1 , and is continuous, increasing,
and concave upward. The points on this curve are determined
by ranking all households by income and then computing the
percentage of households whose income is less than or equal
to a given percentage of the total income of the country. For
example, the point a 100, b 100 is on the Lorenz curve if the
bottom a% of the households receive less than or equal to b%
of the total income. Absolute equality of income distribution
would occur if the bottom a% of the households receive a% of
the income, in which case the Lorenz curve would be the line
y x. The area between the Lorenz curve and the line y x
measures how much the income distribution differs from
absolute equality. The coefﬁcient of inequality is the ratio of
the area between the Lorenz curve and the line y x to the
area under y x.
y
(1, 1) 1 y=x
y=L (x) 60. Suppose that a volcano is erupting and readings of the rate r t at which solid materials are spewed into the atmosphere are
given in the table. The time t is measured in seconds and the
units for r t are tonnes (metric tons) per second.
t 0 1 2 3 4 5 6 rt 2 10 24 36 46 54 60 0 (a) Show that the coefﬁcient of inequality is twice the area
between the Lorenz curve and the line y x, that is, show
that
coefficient of inequality (a) Give upper and lower estimates for the quantity Q 6 of
erupted materials after 6 seconds.
(b) Use the Midpoint Rule to estimate Q 6 .
61. The marginal cost of manufacturing x yards of a certain fabric
2 is C x
3 0.01x 0.000006 x (in dollars per yard). Find
the increase in cost if the production level is raised from
2000 yards to 4000 yards. x 1 1 2y x
0 L x dx (b) The income distribution for a certain country is represented
by the Lorenz curve deﬁned by the equation
Lx 5
12 x2 7
12 x What is the percentage of total income received by the
bottom 50% of the households? Find the coefﬁcient of
inequality. 5E05(pp 354363) 1/17/06 3:48 PM Page 359 ❙❙❙❙ WRITING PROJECT NEWTON, LEIBNIZ, AND THE INVENTION OF CALCULUS ; 64. On May 7, 1992, the space shuttle Endeavour was launched on
mission STS49, the purpose of which was to install a new
perigee kick motor in an Intelsat communications satellite. The
table gives the velocity data for the shuttle between liftoff and
the jettisoning of the solid rocket boosters.
Event Time (s) Launch Velocity (ft s) 0
10 185 End roll maneuver 15 319 Throttle to 89% 20 447 Throttle to 67% 32 742 Throttle to 104% 59 1325 Maximum dynamic pressure 62 1445 125 ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ The following exercises are intended only for those who have
already covered Chapter 7.
65–67 65. y 67.  x e y Evaluate the integral. 2 x
x 1 1 66. dx 0 Begin roll maneuver ■ 359 y 9 1
sx sx 4 2 dx 4151 Solid rocket booster separation (a) Use a graphing calculator or computer to model these data
by a thirddegree polynomial.
(b) Use the model in part (a) to estimate the height reached by
the Endeavour, 125 seconds after liftoff. ■ x2 1 1 ■ x2 ■ ■ 1 dx ■ ■ ■ ■ ■ ■ ■ ■ 68. The area labeled B is three times the area labeled A. Express b in terms of a.
y y y=´ y=´ B A
0 a x 0 b x WRITING PROJECT
Newton, Leibniz, and the Invention of Calculus
We sometimes read that the inventors of calculus were Sir Isaac Newton (1642–1727) and
Gottfried Wilhelm Leibniz (1646–1716). But we know that the basic ideas behind integration
were investigated 2500 years ago by ancient Greeks such as Eudoxus and Archimedes, and methods for ﬁnding tangents were pioneered by Pierre Fermat (1601–1665), Isaac Barrow (1630–1677),
and others. Barrow, Newton’s teacher at Cambridge, was the ﬁrst to understand the inverse relationship between differentiation and integration. What Newton and Leibniz did was to use this
relationship, in the form of the Fundamental Theorem of Calculus, in order to develop calculus
into a systematic mathematical discipline. It is in this sense that Newton and Leibniz are credited
with the invention of calculus.
Read about the contributions of these men in one or more of the given references and write a
report on one of the following three topics. You can include biographical details, but the main
thrust of your report should be a description, in some detail, of their methods and notations. In
particular, you should consult one of the sourcebooks, which give excerpts from the original
publications of Newton and Leibniz, translated from Latin to English.
The Role of Newton in the Development of Calculus
The Role of Leibniz in the Development of Calculus
The Controversy between the Followers of Newton and Leibniz over
Priority in the Invention of Calculus
References 1. Carl Boyer and Uta Merzbach, A History of Mathematics (New York: Wiley, 1987),
Chapter 19. 5E05(pp 354363) 360 ❙❙❙❙ 1/17/06 3:49 PM Page 360 CHAPTER 5 INTEGRALS 2. Carl Boyer, The History of the Calculus and Its Conceptual Development (New York: Dover,
1959), Chapter V.
3. C. H. Edwards, The Historical Development of the Calculus (New York: SpringerVerlag,
1979), Chapters 8 and 9.
4. Howard Eves, An Introduction to the History of Mathematics, 6th ed. (New York: Saunders,
1990), Chapter 11.
5. C. C. Gillispie, ed., Dictionary of Scientiﬁc Biography (New York: Scribner’s, 1974).
See the article on Leibniz by Joseph Hofmann in Volume VIII and the article on Newton by
I. B. Cohen in Volume X.
6. Victor Katz, A History of Mathematics: An Introduction (New York: HarperCollins, 1993),
Chapter 12.
7. Morris Kline, Mathematical Thought from Ancient to Modern Times (New York: Oxford
University Press, 1972), Chapter 17.
Sourcebooks 1. John Fauvel and Jeremy Gray, eds., The History of Mathematics: A Reader (London:
MacMillan Press, 1987), Chapters 12 and 13.
2. D. E. Smith, ed., A Sourcebook in Mathematics (New York: Dover, 1959), Chapter V.
3. D. J. Struik, ed., A Sourcebook in Mathematics, 1200–1800 (Princeton, N.J.: Princeton
University Press, 1969), Chapter V.  5.5 The Substitution Rule
Because of the Fundamental Theorem, it’s important to be able to ﬁnd antiderivatives. But
our antidifferentiation formulas don’t tell us how to evaluate integrals such as y 2 x s1 1  Differentials were deﬁned in Section 3.10.
If u f x , then du f x dx x 2 dx To ﬁnd this integral we use the problemsolving strategy of introducing something extra.
Here the “something extra” is a new variable; we change from the variable x to a new variable u. Suppose that we let u be the quantity under the root sign in (1), u 1 x 2. Then
the differential of u is du 2 x dx. Notice that if the dx in the notation for an integral were
to be interpreted as a differential, then the differential 2 x dx would occur in (1) and, so,
formally, without justifying our calculation, we could write y 2 x s1 2 x 2 dx y s1
2
3 x 2 2 x dx u3 2 2
3 C y su du x2 1 32 C But now we can check that we have the correct answer by using the Chain Rule to differentiate the ﬁnal function of Equation 2:
d
dx [2
3 x2 1 32 C] 2
3 3
2 x2 1 12 2x 2x sx 2 1 In general, this method works whenever we have an integral that we can write in the
form x f t x t x d x. Observe that if F
f , then
3 yF t x t x dx Ftx C 5E05(pp 354363) 1/17/06 3:49 PM Page 361 S ECTION 5.5 THE SUBSTITUTION RULE ❙❙❙❙ 361 because, by the Chain Rule,
d
Ftx
dx F tx t x If we make the “change of variable” or “substitution” u
we have yF
or, writing F t x t x dx Ftx C Fu t x , then from Equation 3 C yF u du f , we get yf t x t x dx yf u du Thus, we have proved the following rule.
t x is a differentiable function whose range is an
4 The Substitution Rule If u
interval I and f is continuous on I , then yf t x t x dx yf u du Notice that the Substitution Rule for integration was proved using the Chain Rule for
differentiation. Notice also that if u t x , then du t x d x, so a way to remember the
Substitution Rule is to think of dx and du in (4) as differentials.
Thus, the Substitution Rule says: It is permissible to operate with dx and du after
integral signs as if they were differentials. EXAMPLE 1 Find yx 3 cos x 4 2 d x. x 4 2 because its differential is du 4 x 3 dx,
which, apart from the constant factor 4, occurs in the integral. Thus, using x 3 dx du 4
and the Substitution Rule, we have
SOLUTION We make the substitution u  Check the answer by differentiating it. cos x 4 2 dx 1
4 sin u C 1
4 3 y cos u
1
4 yx sin x 4 du 2 1
4 y cos u du C Notice that at the ﬁnal stage we had to return to the original variable x.
The idea behind the Substitution Rule is to replace a relatively complicated integral
by a simpler integral. This is accomplished by changing from the original variable x to
a new variable u that is a function of x. Thus, in Example 1 we replaced the integral
x x 3 cos x 4 2 d x by the simpler integral 1 x cos u du.
4
The main challenge in using the Substitution Rule is to think of an appropriate substitution. You should try to choose u to be some function in the integrand whose differential
also occurs (except for a constant factor). This was the case in Example 1. If that is not 5E05(pp 354363) 362 ❙❙❙❙ 1/17/06 3:49 PM Page 362 CHAPTER 5 INTEGRALS possible, try choosing u to be some complicated part of the integrand. Finding the right
substitution is a bit of an art. It’s not unusual to guess wrong; if your ﬁrst guess doesn’t
work, try another substitution.
EXAMPLE 2 Evaluate
SOLUTION 1 Let u y s2 x 2x 1 d x. 1. Then du 2 dx, so dx du 2. Thus, the Substitution Rule gives y s2 x du
2 y su 1 dx u3 2
32 1
2
1
3 2x 1 SOLUTION 2 Another possible substitution is u dx
s2 x 1 du
(Or observe that u 2 2x 1 y s1 SOLUTION Let u 1 4x 2 x
s1 4x 2 dx ©= ƒ dx
_1 FIGURE 1 ƒ= x
14≈
œ„„„„„„ 1
©=j ƒ dx=_ 4 œ„„„„„„
14≈ C C
1 . Then s2 x u du
C 1 du yu
1
3 2 u du du 2x 1 32 1
8 8 x dx, so x dx
du 1
8 y su 1
8 y u3 2 C d x. 4 x 2. Then du 1 f
_1 x 32 du 2 dx.) Therefore u3
3
EXAMPLE 3 Find 1
3 C dx yu 1 dx 12 yu s2 x so 1, so 2u du y s2 x 1
2 1
8 (2 su ) yu 12 1
4 C du and du
4x 2 s1 C The answer to Example 3 could be checked by differentiation, but instead let’s check
it with a graph. In Figure 1 we have used a computer to graph both the integrand
1
fx
x s1 4 x 2 and its indeﬁnite integral t x
4 x 2 (we take the case
4 s1
C 0). Notice that t x decreases when f x is negative, increases when f x is positive,
and has its minimum value when f x
0. So it seems reasonable, from the graphical evidence, that t is an antiderivative of f .
EXAMPLE 4 Calculate
SOLUTION If we let u y cos 5x dx.
5x, then du y cos 5x dx 1
5 5 dx, so dx y cos u du 1
5 1
5 sin u du. Therefore
C 1
5 sin 5x C 5E05(pp 354363) 1/17/06 3:50 PM Page 363 SECTION 5.5 THE SUBSTITUTION RULE E XAMPLE 5 Find y s1 1 x 2. Then du y s1 363 x 2 x 5 dx. SOLUTION An appropriate substitution becomes more obvious if we factor x 5 as x 4 u ❙❙❙❙ du 2. Also x 2 2 x dx, so x dx y s1 x 2 x 5 dx y su
1
2 ( 12
27
1
7 u 1 du
2 2 1
2 2u 3 2 u7 2 2
5 2 x2 1 u x. Let
1 2: x 2 x 4 x dx u5 2 y 1, so x 4 u u2 2u 1 du u 1 2 du
2
3 u5 2
2
5 72 y su u3 2) x2 1 C
1
3 52 1 x2 32 C Definite Integrals
When evaluating a deﬁnite integral by substitution, two methods are possible. One method
is to evaluate the indeﬁnite integral ﬁrst and then use the Fundamental Theorem. For
instance, using the result of Example 2, we have y 4 1 dx 1 dx
1
3 9 4 y s2 x
1
3 0 s2 x 32 1 1
3 0 1
3 32 2x 1 27 4 32 1 0 26
3 Another method, which is usually preferable, is to change the limits of integration when
the variable is changed.
5  This rule says that when using a substitution
in a deﬁnite integral, we must put everything in
terms of the new variable u, not only x and dx
but also the limits of integration. The new limits
of integration are the values of u that correspond
to x a and x b. The Substitution Rule for Definite Integrals If t is continuous on a, b and f is continuous on the range of u y b a t x , then y f t x t x dx tb ta f u du Proof Let F be an antiderivative of f . Then, by (3), F t x is an antiderivative of
f t x t x , so by Part 2 of the Fundamental Theorem, we have y b a f t x t x dx Ftx b
a F tb F ta But, applying FTC2 a second time, we also have y tb ta EXAMPLE 6 Evaluate y 4 0 f u du s2 x Fu tb
ta F tb F ta 1 d x using (5). SOLUTION Using the substitution from Solution 1 of Example 2, we have u dx du 2. To ﬁnd the new limits of integration we note that
when x 0, u 1 and when x 4, u 9 2x 1 and 5E05(pp 364371) 364 ❙❙❙❙ 1/17/06 3:53 PM Page 364 CHAPTER 5 INTEGRALS y Therefore 4 0 s2 x 91
2
1 y 1 dx  The geometric interpretation of Example 6 is
shown in Figure 2. The substitution u 2 x 1
stretches the interval 0, 4 by a factor of 2 and
translates it to the right by 1 unit. The Substitution Rule shows that the two areas are equal. su du 2
3 1
3 32 9 9 1
2 u3 2 1
26
3 13 2 Observe that when using (5) we do not return to the variable x after integrating. We simply evaluate the expression in u between the appropriate values of u.
y y
3 3 y=œ„„„„„
2x+1
2 2 1 œ„
u
2 1 0 0 x 4 y= 1 9 u FIGURE 2 dx
.
5x 2 2 EXAMPLE 7 Evaluate
 The integral given in Example 7 is an
abbreviation for y 1 2 1 3 5x 2 y SOLUTION Let u 5x. Then du
7. Thus when x 3 2, u 3 1 dx y dx
3 5x 2 1 5 dx, so dx
1
5 2 y 7
2 1
5 du 5. When x 2 and du
u2
7 1
u 1
5 1, u 1
7 1
5u 2 1
2 7 2 1
14 Symmetry
The next theorem uses the Substitution Rule for Deﬁnite Integrals (5) to simplify the calculation of integrals of functions that possess symmetry properties.
6 Integrals of Symmetric Functions Suppose f is continuous on (a) If f is even f
(b) If f is odd f a f x , then x x a f x dx 2 x f x dx. f x , then xa a f x dx x a, a . a
0 0. Proof We split the integral in two:
7 y a
a f x dx y 0
a f x dx y a 0 f x dx y 0 a f x dx In the ﬁrst integral on the far right side we make the substitution u y a 0 f x dx
x. Then 5E05(pp 364371) 1/17/06 3:53 PM Page 365 S ECTION 5.5 THE SUBSTITUTION RULE du d x and when x a, u
a y 0 ❙❙❙❙ 365 a. Therefore y f x dx a f 0 u f y du u du a 0 f u du and so Equation 7 becomes (a) If f is even, then f y a
a 0 a a x _a a 0 x6 EXAMPLE 8 Since f x 0 y
a y 2
2 x6 0 f x dx 0 tan x f x , it is even and so
1 dx 0 x2 1 2 x 2(128
7 0 x 4 satisﬁes f 2) 284
7 x f x , it is odd and so
tan x
dx
x2 x4 1
1 1 0 Exercises
sin sx
dx ,
sx x
3 u 2 10 dx, 1 dx, 4. y
y 6. yx4
sx x [ y y cos 3x dx, yx a 0 2 5. 3. y 2 y x6 1 dx Evaluate the integral by making the given substitution. 2 a 2 y f x dx f x dx f u du 1 satisﬁes f FIGURE 3 2. a 0 2 1 x7
7
EXAMPLE 9 Since f x a _a 1. a 0 x (b) ƒ odd, j ƒ dx=0  f x dx f u and so Equation 8 gives 0  5.5 y f u du f x dx a 0 Theorem 6 is illustrated by Figure 3. For the case where f is positive and even, part (a)
says that the area under y f x from a to a is twice the area from 0 to a because of
symmetry. Recall that an integral xab f x d x can be expressed as the area above the xaxis
and below y f x minus the area below the axis and above the curve. Thus, part (b) says
the integral is 0 because the areas cancel. y 1–6 0 a y a (a) ƒ even, j ƒ dx=2 j ƒ dx _a y u
a a f u so Equation 8 gives f x dx y _a a y f x dx u (b) If f is odd, then f y a y 8 y cos 3x
u
u 4
x 3 x 4
1 2 sx dx, u 1 sin d , u cos 2x
4 u 3 2x 1
■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 5E05(pp 364371) ❙❙❙❙ 366 7–32 1/17/06 3:54 PM Page 366 CHAPTER 5 INTEGRALS Evaluate the indeﬁnite integral.  9.
11.
13. y 2x x
y 2 3 4 dx 3x 20 2
1 y s1 4x
x 2y 15. 2x y sin 19. y 21. y cos 2 12. dx 14. dy 5 1 y s4 17. 10. dx 3 y 8. yx
y 45. x 2 1 5t 4 3 y sec 2 20. y sx sin 1 22. cos st
dt
st yy 51. y 53. y y dt 2.7 4 18. t dt tan3 d 44. y sin
d
cos2 46. 2 48. y 1 dx 50. y 3 52. y 54. y 12 y 6 s2y 3 0 1 dy tan 2 d
x3 2 dx 1 5 tan dx 13
3
s1 0 2 2x x sx 1 dx 4 x 0 a 2 x sx 2 0 ■ sin6 d y 6 dx 2 1 16. t dt y 42. csc t cot t dt 16 dx 2 2x 0 x 6 dx
x y y 0 5 9 dx 2 y y 47. x3 y 49. 2 sec 2 t 4 dt y 43. 7. 41. ■ a 2 dx
■ ■ a 0 ■ ■ ■ 2
2 2 0 3 2 x 2 sin x
dx
1 x6 cos x sin sin x d x
x 4 s1 0 a 2x x sa 2 0 a
a
■ x sx 2
■ dx x 2 dx a 2 dx
■ ■ ■ sec 2 d ; 55–56
23. 25. y s1 z2 3 24. dz z3 2 y scot x csc x dx y sax ax b
2bx 2 c  Use a graph to give a rough estimate of the area of the
region that lies under the given curve. Then ﬁnd the exact area. dx 55. y
56. y cos x
dx
x2 26. y 28. y sx s2 x
2 sin x ■ 3 27. y sec x tan x dx 29. yx 30. y sin t sec 31. y sx a cx a sb x 4 ■ 2 ■ 2 1 dx c 0, a 3 3 1 x dx 35. 3x
3x 2 1 ■ ■ x2 y s1 ■ ■ ■ ■ ■ y 2 0 y 1 0 ■  x 34. dx y sin x cos x dx 37–54 39. 4 ■ ■ ■ ■ ■ ■ ■ ■ 59. Breathing is cyclic and a full respiratory cycle from the begin dx x ■ ■ y sx 36. 1 3 ■ 37. 1 ■ x 4 dx by making a substitution and
interpreting the resulting integral in terms of an area. 32. 2x ■ x 3 s4 x 2 dx by writing it as a sum of
two integrals and interpreting one of those integrals in terms
of an area. ■ ■  Evaluate the indeﬁnite integral. Illustrate and check that
your answer is reasonable by graphing both the function and its
antiderivative (take C 0). y sin 2 x, 0 57. Evaluate x 2 2 x ; 33–36 33. 1 1
58. Evaluate x0 x s1 dx
■ ■ x 5 cos t dt ■ ■ 1, 0 2 2 y tan ■ 1
2 ■ manufacture a new calculator. The rate of production of these
calculators after t weeks is d ■ ■ 60. Alabama Instruments Company has set up a production line to dx sec ning of inhalation to the end of exhalation takes about 5 s. The
maximum rate of air ﬂow into the lungs is about 0.5 L s. This
1
explains, in part, why the function f t
2 sin 2 t 5 has
often been used to model the rate of air ﬂow into the lungs.
Use this model to ﬁnd the volume of inhaled air in the lungs at
time t. ■ ■ dx
dt 5000 1 t 100
10 2 calculators week Evaluate the deﬁnite integral, if it exists. x
x2 1 1 25 38. dx 2x 3 5 dx y 40. y 7 0
s 0 s4 3x dx x cos x 2 dx (Notice that production approaches 5000 per week as time goes
on, but the initial production is lower because of the workers’
unfamiliarity with the new techniques.) Find the number of calculators produced from the beginning of the third week to the
end of the fourth week. 5E05(pp 364371) 1/17/06 3:55 PM Page 367 S ECTION 5.5 THE SUBSTITUTION RULE
4 2 61. If f is continuous and y f x d x 10, ﬁnd y f 2 x d x. 0 0 9 3 4, ﬁnd y x f x 2 d x. 62. If f is continuous and y f x d x
0 63. Suppose f is continuous on 0 . y a f x dx y a
b y a fx y c dx bc ac f x dx For the case where f x
0, draw a diagram to interpret this
equation geometrically as an equality of areas.
64. Show that the area under the graph of y sin sx from 0 to 4
is the same as the area under the graph of y 2 x sin x from
0 to 2. 65. If a and b are positive numbers, show that y 1 0 xa 1 x b dx y 1 0 xb 1 y 0 x f sin x d x x a dx x to show that 2 y 0 ■ ■ ■ ■ ■ ■ f sin x d x ■ ■ 367 ■ The following exercises are intended only for those who have
already covered Chapter 7.
67–82  Evaluate the integral.
dx y5 69. y ln x
x 71. ye x 73. x 68. y 70. y1 72. ye y x ln x 74. ye 75. y cot x d x 76. y 77. y1 78. y y e1 x
dx
x2 80. y dx
x sln x 82. 81.
■ 66. Use the substitution u ■ 79. f x dx 0 and 0 a b, draw a diagram
For the case where f x
to interpret this equation geometrically as an equality of areas.
(b) Prove that
b ■ 67. (a) Prove that
b ■ ❙❙❙❙ 3x
2 dx
e x dx s1 dx 1 2 1 y e4 e
■ x2 x
dx
x2 ■ ■ ■ y 0 1 cos t sin t dt ex
x 1 dx sin x
dx
1 cos2x
x
x4 1
1 xe 12 0 ■ ■ 83. Use Exercise 66 to evaluate the integral y dx tan 1x
dx
x2 0 ■ 1 x sin x
dx
cos2x x2 dx dx sin 1x
dx
s1 x 2
■ ■ ■ ■ 5E05(pp 364371) ❙❙❙❙ 368  1/17/06 3:55 PM Page 368 CHAPTER 5 INTEGRALS 5 Review ■ CONCEPT CHECK 1. (a) Write an expression for a Riemann sum of a function f . Explain the meaning of the notation that you use.
(b) If f x
0, what is the geometric interpretation of a
Riemann sum? Illustrate with a diagram.
(c) If f x takes on both positive and negative values, what is
the geometric interpretation of a Riemann sum? Illustrate
with a diagram. (b) If r t is the rate at which water ﬂows into a reservoir, what
does xtt r t d t represent?
2 1 5. Suppose a particle moves back and forth along a straight line
with velocity v t , measured in feet per second, and accelera tion a t .
120
(a) What is the meaning of x60 v t dt ?
120
(b) What is the meaning of x60 v t dt ? 2. (a) Write the deﬁnition of the deﬁnite integral of a continuous function from a to b.
(b) What is the geometric interpretation of xab f x d x if
fx
0?
(c) What is the geometric interpretation of xab f x d x if f x
takes on both positive and negative values? Illustrate with a
diagram. 120
(c) What is the meaning of x60 a t dt ? 6. (a) Explain the meaning of the indeﬁnite integral x f x d x. (b) What is the connection between the deﬁnite integral
xab f x d x and the indeﬁnite integral x f x d x ?
7. Explain exactly what is meant by the statement that “differen tiation and integration are inverse processes.” 3. State both parts of the Fundamental Theorem of Calculus.
4. (a) State the Net Change Theorem. 8. State the Substitution Rule. In practice, how do you use it? ■ TRUEFALSE QUIZ Determine whether the statement is true or false. If it is true, explain why.
If it is false, explain why or give an example that disproves the statement. b a fx y t x dx b a y f x dx b a y a f x t x dx y b a f x dx b a 5 f x dx b a x f x dx b a sf x d x y a 1 y 10. y 11. b a y a 12. 0, then a b a t x for a t x for a x b, then x t x for a x b, b. sin x
1 x4 6x 9 1 5
1
2 ax 2
1
dx
x4 bx c dx dx 2 5 2 y ax 2
0 0 c dx 3
8 x x 3 d x represents the area under the curve y
from 0 to 2. 13. All continuous functions have derivatives. f x dx f 1. t x dx x02 x5 5 b x y f x dx b y f x dx f3 t x dx 5 y f x dx y b a 9. 5. If f is continuous on a, b and f x y y
then f x 4. If f is continuous on a, b , then y 1 8. If f and t are differentiable and f x
b 3. If f is continuous on a, b , then y 3 6. If f is continuous on 1, 3 , then y f v dv t x dx 2. If f and t are continuous on a, b , then
b ■ 7. If f and t are continuous and f x 1. If f and t are continuous on a, b , then y ■ 14. All continuous functions have antiderivatives. x x3 5E05(pp 364371) 1/17/06 3:56 PM Page 369 C HAPTER 5 REVIEW ■ EXERCISES ❙❙❙❙ 369 ■ 7. The ﬁgure shows the graphs of f, f , and x0x f t dt. 1. Use the given graph of f to ﬁnd the Riemann sum with six subintervals. Take the sample points to be (a) left endpoints and
(b) midpoints. In each case draw a diagram and explain what
the Riemann sum represents. Identify each graph, and explain your choices.
y b y c
y=ƒ 2 x a
0 2 6 x 8. Evaluate: x2 fx 0 x x with four subintervals, taking the sample points to be right
endpoints. Explain, with the aid of a diagram, what the
Riemann sum represents.
(b) Use the deﬁnition of a deﬁnite integral (with right endpoints) to calculate the value of the integral y 2 0 x2 (b) 9–28 s1 11. y 13. y 15. y 17. y 19. x 2 ) dx by interpreting it in terms of areas. y 0 y x 8x 3 y y 21. y sx 23. y sin 25. y 27. y 3x 2 dx T x4 8x 1 1 x 9 dx 10. y 12. y du 14. y (su 1 5 dy 16. y 2 18. y 20. y 2
dx
4x 22. y csc t cos t dt 24. y sin x cos cos x sec 2 tan 2 d 26. y x2 1 28. y 1 9 x 9 dx su 0 2u2
u 1 3. Evaluate y0 ( x 2 x
x
cos
dx
2
3
2
x
x
sin cos d x
2
3
t
2
t
sin cos dt
2
3
sin Evaluate the integral, if it exists.  9. x dx (c) Use the Fundamental Theorem to check your answer to
part (b).
(d) Draw a diagram to explain the geometric meaning of the
integral in part (b).
1 0 d
dx
d
(c)
dx 2 d
dx 2 y (a) 2. (a) Evaluate the Riemann sum for 1 y y2 0 dt 5 t 1 4 0
1 0
1 2 y 2s1 1 n sin xi x nl i1 as a deﬁnite integral on the interval 0,
the integral.
6
5. If x0 f x d x CAS 5
6. (a) Write x1 x 10 and x04 f x d x and then evaluate 7, ﬁnd x46 f x d x. 2 x 5 dx as a limit of Riemann sums,
taking the sample points to be right endpoints. Use a
computer algebra system to evaluate the sum and to
compute the limit.
(b) Use the Fundamental Theorem to check your answer to
part (a). ■ 1 v 2 cos v 3 dv 0 x 8 0
3 0
■ 2 4 dx
■ ■ ■ ■ ■ y 3 dy sin 3 t dt 0 4. Express lim 1 2 du 4 0 0 7 dx sin x
dx
1 x2 1
1 2 4 0
4 0
■ 3 t dt
dx tan t 3 sec2t dt 1
sx 1 dx
■ ■ ■ ■ 5E05(pp 364371) ❙❙❙❙ 370 1/17/06 3:56 PM Page 370 CHAPTER 5 INTEGRALS 43. Use the Midpoint Rule with n ; 29–30  Evaluate the indeﬁnite integral. Illustrate and check that
your answer is reasonable by graphing both the function and its
antiderivative (take C 0). x01 s1 44. A particle moves along a line with velocity function
vt
t 2 t, where v is measured in meters per second. cos x
dx
s1 sin x 29. y 30. y sx ■ x3
2 ■ Find (a) the displacement and (b) the distance traveled by the
particle during the time interval 0, 5 . dx 1
■ 5 to approximate x 3 d x. 45. Let r t be the rate at which the world’s oil is consumed, where ■ ■ ■ ■ ■ ■ ■ ■ ■ ; 31. Use a graph to give a rough estimate of the area of the region
that lies under the curve y
exact area. x s x, 0 x 4. Then ﬁnd the t is measured in years starting at t 0 on January 1, 2000,
and r t is measured in barrels per year. What does x03 r t d t
represent?
46. A radar gun was used to record the speed of a runner at the times given in the table. Use the Midpoint Rule to estimate the
distance the runner covered during those 5 seconds. cos2 x sin3 x and use the graph to
guess the value of the integral x02 f x d x. Then evaluate the
integral to conﬁrm your guess. ; 32. Graph the function f x t (s) 33–38 33. F x y 34. F x y 35. t x y x x y tan s 2 ds
t x3 0 36. t x s1 y x cos sx y 38. y 3x 1 2x ■ ■ 39– 40  d y 40. y 3 ■ ■ ■ ■ ■ ■ ■ ■ 8000 ■ 4000 ■ x
■ 41– 42 41. y 42. y ■ 3 dx 1 5 3 2
4
■ ■ ■ ■ ■ ■ ■ ■ 8 12 16 20 t
24
(weeks) ■ fx Use the properties of integrals to verify the inequality.
1
3 sin x
dx
x
■ 4 48. Let
■ x 2 cos x dx 0 0 dx ■  1 1 10.51
10.67
10.76
10.81
10.81 r Use Property 8 of integrals to estimate the value of the sx 2 1 3.0
3.5
4.0
4.5
5.0 12000 integral.
39. 0
4.67
7.34
8.86
9.73
10.22 week, where the graph of r is as shown. Use the Midpoint Rule
with six subintervals to estimate the increase in the bee population during the ﬁrst 24 weeks. sin t 4 dt ■ v (m s) 47. A population of honeybees increased at a rate of r t bees per t 2 dt s1 t (s) dt t3 cos x 3 1 37. y t 4 dt s1 1 v (m s) 0
0.5
1.0
1.5
2.0
2.5 Find the derivative of the function.  if 3 x 0
if 0 x 1 Evaluate x1 3 f x d x by interpreting the integral as a difference
of areas. s2
2
■ x1
s1 x 2 49. Find an antiderivative F of f x
■ ■ ■ ■ ■ ■ ■ ■ F1 0. x 2 sin x 2 such that 5E05(pp 364371) 1/17/06 3:57 PM Page 371 C HAPTER 5 REVIEW x0x sin( 1 t 2) dt was introduced in
2
Section 5.3. Fresnel also used the function 50. The Fresnel function S x Cx CAS y x 0 52. Find a function f and a value of the constant a such that
x 2 y f t dt cos ( 1 t 2) dt
2 b 2 y f x f x dx x CAS cos ( 1 t 2) dt
2 0.7 54. Find lim 1
h y 2h 2 y 0 f t dt x sin x y 0 y 1 0 51. If f is a continuous function such that fa 2 t 3 dt. y f x dx 1 0 f1 x dx 56. Evaluate ft x s1 2 55. If f is continuous on 0, 1 , prove that (d) Plot the graphs of C and S on the same screen. How are
these graphs related? x fb a hl0 0 1 53. If f is continuous on a, b , show that in his theory of the diffraction of light waves.
(a) On what intervals is C increasing?
(b) On what intervals is C concave upward?
(c) Use a graph to solve the following equation correct to two
decimal places: y 2 sin x a 1 for all x, ﬁnd an explicit formula for f x . t2 dt
lim nl 1
n 1
n 9 2
n 9 3
n 9 n
n 9 ❙❙❙❙ 371 5E05(pp 372373) 1/17/06 3:58 PM PROBLEMS
PLUS Page 372 Before you look at the solution of the following example, cover it up and ﬁrst try to solve
the problem yourself.
x EXAMPLE 1 Evaluate lim y
3 x x l3 x 3 sin t
dt .
t SOLUTION Let’s start by having a preliminary look at the ingredients of the function. What
happens to the ﬁrst factor, x x 3 , when x approaches 3? The numerator approaches 3
and the denominator approaches 0, so we have x
x  The principles of problem solving are
discussed on page 58. 3 l as x and xl3 x 3 as l xl3 The second factor approaches x33 sin t t dt, which is 0. It’s not clear what happens to the
function as a whole. (One factor is becoming large while the other is becoming small.)
So how do we proceed?
One of the principles of problem solving is recognizing something familiar. Is there a
part of the function that reminds us of something we’ve seen before? Well, the integral y x 3 sin t
dt
t has x as its upper limit of integration and that type of integral occurs in Part 1 of the
Fundamental Theorem of Calculus:
d
dx y x a f t dt fx This suggests that differentiation might be involved.
Once we start thinking about differentiation, the denominator x 3 reminds us of
something else that should be familiar: One of the forms of the deﬁnition of the derivative in Chapter 3 is
Fx
Fa
Fa
lim
xla
xa
and with a 3 this becomes
F3 lim
x l3 Fx
x F3
3 So what is the function F in our situation? Notice that if we deﬁne
Fx y x 3 sin t
dt
t then F 3
0. What about the factor x in the numerator? That’s just a red herring, so
let’s factor it out and put together the calculation: lim
x l3  Another approach is to use l’Hospital’s Rule. x
x 3 y x 3 sin t
dt
t y
lim x lim
x l3 3 lim
x l3 x l3 Fx
x 3F 3
sin 3
3
3
sin 3
372 x 3 sin t
dt
t
x3
F3
3 (FTC1) 5E05(pp 372373) 1/17/06 3:59 PM Page 373 P RO B L E M S 1. If x sin y x x2 0 f t dt, where f is a continuous function, ﬁnd f 4 . 2. In this problem we approximate the sine function on the interval 0, ; by three quadratic
functions, each of which has the same zeros as the sine function on this interval.
(a) Find a quadratic function f such that f 0
0 and which has the same maximum
f
value as sin on 0, .
(b) Find a quadratic function t such that t 0
t
0 and which has the same rate of
change as the sine function at 0 and .
(c) Find a quadratic function h such that h 0
h
0 and the area under h from 0 to is
the same as for the sine function.
(d) Illustrate by graphing f, t, h, and the sine function in the same viewing rectangle 0,
by
0, 1 . Identify which graph belongs to each function. 3. Show that 1
17 y 1 2 1 1 4. Suppose the curve y x4 dx 7
.
24 f x passes through the origin and the point 1, 1 . Find the value of the integral x01 f x d x.
5. Find a function f such that f 1 1, f 4 7, and f x 3 for all x, or prove that such a function cannot exist.
2cx x 2 c 3 for c 0 and
look at the regions enclosed by these curves and the xaxis. Make a conjecture about how
the areas of these regions are related.
Prove your conjecture in part (a).
Take another look at the graphs in part (a) and use them to sketch the curve traced out by
the vertices (highest points) of the family of functions. Can you guess what kind of curve
this is?
Find the equation of the curve you sketched in part (c).
tx
1
cos x
x
y0 s1 t 3 dt, where t x y0 1 sin t 2 dt, ﬁnd f 2 . ; 6. (a) Graph several members of the family of functions f x
(b)
(c) (d)
7. If f 8. If f x x0x x 2 sin t 2 dt, ﬁnd f x . b
9. Find the interval a, b for which the value of the integral xa 2 x 2 d x is a maximum. x 10000 10. Use an integral to estimate the sum si.
i1 n
11. (a) Evaluate x0 x d x, where n is a positive integer. (b) Evaluate xab x d x, where a and b are real numbers with 0
12. Find d2
dx 2 x yy
0 sin t 1 s1 b. u 4 du dt.
x 13. If f is a differentiable function such that y f t dt 2 a 0 fx 2 for all x, ﬁnd f . 14. A circular disk of radius r is used in an evaporator and is rotated in a vertical plane. If it is to 2 2 be partially submerged in the liquid so as to maximize the exposed wetted area of the disk,
2 above the surface
show that the center of the disk should be positioned at a height r s1
of the liquid.
x 15. Prove that if f is continuous, then y f u x
0 u du x yy
0 u 0 f t dt du. 16. The ﬁgure shows a region consisting of all points inside a square that are closer to the center
2 FIGURE FOR PROBLEM 16 than to the sides of the square. Find the area of the region.
17. Evaluate lim
nl 1
sn sn 1 1
sn sn 2 1
sn sn n . 373 ...
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 Fall '09
 hamrick

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