**Unformatted text preview: **(c) 0 (c) Are the answers to parts (a) and (b) the same? Explain.
ins5-2h22-28 2 (d) f (x) dx 5 f (x) dx 2
In Problems ??, ﬁnd the area of the regions between the curve
Math for Econ II, Written Assignment 8 (28 points)
f (x)
and the horizontal axis
1 5-2h22
5-2h23
5-2h24
5-2h25
5-2h26
5-2h27
5-2h28
5-2h29 Due Friday, November 7 x
22. Under y = 6x3 − 2 for 5 ≤ x ≤ 10.
Jankowski, Fall 2014
−1
23. Please the curve y = solutions ≤ t ≤ π/2.
Under write neat cos t for 0 for the problems below. Show all your work. If you only write the answer with no work,
5-2h32fig −2
24. you will = ln be given x ≤ 4.
Under y not x for 1 ≤ any credit.
2
4
6
8
10
25. Under y = 2 cos(t/10) for 1 ≤ t ≤ 2.
√
• Write your name and recitation section number. Figure 5.39: Graph consists of a semicircle and
26. Under the curve y = cos x for 0 ≤ x ≤ 2.
line segments
27. Under the curve y =homework above the x-axis.
• Staple your 7 − x2 and if you have multiple pages!
28. Above the curve y = x4 − 8 and below the x-axis.
1. (2 pts total; 1 pt each) Use the ﬁgure below to ﬁnd the values of
29. Use Figure ?? to ﬁnd the values of
5-2h33 33. (a) Graph f (x) = x(x + 2)(x − 1).
b
c
f bf
(b)
f (x) dx
(a)
(b) Find the total area between the graph and the x-axis
(a)(x) dx(x) dx
a
b
c
c
c
between x = −2 and x = 1.
(c)
f (x)c |f (x)| dx (d)
dx
|f (x)| dx
a
a
1
(b) a
(c) Find −2 f (x) dx and interpret it in terms of areas.
f (x) 5-2h34 Area = 13 34. Compute the deﬁnite integral
the result in terms of areas. © 4
0 cos √ x dx and interpret 5-2h35 a b 5-2h29fig 35. Without computation, decide if 0 e−x sin x dx is positive or negative. [Hint: Sketch e−x sin x.] 5-2h36 36. Estimate 2π x c sArea = 2 ! (a) Figure 5.36
5-2h30 5-4h43 1
0 2 e−x dx using n = 5 rectangles to form a
(b) Left-hand sum Right-hand sum 5-2h37 37. (a) On a sketch of y = ln x, represent the left Riemann
2
sum with n = 2 approximating 1 ln x dx. Write
2
2
out the terms in the sum, but do not evaluate it.
f b
(b) c
(a)
0
(a)(x)dx (x) dx = − b f−2 f (x)dx −(−2) = 2. (b) On another sketch, represent the right Riemann sum
f
(x) dx =
c
(c) The total shaded area.
2
(b) The graph of |f (x)| is the same as the graph of with n except that the part belowWrite x-axis is reﬂected to be
f (x) = 2 approximating 1 ln x dx. the out the
f (x)
terms in the sum, but do not evaluate it.
above it. Thus
c
(c) Which sum is an overestimate? Which sum is an un2
5.4 THEOREMS 2 = 15.
311
|f (x)| dx = 13 +ABOUT DEFINITE INTEGRALS
derestimate?
a
0 30. GivenSolution: = 4 and Figure ??, estimate:
f (x)dx
−2 x 5-2h38
2
43. Using the graph −2 f in Figure ??, arrange the ins5-4h47-48 In 38. (a) Draw the rectangles that give the left-hand sumor
of
following
Problems ??, evaluate the expression, if possible, ap2. (4 pts) Using the graph of f in Figure, arrange the following quantities in increasing order, from least to greatest.
π
−2
quantities in increasing order, from least to greatest.
say what proximationinformation dx with n = 2.
additional to
sin x is needed, given that
0
1
2
2
3
2
5-2h30fig
4
1 i)
2
f (x) dx ii) 1 f (x)(x) dx iii) 0 f (x) dx g(x) dx 2 f (x) dx(a) for − 1 f x dx.
dx
iv) Repeat part
v) 0 sin (x) dx vi) The number 0
= 12.
(b)
f (x) dx
(ii)
f
(i)
0
−4
0 (iii) 2 0 Figure 5.37
(iv) f (x) dx 1 3 2 −π (c) From your answers to parts (a) and (b), what is
4
the value of the left-hand 4
sum approximation to
5-4h48 48.
π
g(x) dx x dx with n = 4?
g(−x) dx
sin
−π f (x) dx 20 viii) The 0 5-4h47 47.
(v) − vii) (x) dx
f The number (vi) The numbernumber −10
0
1
31. (a) Using Figure ??, ﬁnd −3 f (x) dx.
0
−4
(vii) The number 20
(viii) The number −10
(b) If the area of the shaded region is A, estimate
5-2h39 39. (a) Use a calculator or computer to ﬁnd 6 (x2 + 1) dx.
4
0
f (x) dx.
−3
f (x)
Represent this value as the area under a curve.
10
ins5-4h49-52
6
In Problems ??, evaluate 2the 1) dx using if left-hand sumsay
(b) Estimate 0 (x + expression 7a possible, or with
1
x
f (x)
what extra information is needed, given graphically on 25.
1
2
3
n = 3. Represent this sum 0 f (x) dx = a sketch
4
x
5-4h43fig −10
of f (x) = x2 + 1. Is this sum an overestimate or
−4 −3 −2 −1
1
2
3 !
5
underestimate of the true value found in part (a)?
7
3.5
5-2h31fig
−1
6 5-4h50
Figure 5.71
5-4h49 49.
f (x) dx
50.
(c) Estimate 0 (x2 +1) dx using a f (x) dx sum with
right-hand
0 n = 3. Represent this sum on your sketch. Is this
0
Figure 5.38
5-4h44 44. (a) Using Figures ?? and ??, ﬁnd the average value on
sum an overestimate or underestimate?
Solution:
5
7
0 ≤ x ≤ 2 of
5-4h51 51.
5-4h52 52.
1
2
3
f (x + 2)dx = A , we know (f (x) + 2) dx
Since 0 f (x) dxg(x) 1 and 1 ff (x)·g(x) −A2 and 2 f (x) dx
=A
(x) dx =
that
3
(i) f (x)
(ii)
(iii)
−2
0
(b) Is the following statement true? Explain your an1
2
3
swer.
0 < 5-4h53f (x) dx < −
f (x) dx <
f (x) dx.
53. (a) Sketch a graph√ f√2 = sin(x2 ) and mark on it
of (x) √
1
√
Average(f ) · Average(g) = Average(f · g) 0
the points x = π, 2π, 3π, 4π.
2
(b) Use your graph to in magnitude than 2 f (x)
1
1
In addition, 0 f (x) dx = A1g(x) 2 , which is negative, but smaller decide which of the four numbersdx. Thus
−A
f (x)
1
5-2h31 2 2 5-4h44figa x 1 5-4h44figb Figure 5.72
5-4h45 5-4h46 2 f (x) dx < x
1 2 Figure 5.73 45. (a) Without computing any integrals, explain why the
ins5-4h54-56
average value of f (x) = sin x on [0, π] must be between 0.5 and 1.
(b) Compute this average.
46. Figure ?? shows the standard normal distribution from 2 1 0 √
nπ
2
f (x) sin(x 0.
dx < ) dx n = 1, 2, 3, 4 0 is largest. Which is smallest? How many of the numbers are positive?
For Problems ??, assuming F = f , mark the quantity on a
copy of Figure ??.
F (x) The area A3 lies inside a rectangle of height 20 and base 1, so A3 < 20. The area A2 lies inside a rectangle below
the x-axis of height 10 and width 1, so −10 < A2 . Thus:
(viii) < (ii) < (iii) < (vi) < (i) < (v) < ( iv) < (vii).
−1
f (x) dx
−2
4
f (x) dx.
1 3. (3 pts) Suppose f is even,
enough information to ﬁnd −1
−2 Solution: Because f is even, 4
2 = 3, and f (x) dx = 4 2
1 f (x) dx = 5. Either ﬁnd 4 f (x) dx + 1 f (x) dx, or show there is not f (x) dx, so 2 f (x) dx = 4
1 f (x) dx = 3 + 5 = 8 . 1 2 4. (5 pts) Evaluate the following integrals:
1
(f (x)
0 (a) (3 pts) If 1
(2f (x)
0 − 2g(x)) dx = 6 and + 2g(x)) dx = 9, ﬁnd 1
(f (x)
0 − g(x)) dx. 3 (x9 − 4x3 + x) dx (b) (2 pts)
−3 Solution:
(a) Adding the ﬁrst 2 integrals gives us 3
1 1
0 1 f (x) dx −
0 1
0 f (x) = 15, so f (x) = 5. We use the ﬁrst integral again: 1 5 − 2 2g(x) dx = 6, 1 −2 g(x) dx = 6, 0 0 1 g(x) dx = 1,
0 0 1
g(x) dx = − .
2 Thus,
1 1 (f (x) − g(x)) dx =
0 1 f (x) − g(x) dx = 5 − (−1/2) = 0 0 11
.
2 3 (x9 − 4x3 + x) dx = 0 because f (x) = x9 − 4x3 + x is an odd continuous function and the interval is (b)
−3 symmetric.
(check f is odd: f (−x) = (−x)9 − 4(−x)3 + (−x) = −x9 + 4x3 − x = −(x9 − 4x3 + x) = −f (x)).
5. (2 pts) Explain what is wrong with the following calculation of the area under the curve
1
−1 1
1
dx = − 3
x4
3x 1
−1 1
x4 from x = −1 to x = 1: 1 1
2
=− − =− .
3 3
3 1
Solution: We cannot apply the evaluation theorem because x4 is not continuous on [−1, 1] (it has an inﬁnite
discontinuity when x = 0). Whatever the integral is (if it even exists!), it can’t be zero, because 1/x4 is a strictly
positive function with plenty of positive area. 6. (3 pts) Suppose the area under the curve ex from x = 0 to x = a is six times the area under the curve 2e2x from
x = 0 to x = b. Solve for a in terms of b. (in other words, write a =(some formula involving b))
Solution: We use the Evaluation Theorem.
a b ex dx = 6 2e2x dx. 0
a ex dx = ex
0 a 0
b = ea − 1, 0 2e2x dx = 6 e2x 6
0 b = 6e2b − 6. 0 Thus
ea − 1 = 6e2b − 6, ea = 6e2b − 5, a = ln(6e2b − 5) . 5 1
dx = ln(5). Now ﬁnd a fraction which approximates ln(5), by
x
1
5
1
using M4 (midpoint sum with 4 rectangles) to approximate
dx.
1 x
(The actual value of ln(5) is 1.6094 . . .. For fun, plug your approximation into a calculator and compare) 7. (4 pts) Use the Evaluation Theorem to show 5 Solution: By the Evaluation Theorem,
1 1
dx = ln(x)
x 5 = ln(5) − ln(1) = ln(5).
1 (the formulas for ln |x| and ln(x) are identical since x > 0 on the interval [1, 5]).
M4 = 1 · f 3
5
7
9
2 2 2 2
496
+1·f
+1·f
+1·f
= + + + =
2
2
2
2
3 5 7 9
315 How good is our approximation? ln(5) = 1.6094.... Our approximating fraction is 496
315 = 1.5746... 6000
and the supply curve is given by P = Q + 10. Find
Q + 50
the equilibrium price and quantity, and compute the consumer and producer surplus. 8. (5 pts) Suppose the demand curve is given by P = Solution: To ﬁnd Q∗ we set both values of P equal:
6000
= Q∗ + 10,
Q∗ + 50 6000 = (Q + 50)(Q + 10), (Q∗ )2 + 60Q∗ − 5500 = 0, 6000 = (Q∗ )2 + 60Q∗ + 500, (Q∗ − 50)(Q∗ + 110) = 0, Q∗ = 50 , Q∗ =-110. For practical purposes we ignored the negative equilibrium quantity Q∗ = −110 above.
Thus P ∗ = Q∗ + 10 = 50 + 10 = 60, so P ∗ = 60 .
Q∗ 50 (f (Q) − P ∗ ) dQ = CS =
0 (
0 6000
− 60) dQ = [6000 ln(Q + 50) − 60Q]
Q + 50 50
0 = (6000 ln(100) − 60(50)) − 6000 ln(50) = 6000 ln(2) − 3000 .
Also,
Q∗ 50 (P ∗ − g(Q)) dQ = PS =
0 50 (60 − (Q + 10)) dQ =
0 (50 − Q) dQ = [50Q −
0 Q2
2 50 =
0 2500
= 1250 .
2 Some extra practice (not to be handed in)
3 1. Estimate f (x) dx using R5 and L5 . −2 −1 1 2 3 −5 −4 −3 −2 −3 −1 1 2 3 4 5 −2 2. Suppose h is a function such that h(1) = −2, h (1) = 3, h (1) = 4, h(2) = 6, h (2) = 5, h (2) = 13, and h is
2
continuous everywhere. Find 1 h (u) du.
Solution: By the Evaluation Theorem, 2
1 h (u) du = h (2) − h (1) = 5 − 3 = 2. ...

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