**Unformatted text preview: **ins5-2h22-28 In Problems ??, find the area of the regions between the curve
and the horizontal axis 5-2h22
5-2h23
5-2h24
5-2h25
5-2h26
5-2h27
5-2h28
5-2h29 2 f (x) 1 x
22. Under y = 6x3 − 2 for 5 ≤ x ≤ 10.
−1 Assignment 8 (28 points)
for Econ II, Written
23. Under the curve y = cos t for 0 ≤Math
t ≤ π/2.
Please
write
neat
solutions
for
the
problems
below.
Show
5-2h32fig −2 all your work. If you only write the answer with no work,
24. Under y = ln x for 1 ≤ x ≤ 4.
2
4
6
8
10
you
will
not
be
given
any
credit.
25. Under y = 2 cos(t/10) for 1 ≤ t ≤ 2.
Figure
5.39:
Graph
consists
of
a
semicircle
and
√
26. Under
the curve
y =name
cos and
x forrecitation
0 ≤ x ≤ 2. section number.
• Write
your
line segments
27. Under the curve y = 7 − x2 and above the x-axis.
• Staple your homework if you have multiple pages!
28. Above the curve y = x4 − 8 and below the x-axis.
1. Figure
(2 pts??total;
ptvalues
each)ofUse the figure below
to find the values of
29. Use
to find1the
5-2h33 33. (a) Graph f (x) = x(x + 2)(x − 1).
"b R
"c
f (x)bdx
(b)
f (x) dx
(a)
(b) Find the total area between the graph and the x-axis
f (x) dx
" a(a)
"bc
c
c
R
between
(c)
f (x)cdx
(d)
|f (x)| dx
" 1 x = −2 and x = 1.
a
a
(b) a |f (x)| dx
(c) Find −2 f (x) dx and interpret it in terms of areas.
f (x) Area = 13 !
a b 5-2h29fig c √ 34. Compute the definite integral
the result in terms of areas. 5-2h35 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 x "Area = 2 "4 5-2h34 ! (a) Figure 5.36 "1
0 0 cos x dx and interpret " 2π 2 e−x dx using n = 5 rectangles to form a Left-hand sum (b) Right-hand sum 5-2h37 "0 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 Rb
"
R 2
out the terms in the sum, but do not evaluate it.
f (x)dx
(b) c f−2
(a)
(a)
f (x) dx = −
(x)f (x)dx
dx = −(−2) = 2.
0
c
b
(b) On another sketch, represent
" 2the right Riemann sum
(c) The total shaded area.
(b) The graph of |f (x)| is the same as the graph of with
f (x)n except
that the part
the x-axis
lnbelow
x dx. Write
out the is reflected to be
= 2 approximating
1
above it. Thus
f (x)
Z c
terms in the sum, but do not evaluate it.
5.4 THEOREMS
DEFINITE INTEGRALS
311
(c)(x)|
Which
sum
Which sum is an
un|f
dx =
13is+anABOUT
2overestimate?
= 15.
2
derestimate?
a
5-4h43 43. Using the graph of f in Figure ??, arrange the
x ins5-4h47-48
following In Problems ??, evaluate the expression, if possible, or
−2
2 of f in Figure, 5-2h38
2.
(4
pts)
Using
the
graph
in give
increasing
order,
from
38.the
(a)following
Draw
the quantities
rectangles
the left-hand
ap- least to greatest.
quantities in increasing−2
order, from least to greatest. arrange
say
additional
information
is needed,
givensum
that
" π that
! 4 what proximation
R2!2
R2
R3
Rx2dx with n = 2.
!1 R1
sin
to
g(x)
0 "−
i)f (x)0 dx
f (x) dx ii)(ii) 1 f (x)
dxdx iii) 0 f (x) dx
iv)dx 2=f12.
(x) dx v)
f (x) dx vi) The number 0
f (x)
(i)5-2h30fig
−4
0 1
0
!2
!13
(b) Repeat part (a) for −π sin x dx.
(iii)
f (x) dx
(iv)
f (x) dx
" (c)
" 4 (a) and (b), what is
From your answers to parts
4
0
Figure 5.37 2
!2
5-4h47
5-4h48
Thedxnumber (vi)
20 The
viii)number
The number
−10
the
sum
approximation
to
47.
g(x)
48.
g(−x)
dx
(v) − vii)
f (x)
0
" πdxvalue of the left-hand
1
"0
0
sin x dx with n = 4? −4
5-2h31 31.(vii)
The number
Thedx.
number −10
(a) Using
Figure20
??, find (viii)
f
(x)
−π
−3
"
(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
f (x)
0
f
(x)
dx.
10
−3
ins5-4h49-52
value
as the areaifunder
a curve.
In ProblemsRepresent
??, evaluate
the expression
possible,
or say
"this
!
6
2
x
+ 1) dx
using7afleft-hand
sum
(b) information
Estimate 0 is(xneeded,
(x)
dx
=
25. with
what extra
given
11
2
3f (x)
0
n = 3. Represent this sum graphically on a sketch
5-4h43fig −10
4
#" of f (x) = x2 + 1. Is this
x
an overestimate or
!
" sum
7
3.5
−4 −3 2 −1
1
2
3
5
underestimate
of
the
true
value
found
in part (a)?
Figure 5.71
5-4h49 49.
5-4h50
5-2h31fig
f (x) dx " 6
50.
f (x) dx
−1
(c)0 Estimate 0 (x2 +1) dx using
0 a right-hand sum with
5-4h44 44. (a) Using Figures ?? and ??, find the average value on
" 5 n = 3. Represent this sum
" 7on your sketch. Is this
Solution: Figure 5.38
0 ≤ x ≤ 2Rof
R2
R 3sum an overestimate
or underestimate?
5-4h51 51.
5-4h52
1
2)dx
dx = A3 , we52.
(x) + 2) dx
f (x)
dx = A and(iii)1 ff(x)
dx = −A2 and f2(xf+
(x)
know (f
that
Since
(i) f (x) 0
(ii) g(x) 1
(x)·g(x)
−2
0
Z
Z 2
Z 3
(b) Is the following statement true? Explain your an-1
0
<
f
(x)
dx
<
−
f
(x)
dx
<
f (x) dx.
swer.
5-4h53
0
1 a graph of f (x)
2 = sin(x2 ) and mark on it
53. (a) Sketch
√
√
√
√
Average(f ) · Average(g)
= Average(f · g)
R2
R2
the points x = π, 2π, 3π, 4π.
In addition, 0 f (x) dx = A1 − A2 , which is negative, but smaller in magnitude than 1 f (x) dx. Thus
(b)
Use
your
graph
to
decide
which
of
the
four
numbers
1
1
f (x)
g(x)
5-2h30 30. GivenSolution:
f (x)dx = 4 and Figure ??, estimate:
−2 Z Z 2 f (x) dx <
5-4h44figa The1area x 5-4h44figb
A32 lies 1
x
1
2of height
rectangle 0 2
" √
nπ f (x) dx <2 0. sin(x ) dx n = 1, 2, 3, 4 0 inside a
20 and base 1, so A3 < 20. The area A2 lies inside a rectangle below
largest. Which is smallest? How many of the numthe
x-axis
of
height
10
and
width
1,
so
−10
<
A2 . isThus:
Figure 5.72
Figure 5.73
bers are positive? 5-4h45 5-4h46 (viii) < (ii) < (iii) < (vi) < (i) < (v) < ( iv) < (vii). 45. (a) Without computing any integrals, explainins5-4h54-56
why the
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
statistics, which is given by
2
1
√ e−x /2 .
2π For Problems ??, assuming F # = f , mark the quantity on a
copy of Figure ??.
F (x) R −1 3. (3 pts) Suppose f is even,
enough information to find R −1 Solution: Because f is even, −2 Z R4 f (x) dx = 3, and R−2
4
f (x)
1 2 f (x) dx = 5. Either find R4
1 f (x) dx, or show there is not dx. f (x) dx = R2 Z 4 f (x) dx, so 1 Z 2 f (x) dx = 4 f (x) dx + 1 f (x) dx = 3 + 5 = 8 . 1 2 4. (5 pts) Evaluate the following integrals:
R1
R1
R1
(a) (3 pts) If 0 (f (x) − 2g(x)) dx = 6 and 0 (2f (x) + 2g(x)) dx = 9, find 0 (f (x) − g(x)) dx.
Z 3
(b) (2 pts)
(x9 − 4x3 + x) dx
−3 Solution:
(a) Adding the first 2 integrals gives us 3
Z Z 1 f (x) dx −
0 R1
0 f (x) = 15, so
Z 1 f (x) = 5. We use the first integral again:
Z
−2 g(x) dx = 6, 0 0 Thus, Z Z 1 (f (x) − g(x)) dx =
0 Z 0 1 5 − 2 2g(x) dx = 6, R1 Z 1 g(x) dx = 1,
0 Z 1 0 1 f (x) − g(x) dx = 5 − (−1/2) = 0 0 1 1
g(x) dx = − .
2 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
Z 1 −1 1
x4 from x = −1 to x = 1: h
1 1
1
1 i 1
2
dx = − 3 = − − = − .
4
x
3x −1
3 3
3 Solution: We cannot apply the evaluation theorem because x14 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.
Z Z a ex dx = 6 2e2x dx. 0 Z a 0 0 a
e dx = e = ea − 1,
x b x 0 Z
6 b h ib
2e2x dx = 6 e2x = 6e2b − 6. 0 0 Thus
ea − 1 = 6e2b − 6, ea = 6e2b − 5,
Z 5 a = ln(6e2b − 5) . 1
dx = ln(5). Now find a fraction which approximates ln(5), by
1 x
Z 5
1
using M4 (midpoint sum with 4 rectangles) to approximate
dx.
x
1 7. (4 pts) Use the Evaluation Theorem to show (The actual value of ln(5) is 1.6094 . . .. For fun, plug your approximation into a calculator and compare)
Z 5 Solution: By the Evaluation Theorem,
1 5
1 dx = ln(x) = ln(5) − ln(1) = ln(5).
x
1 (the formulas for ln |x| and ln(x) are identical since x > 0 on the interval [1, 5]).
M4 = 1 · f 3
2 +1·f 5
2 +1·f 7
2 +1·f 9
2 = 2 2 2 2
496
+ + + =
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 .
Z Z Q∗ (f (Q) − P ∗ ) dQ = CS =
0 50 (
0 50
6000 − 60) dQ = [6000 ln(Q + 50) − 60Q] Q + 50
0 = (6000 ln(100) − 60(50)) − 6000 ln(50) = 6000 ln(2) − 3000 .
Also,
Z Z Q∗ (P ∗ − g(Q)) dQ = PS =
0 Z 50 (60 − (Q + 10)) dQ =
0 50 (50 − Q) dQ = [50Q −
0 2500
Q2 50 =
= 1250 .
2 0
2 Some extra practice (not to be handed in)
Z 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, h0 (1) = 3, h00 (1) = 4, h(2) = 6, h0 (2) = 5, h00 (2) = 13, and h00 is
R2
continuous everywhere. Find 1 h00 (u) du.
Solution: By the Evaluation Theorem, R2
1 h00 (u) du = h0 (2) − h0 (1) = 5 − 3 = 2. ...

View
Full Document