Math 241 Homework 11 Solutions

Math 241 Homework 11 Solutions

Section 5.2

Problem 1. Write the sum2

∑k=1

6k

k + 1

without sigma notation. Then evaluate it.

Solution

2

∑k=1

6k

k + 1=

6(1)

(1) + 1+

6(2)

(2) + 1

=6

2+

12

3

= 3 + 4

= 7

Problem 5. Write the sum3

∑k=1

(−1)k+1 sinπ

k

without sigma notation. Then evaluate it.

Solution

3

∑k=1

(−1)k+1 sinπ

k= (−1)(1)+1 sin

π

(1)+ (−1)(2)+1 sin

π

(2)+ (−1)(3)+1 sin

π

(3)

= sinπ − sinπ

2+ sin

π

3

= 0 − 1 +

√3

2

=

√3 − 2

2

1

Problem 8. Which of the following express 1 − 2 + 4 − 8 + 16 − 32 in sigma notation?

(a)6

∑k=1

(−2)k−1

(b)5

∑k=0

(−1)k2k

(c)3

∑k=−2

(−1)k+12k+2

Solution (a) and (b)

Problem 11. Express the sum 1+2+3+4+5+6 in sigma notation. The form of your answer willdepend on your choice of the lower limit of summation.

Solution6

∑k=1

k

Problem 13. Express the sum1

2+

1

4+

1

8+

1

16in sigma notation. The form of your answer will

depend on your choice of the lower limit of summation.

Solution4

∑k=1

1

2k

2

Section 5.3

Problem 1. Express the limit

lim∥P ∥→0

n

∑k=1

c2k∆xk, where P is a partition of [0,2]

as a definite integral.

Solution From the book we have

lim∥P ∥→0

n

∑k=1

f(ck)∆xk = ∫b

af(x) dx, where P a partition of [a, b]

Thus we have

lim∥P ∥→0

n

∑k=1

c2k∆xk = ∫2

0x2 dx

Problem 5. Express the limit

lim∥P ∥→0

n

∑k=1

1

1 − ck∆xk, whereP is a partition of [2,3]

as a definite integral.

Solution From the book we have

lim∥P ∥→0

n

∑k=1

f(ck)∆xk = ∫b

af(x) dx, where P a partition of [a, b]

Thus we have

lim∥P ∥→0

n

∑k=1

1

1 − ck∆xk = ∫

3

2

1

1 − xdx

3

Problem 9. Suppose that f and g are integrable and that

∫

2

1f(x) dx = −4, ∫

5

1f(x) dx = 6, ∫

5

1g(x) dx = 8

Use the rules in Table 5.3 to find

(a) ∫2

2g(x) dx

(b) ∫1

5g(x) dx

(c) ∫2

13f(x) dx

(d) ∫5

2f(x) dx

(e) ∫5

1[f(x) − g(x)] dx

(f) ∫5

1[4f(x) − g(x)] dx

Solution

(a) 0

(b)

∫

1

5g(x) dx = −∫

5

1g(x) dx = −8

(c)

∫

2

13f(x) dx = 3∫

2

1f(x) dx = 3(−4) = −12

(d)

∫

5

2f(x) dx = ∫

5

1f(x) dx − ∫

2

1f(x) dx = 6 − (−4) = 10

(e)

∫

5

1f(x) dx − ∫

5

1g(x) dx = 6 − 8 = −2

(f)

∫

5

1f(x) dx = 4∫

5

1f(x) dx − ∫

5

1g(x) dx = 4(6) − 8 = 16

4

Problem 11. Suppose that ∫2

1f(x) dx = 5. Find

(a) ∫2

1f(u) du

(b) ∫2

1

√3f(z) dz

(c) ∫1

2f(t) dt

(d) ∫2

1[−f(x)] dx

Solution

(a)

∫

2

1f(u) du = ∫

2

1f(x) dx = 5

(b)

∫

2

1

√3f(z) dz =

√3∫

2

1f(x) dx = 5

√3

(c)

∫

1

2f(t) dt = −∫

2

1f(x) dx = −5

(d)

∫

2

1[−f(x)] dx = −∫

2

1f(x) dx = −5

5

Problem 13. Suppose that f is integrable and that ∫3

0f(z) dz = 3 and ∫

4

0f(z) dz = 7. Find

(a) ∫4

3f(z) dz

(b) ∫3

4f(t) dt

Solution

(a)

∫

4

3f(z) dz = ∫

4

0f(z) dz − ∫

3

0f(z) dz

= 7 − 3

= 4

(b)

∫

3

4f(t) dt = −∫

4

3f(z) dz = −4

Problem 15. Graph the integrand and use areas to evaluate the integral: ∫4

−2(x

2+ 3) dx

Solution

The area can be found by subtracting the area of the smaller triangle from the area of the largertriangle. (Or by using the area of a trapezoid, or by adding the area of the triangle to the area ofthe rectangle, etc)

1

2(10)(5) −

1

2(2)(4) = 25 − 4 = 21

6

Problem 25. Use areas to evaluate ∫ba 2s ds, 0 < a < b

Solution The graph looks like

Using the area of a trapezoid we have

∫

b

a2s ds =

1

2(2b + 2a)(b − a) = (b + a)(b − a) = b2 − a2

Problem 41. Use the rules in Table 5.3 and Equations (1)-(3) to evaluate the integral ∫2

05x dx

Solution

∫

2

05x dx = 5∫

2

0x dx

= 5(22

2−

02

2)

= 5(2)

= 10

Problem 43. Use the rules in Table 5.3 and Equations (1)-(3) to evaluate the integral ∫2

0(2t−3) dt

Solution

∫

2

0(2t − 3) dt = 2∫

2

0t dt − ∫

2

03 dt

= 2(22

2−

02

2) − 3(2 − 0)

= 2(2) − 6

= −2

7

Problem 51. Use a definite integral to find the area of the region between the given curve and thex-axis on the interval [0, b]: y = 3x2

Solution

Area = ∫b

03x2 dx

= 3∫b

0x2 dx

= 3(b3

3−

03

3)

= 3 ⋅b3

3

= b3

Problem 61. Graph the function and find its average value over the given interval: g(x) = ∣x∣ − 1on

(a) [−1,1]

(b) [1,3]

(c) [−1,3]

Solution The graph looks like

(a) The average value of g(x) over [-1,1] is found by

1

1 − (−1)∫

1

−1( ∣x∣ − 1) dx

8

The integral is given by the area of the following shaded region (with below the x-axis con-sidered negative)

∫

1

−1( ∣x∣ − 1) dx =

1

2(1 − (−1))(−1)

=1

2(2)(−1)

= −1

This the average value is1

1 − (−1)(−1) =

1

2(−1) = −1 2

(b) The average value of g(x) over [1,3] is found by

1

3 − 1∫

1

−1( ∣x∣ − 1) dx


∫

3

1( ∣x∣ − 1) dx =

1

2(3 − 1)(2)

=1

2(2)(2)

= 2

Thus the average value is1

3 − 1(2) =

1

2(2) = 1

9

(c) The average value of g(x) over [-1,3] is found by

1

3 − (−1)∫

1

−1( ∣x∣ − 1) dx


∫

3

−1( ∣x∣ − 1) dx =

1

1 − (−1)(2)(−1) +

1

2(3 − 1)(2)

=1

2(2)(−1) +

1

2(2)(2)

= −1 + 2

= 1

Thus the average value is1

3 − (−1)(1) =

1

4(1) =

1

4

10

Problem 77. (a) Suppose the graph of a continuous function f(x) rises steadily as x movesfrom left to right across an interval [a, b]. Let P be a partition of [a, b] into n subintervalsof length ∆x = (b − a)/n. Show by referring to the accompanying figure that the differencebetween the upper and lower sums for f on this partition can be represented graphically asthe area of a rectangle R whose dimensions are [f(b) − f(a)] by ∆x. ( Hint: The differenceU −L is the sum of areas of rectangles whose diagonals Q0Q1,Q1Q2, ...,Qn−1Qn lie along thecurve. There is no overlapping when these rectangles are shifted horizontally onto R)

(b) Suppose that instead of being equal, the lengths ∆xk of the subintervals of the partition of[a.b] vary in size. Show that

U −L ≤ ∣f(b) − f(a)∣∆xmax

where ∆xmax is the norm of P , and hence that lim∥P ∥→0

(U −L) = 0

x

y

0 x0 = a xn = bx1

Q1

Q2

Q3

x2

y = f (x)

f (b) − f (a)

R

Δx

84. Upper and lower sums for decreasing functions (Continua-tion of Exercise 83.)

a. Draw a figure like the one in Exercise 83 for a continuous function ƒ(x) whose values decrease steadily as x moves from left to right across the interval 3a, b4 . Let P be a partition of 3a, b4 into subintervals of equal length. Find an expression for U - L that is analogous to the one you found for U - L in Exercise 83a.

b. Suppose that instead of being equal, the lengths ∆xk of the subintervals of P vary in size. Show that the inequality

U - L … 0 ƒ(b) - ƒ(a) 0 ∆xmax

of Exercise 83b still holds and hence that lim}P}S0

(U - L) = 0.

85. Use the formula

sin h + sin 2h + sin 3h + g+ sin mh

=cos (h>2) - cos ((m + (1>2))h)

2 sin (h>2)

to find the area under the curve y = sin x from x = 0 to x = p>2 in two steps:

a. Partition the interval 30, p>24 into n subintervals of equal length and calculate the corresponding upper sum U; then

b. Find the limit of U as n S q and ∆x = (b - a)>n S 0.

86. Suppose that ƒ is continuous and nonnegative over 3a, b4 , as in the accompanying figure. By inserting points

x1, x2,c, xk - 1, xk,c, xn - 1

as shown, divide 3a, b4 into n subintervals of lengths ∆x1 = x1 - a, ∆x2 = x2 - x1,c, ∆xn = b - xn - 1, which need not be equal.

a. If mk = min 5ƒ(x) for x in the kth subinterval6 , explain the connection between the lower sum

L = m1 ∆x1 + m2 ∆x2 + g+ mn ∆xn

and the shaded regions in the first part of the figure.

b. If Mk = max 5ƒ(x) for x in the kth subinterval6 , explain the connection between the upper sum

U = M1 ∆x1 + M2 ∆x2 + g+ Mn ∆xn

and the shaded regions in the second part of the figure.

c. Explain the connection between U - L and the shaded regions along the curve in the third part of the figure.

x

y

0 a bx1 x2 x3 xk−1 xn−1xk

y = f (x)

x

y

0 a bxk+1xk

x

y

0 a bxk+1xk

b − a

P

87. We say ƒ is uniformly continuous on 3a, b4 if given any P 7 0, there is a d 7 0 such that if x1, x2 are in 3a, b4 and 0 x1 - x2 0 6 d, then 0 ƒ(x1) - ƒ(x2) 0 6 P. It can be shown that a continuous function on 3a, b4 is uniformly continuous. Use this and the figure for Exercise 86 to show that if ƒ is continuous and P 7 0 is given, it is possible to make U - L … P # (b - a) by making the largest of the ∆xk>s sufficiently small.

88. If you average 30 mi >h on a 150-mi trip and then return over the same 150 mi at the rate of 50 mi >h, what is your average speed for the trip? Give reasons for your answer.

COMPUTER EXPLORATIONSIf your CAS can draw rectangles associated with Riemann sums, use it to draw rectangles associated with Riemann sums that converge to the integrals in Exercises 89–94. Use n = 4, 10, 20, and 50 subinter-vals of equal length in each case.

89. L1

0(1 - x) dx = 1

2

5.3 The Definite Integral 311

Solution

(a)

Area of all rectangles = (x1 − a)(f(x1) − f(a)) + (x2 − x1)(f(x1) − f(x2))

+ (x3 − x2)(f(x3) − f(x2))... + (b − xn−1)(f(b) − f(xn−1))

(Since all widths are ∆x we have)

= ∆x(f(x1) − f(a)) +∆x(f(x2) − f(x1))+

∆x(f(x3) − f(x2))... +∆x(f(b) − f(xn−1))

= ∆x(��f(x1) − f(a) +��f(x2) −��f(x1) +��f(x3) −��f(x2) + ... + f(b) −��f(xn−1))

(All middle terms cancel out to give)

= ∆x(f(b) − f(a))

= Area of the right rectangle

11

(b)

Area of all rectangles = (x1 − a)(f(x1) − f(a)) + (x2 − x1)(f(x1) − f(x2))

+ (x3 − x2)(f(x3) − f(x2))... + (b − xn−1)(f(b) − f(xn−1))

(Letting ∆xmax be the largest subinterval length we have this is)

≤ ∆xmax(f(x1) − f(a)) +∆xmax(f(x2) − f(x1))

+∆xmax(f(x3) − f(x2))... +∆xmax(f(b) − f(xn−1))

= ∆xmax(��f(x1) − f(a) +��f(x2) −��f(x1) +��f(x3) −��f(x2) + ... + f(b) −��f(xn−1))

(All middle terms cancel out to give)

= ∆xmax(f(b) − f(a))

12

Section 5.4

Problem 1. Evaluate ∫0

−2(2x + 5) dx

Solution

∫

0

−2(2x + 5) dx = (x2 + 5x)∣

0

−2

= (02 + 5(0)) − ((−2)2 + 5(−2))

= −(4 − 10)

= 6


0(3x −

x3

4)

Solution

∫

4

0(3x −

x3

4) = (

3x2

2−x4

16)

RRRRRRRRRRR

4

0

= (3(16)

2−

256

16) − (0 − 0)

= 24 − 16

= 8


0(x2 +

√x) dx

Solution

∫

1

0(x2 +

√x) dx = (

x3

3+

2

3x3/2)

RRRRRRRRRRR

1

0

= (1

3+

2

3) − (0 − 0)

= 1

13

Problem 11. Evaluate ∫π/3

02 sec2 x dx

Solution

∫

π/3

02 sec2 x dx = 2 tanx∣

π/30

= 2 tanπ

3− 2 tan 0

= 2√

3


π/2

1 + cos 2t

2dt

Solution

∫

0

π/2

1 + cos 2t

2dt = ∫

0

π/2(

1

2+

cos 2t

2) dt

= (1

2t +

sin 2t

4)

RRRRRRRRRRR

0

π/2

= (0 + 0) − (π

4+

sinπ

4)

= −π

4

14

Problem 19. Evaluate ∫−1

1(r + 1)2 dr

Solution

∫

−1

1(r + 1)2 dr = ∫

−1

1(r2 + 2r + 1) dr

= (r3

3+ r2 + r)

RRRRRRRRRRR

−1

1

= (−1

3+ 1 − 1) − (

1

3+ 1 + 1)

= −1

3−

1

3− 2

= −2

3−

6

3

= −8

3


√2(u7

2−

1

u5) du

Solution

∫

1

√2(u7

2−

1

u5) du = (

u8

16+

1

4u4)

RRRRRRRRRRR

1

√2

= (1

16+

1

4) − (1 +

1

16)

=1

16+

4

16−

16

16−

1

16

= −12

16

= −3

4

15

Problem 27. Find the derivative by

(a) by evaluating the integral and differentiating the result.

(b) by differentiating the integral directly

d

dx∫

√x

0cos t dt

Solution

(a)

d

dx∫

√x

0cos t dt =

d

dx(sin t)∣

√x

0

=d

dx(sin

√x − sin 0)

=d

dxsin

√x

=1

2√x

cos√x

(b) Let u =√x⇒ du =

1

2√x

d

dx∫

√x

0cos t dt = (

d

du∫

u

0cos t dt) ⋅

du

dx

= cosu ⋅1

2√x

=1

2√x

cos√x

16

Problem 43. Find the areas of the shaded regions

39. ddxL

1x

0 cos t dt 40. d

dxL sin x

13t2 dt

41. ddtL

t4

01u du 42. d

duL tan u

0 sec2 y dy

43. ddx

Lx3

0 e- t dt 44. d

dt L

2t

0 ax4 + 321 - x2

b dx

Find dy>dx in Exercises 45–56.

45. y = Lx

021 + t2 dt 46. y = L

x

1 1t dt, x 7 0

47. y = L01x

sin (t2) dt 48. y = x Lx2

2 sin (t3) dt

49. y = Lx

-1 t2

t2 + 4 dt - L

x

3 t2

t2 + 4 dt

50. y = aL x

0 (t3 + 1)10 dtb3

51. y = L sin x

0 dt21 - t2

, 0 x 0 6 p2

52. y = L0

tan x dt1 + t2 53. y = L

ex2

0 12t

dt

54. y = L1

2x

23 t dt 55. y = Lsin-1 x

0 cos t dt

56. y = Lx1>p

-1 sin-1 t dt

AreaIn Exercises 57–60, find the total area between the region and the x-axis.

57. y = -x2 - 2x, -3 … x … 2

58. y = 3x2 - 3, -2 … x … 2

59. y = x3 - 3x2 + 2x, 0 … x … 2

60. y = x1>3 - x, -1 … x … 8

Find the areas of the shaded regions in Exercises 61–64.

61.

x

y

0

2

p

y = 2

x = p

y = 1 + cos x

62. y

x

1

p6

5p6

y = sin x

63.

u

y

−"

2

"

2

p4

p4− 0

y = sec u tan u

64.

t

y

p4− 0 1

1

2

y = sec2 t

y = 1 − t2

Initial Value ProblemsEach of the following functions solves one of the initial value prob-lems in Exercises 65–68. Which function solves which problem? Give brief reasons for your answers.

a. y = Lx

1 1t dt - 3 b. y = L

x

0 sec t dt + 4

c. y = Lx

-1 sec t dt + 4 d. y = L

x

p

1t dt - 3

65. dydx

= 1x , y(p) = -3 66. y′ = sec x, y(-1) = 4

67. y′ = sec x, y(0) = 4 68. y′ = 1x , y(1) = -3

Express the solutions of the initial value problems in Exercises 69 and 70 in terms of integrals.

69. dydx

= sec x, y(2) = 3

70. dydx

= 21 + x2, y(1) = -2

Theory and Examples71. Archimedes’ area formula for parabolic arches Archimedes

(287–212 b.c.), inventor, military engineer, physicist, and the greatest mathematician of classical times in the Western world, discovered that the area under a parabolic arch is two-thirds the base times the height. Sketch the parabolic arch y = h - (4h>b2)x2, -b>2 … x … b>2, assuming that h and b are positive. Then use calculus to find the area of the region enclosed between the arch and the x-axis.

72. Show that if k is a positive constant, then the area between the x-axis and one arch of the curve y = sin kx is 2>k.

73. Cost from marginal cost The marginal cost of printing a poster when x posters have been printed is

dcdx

= 121x

dollars. Find c(100) - c(1), the cost of printing posters 2–100.

74. Revenue from marginal revenue Suppose that a company’s marginal revenue from the manufacture and sale of eggbeaters is

drdx

= 2 - 2>(x + 1)2,

where r is measured in thousands of dollars and x in thousands of units. How much money should the company expect from a pro-duction run of x = 3 thousand eggbeaters? To find out, integrate the marginal revenue from x = 0 to x = 3.

5.4 The Fundamental Theorem of Calculus 321

Solution We can take the area of the rectangle and subtract the area under the curve

Area = 2π − ∫π

0(1 + cosx) dx

= 2π − (x + sinx)∣π0

= 2π − [(π + sinπ) − (0 + sin 0)]

= 2π − π

= π


39. ddxL

1x

0 cos t dt 40. d

dxL sin x

13t2 dt

41. ddtL

t4

01u du 42. d

duL tan u

0 sec2 y dy

43. ddx

Lx3

0 e- t dt 44. d

dt L

2t

0 ax4 + 321 - x2

b dx


45. y = Lx

021 + t2 dt 46. y = L

x

1 1t dt, x 7 0

47. y = L01x


2 sin (t3) dt

49. y = Lx

-1 t2

t2 + 4 dt - L

x

3 t2

t2 + 4 dt

50. y = aL x

0 (t3 + 1)10 dtb3

51. y = L sin x

0 dt21 - t2

, 0 x 0 6 p2

52. y = L0

tan x dt1 + t2 53. y = L

ex2

0 12t

dt

54. y = L1

2x

23 t dt 55. y = Lsin-1 x

0 cos t dt

56. y = Lx1>p

-1 sin-1 t dt


57. y = -x2 - 2x, -3 … x … 2

58. y = 3x2 - 3, -2 … x … 2

59. y = x3 - 3x2 + 2x, 0 … x … 2

60. y = x1>3 - x, -1 … x … 8


61.

x

y

0

2

p

y = 2

x = p

y = 1 + cos x

62. y

x

1

p6

5p6

y = sin x

63.

u

y

−"

2

"

2

p4

p4− 0

y = sec u tan u

64.

t

y

p4− 0 1

1

2

y = sec2 t

y = 1 − t2


a. y = Lx

1 1t dt - 3 b. y = L

x

0 sec t dt + 4

c. y = Lx

-1 sec t dt + 4 d. y = L

x

p

1t dt - 3

65. dydx

= 1x , y(p) = -3 66. y′ = sec x, y(-1) = 4

67. y′ = sec x, y(0) = 4 68. y′ = 1x , y(1) = -3


69. dydx

= sec x, y(2) = 3

70. dydx

= 21 + x2, y(1) = -2





dcdx

= 121x



drdx

= 2 - 2>(x + 1)2,



Solution We can take the area under the curve from π/6 to 5π/6 and subtract the area of therectangle

Area = ∫5π/6

π/6sinx dx −

1

2⋅2π

3

= − cosx∣5π/6π/6−π

3

= − cos5π

6+ cos

π

6−π

3

=

√3

2+

√3

2−π

3

=√

3 −π

3

17


39. ddxL

1x

0 cos t dt 40. d

dxL sin x

13t2 dt

41. ddtL

t4

01u du 42. d

duL tan u

0 sec2 y dy

43. ddx

Lx3

0 e- t dt 44. d

dt L

2t

0 ax4 + 321 - x2

b dx


45. y = Lx

021 + t2 dt 46. y = L

x

1 1t dt, x 7 0

47. y = L01x


2 sin (t3) dt

49. y = Lx

-1 t2

t2 + 4 dt - L

x

3 t2

t2 + 4 dt

50. y = aL x

0 (t3 + 1)10 dtb3

51. y = L sin x

0 dt21 - t2

, 0 x 0 6 p2

52. y = L0

tan x dt1 + t2 53. y = L

ex2

0 12t

dt

54. y = L1

2x

23 t dt 55. y = Lsin-1 x

0 cos t dt

56. y = Lx1>p

-1 sin-1 t dt


57. y = -x2 - 2x, -3 … x … 2

58. y = 3x2 - 3, -2 … x … 2

59. y = x3 - 3x2 + 2x, 0 … x … 2

60. y = x1>3 - x, -1 … x … 8


61.

x

y

0

2

p

y = 2

x = p

y = 1 + cos x

62. y

x

1

p6

5p6

y = sin x

63.

u

y

−"

2

"

2

p4

p4− 0

y = sec u tan u

64.

t

y

p4− 0 1

1

2

y = sec2 t

y = 1 − t2


a. y = Lx

1 1t dt - 3 b. y = L

x

0 sec t dt + 4

c. y = Lx

-1 sec t dt + 4 d. y = L

x

p

1t dt - 3

65. dydx

= 1x , y(p) = -3 66. y′ = sec x, y(-1) = 4

67. y′ = sec x, y(0) = 4 68. y′ = 1x , y(1) = -3


69. dydx

= sec x, y(2) = 3

70. dydx

= 21 + x2, y(1) = -2





dcdx

= 121x



drdx

= 2 - 2>(x + 1)2,



Solution Since the function has origin symmetry, the shaded region under the x-axis is the sameas the missing section under the curve from 0 to π/4. Thus the shaded region is the same as thearea of the rectangle from x = −π/4 to x = π/4 and from y = 0 to y =

√2 i.e.

Area =

√2π

2

18

Math 241 Homework 11 Solutions

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Transcript of Math 241 Homework 11 Solutions