Answers to Exercises

Appendix F Answers to Exercises

1 Integration
1.1 Definition of the Integral
1.1.8 Exercises

1.1.8.1.

Answer.

The area is between

1.5

and

2.5

square units.

1.1.8.2.

Answer.

The shaded area is between 2.75 and 4.25 square units. (Other estimates are possible, but this is a reasonable estimate, using methods from this chapter.)

1.1.8.3.

Answer.

The area under the curve is a number in the interval

(\frac{3}{8} [\frac{1}{2} + \frac{1}{\sqrt{2}}], \frac{3}{8} [1 + \frac{1}{\sqrt{2}}]) .

1.1.8.4.

Answer.

left

1.1.8.5.

Answer.

Many answers are possible. One example is

f (x) = \sin x,

[a, b] = [0, π],

n = 1 .

Another example is

f (x) = \sin x,

[a, b] = [0, 5 π],

n = 5 .

1.1.8.6.

Answer.

Some of the possible answers are given, but more exist.

$\sum_{i = 3}^{7} i; \sum_{i = 1}^{5} (i + 2)$
$\sum_{i = 3}^{7} 2 i; \sum_{i = 1}^{5} (2 i + 4)$
$\sum_{i = 3}^{7} (2 i + 1); \sum_{i = 1}^{5} (2 i + 5)$
$\sum_{i = 1}^{8} (2 i - 1); \sum_{i = 0}^{7} (2 i + 1)$

1.1.8.7.

Answer.

Some answers are below, but others are possible.

$\sum_{i = 1}^{4} \frac{1}{3^{i}}$ and $\sum_{i = 1}^{4} {(\frac{1}{3})}^{i}$
$\sum_{i = 1}^{4} \frac{2}{3^{i}}$ and $\sum_{i = 1}^{4} 2 {(\frac{1}{3})}^{i}$
$\sum_{i = 1}^{4} (- 1)^{i} \frac{2}{3^{i}}$ and $\sum_{i = 1}^{4} \frac{2}{(- 3)^{i}}$
$\sum_{i = 1}^{4} (- 1)^{i + 1} \frac{2}{3^{i}}$ and $\sum_{i = 1}^{4} - \frac{2}{(- 3)^{i}}$

1.1.8.8.

Answer.

$\sum_{i = 1}^{5} \frac{2 i - 1}{3^{i}}$
$\sum_{i = 1}^{5} \frac{1}{3^{i} + 2}$
$\sum_{i = 1}^{7} i \cdot 10^{4 - i}$ and $\sum_{i = 1}^{7} \frac{i}{10^{i - 4}}$

1.1.8.9.

Answer.

$\frac{5}{2} [1 - {(\frac{3}{5})}^{101}]$
$\frac{5}{2} {(\frac{3}{5})}^{50} [1 - {(\frac{3}{5})}^{51}]$
$270$
$\frac{1 - {(\frac{1}{e})}^{b}}{e - 1} + \frac{e}{4} {[b (b + 1)]}^{2}$

1.1.8.10.

Answer.

$50 \cdot 51 = 2550$
${[\frac{1}{2} (95) (96)]}^{2} - {[\frac{1}{2} (4) (5)]}^{2} = 20, 793, 500$
$- 1$
$- 10$

1.1.8.11.

Answer.

1.1.8.12. (✳).

Answer.

n = 4,

a = 2,

and

b = 6

1.1.8.13.

Answer.

One answer is below, but other interpretations exist.

1.1.8.14.

Answer.

Many interpretations are possible--see the solution to Question 13 for a more thorough discussion--but the most obvious is given below.

1.1.8.15. (✳).

Answer.

Three answers are possible. It is a midpoint Riemann sum for

f

on the interval

[1, 5]

with

n = 4 .

It is also a left Riemann sum for

f

on the interval

[1.5, 5.5]

with

n = 4 .

It is also a right Riemann sum for

f

on the interval

[0.5, 4.5]

with

n = 4 .

1.1.8.16.

Answer.

\frac{25}{2}

1.1.8.17.

Answer.

\frac{21}{2}

1.1.8.18. (✳).

Answer.

\sum_{i = 1}^{50} (5 + (i - \frac{1}{2}) \frac{1}{5})^{8} \frac{1}{5}

1.1.8.19. (✳).

Answer.

54

1.1.8.20. (✳).

Answer.

\int_{- 1}^{7} f (x) d x = lim_{n \to \infty} \sum_{i = 1}^{n} f (- 1 + \frac{8 i}{n}) \frac{8}{n}

1.1.8.21. (✳).

Answer.

f (x) = \sin^{2} (2 + x)

and

b = 4

1.1.8.22. (✳).

Answer.

f (x) = x \sqrt{1 - x^{2}}

1.1.8.23. (✳).

Answer.

\int_{0}^{3} e^{- x / 3} \cos (x) d x

1.1.8.24. (✳).

Answer.

\int_{0}^{1} x e^{x} d x

1.1.8.25. (✳).

Answer.

Possible answers include:

\begin{matrix} \int_{0}^{2} e^{- 1 - x} d x \\ \int_{1}^{3} e^{- x} d x \\ 2 \int_{1 / 2}^{3 / 2} e^{- 2 x} d x and \\ 2 \int_{0}^{1} e^{- 1 - 2 x} d x . \end{matrix}

1.1.8.26.

Answer.

\frac{r^{3 n + 3} - 1}{r^{3} - 1}

1.1.8.27.

Answer.

r^{5} (\frac{r^{96} - 1}{r - 1})

1.1.8.28. (✳).

Answer.

5

1.1.8.29.

Answer.

1.1.8.30.

Answer.

\frac{b^{2} - a^{2}}{2}

1.1.8.31.

Answer.

\frac{b^{2} - a^{2}}{2}

1.1.8.32.

Answer.

4 π

1.1.8.33. (✳).

Answer.

\int_{0}^{3} f (x) d x = 2.5

1.1.8.34. (✳).

Answer.

53 m

1.1.8.35.

Answer.

true

1.1.8.36.

Answer.

3200 km

1.1.8.37. (✳).

Answer.

(a) There are many possible answers. Two are

\int_{- 2}^{0} \sqrt{4 - x^{2}} d x

and

\int_{0}^{2} \sqrt{4 - (- 2 + x)^{2}} d x .

(b)

π

1.1.8.38. (✳).

Answer.

(a)

30

(b)

41 \frac{1}{4}

1.1.8.39. (✳).

Answer.

\frac{56}{3}

1.1.8.40. (✳).

Answer.

6

1.1.8.41. (✳).

Answer.

12

1.1.8.42.

Answer.

f (x) = \frac{3}{10} {(\frac{x}{5} + 8)}^{2} \sin (\frac{2 x}{5} + 2)

1.1.8.43.

Answer.

\frac{1}{\log 2}

1.1.8.44.

Answer.

(a)

\frac{1}{\log 10} (10^{b} - 10^{a})

(b)

\frac{1}{\log c} (c^{b} - c^{a});

yes, it agrees.

1.1.8.45.

Answer.

\frac{π}{4} - \frac{1}{2} \arccos (a) + \frac{1}{2} a \sqrt{1 - a^{2}}

1.1.8.46.

Answer.

$[f (b) - f (a)] \cdot \frac{b - a}{n}$
Choose $n$ to be an integer that is greater than or equal to $100 [f (b) - f (a)] (b - a) .$

1.1.8.47.

Answer.

true (but note, for a non-linear function, it is possible that the midpoint Riemann sum is not the average of the other two)

1.2 Basic properties of the definite integral
1.2.3 Exercises

1.2.3.1.

Answer.

Possible drawings:

1.2.3.2.

Answer.

\sin b - \sin a

1.2.3.3. (✳).

Answer.

(a) False. For example, the function

\begin{array}{r} f (x) = {\begin{cases} 0 & for x < 0 \\ 1 & for x \geq 0 \end{cases} \end{array}

provides a counterexample.

(b) False. For example, the function

f (x) = x

provides a counterexample.

\begin{aligned} f (x) = {\begin{cases} 0 & for x < \frac{1}{2} \\ 1 & for x \geq \frac{1}{2} \end{cases} & and & g (x) = {\begin{cases} 0 & for x \geq \frac{1}{2} \\ 1 & for x < \frac{1}{2} \end{cases} \end{aligned}

provide a counterexample.

1.2.3.4.

Answer.

(a)

- \frac{1}{20},

(b) positive, (c) negative, (d) positive.

1.2.3.5. (✳).

Answer.

- 21

1.2.3.6. (✳).

Answer.

- 6

1.2.3.7. (✳).

Answer.

1.2.3.8.

Answer.

$\frac{π}{4} - \frac{1}{2} \arccos (- a) - \frac{1}{2} a \sqrt{1 - a^{2}} = - \frac{π}{4} + \frac{1}{2} \arccos (a) - \frac{1}{2} a \sqrt{1 - a^{2}}$
$\frac{1}{2} \arccos (a) - \frac{1}{2} a \sqrt{1 - a^{2}}$

1.2.3.9. (✳).

Answer.

5

1.2.3.10.

Answer.

1.2.3.11.

Answer.

1.2.3.12. (✳).

Answer.

20 + 2 π

1.2.3.13. (✳).

Answer.

0

1.2.3.14. (✳).

Answer.

0

1.2.3.15.

Answer.

1.2.3.16.

Answer.

(a)

y = \frac{1}{b} \sqrt{1 - (a x)^{2}}

(b)

\frac{a}{b} \int_{- \frac{1}{a}}^{\frac{1}{a}} \sqrt{\frac{1}{a^{2}} - x^{2}} d x

(c)

\frac{π}{a b}

1.2.3.17.

Answer.

$\times$	even	odd
even	even	odd
odd	odd	even

1.2.3.18.

Answer.

f (0) = 0;

g (0)

can be any real number

1.2.3.19.

Answer.

f (x) = 0

for every

x

1.2.3.20.

Answer.

The derivative of an even function is odd, and the derivative of an odd function is even.

1.3 The Fundamental Theorem of Calculus
1.3.2 Exercises

1.3.2.1. (✳).

Answer.

e^{2} - e^{- 2}

1.3.2.2. (✳).

Answer.

F (x) = \frac{x^{4}}{4} + \frac{1}{2} \cos 2 x + \frac{1}{2} .

1.3.2.3. (✳).

Answer.

(a) True

(b) False

\int_{a}^{b} f (x) d x = \int_{a}^{b} x f (x) d x = 0 .

1.3.2.4.

Answer.

false

1.3.2.5.

Answer.

false

1.3.2.6.

Answer.

\sin (x^{2})

1.3.2.7.

Answer.

\sqrt[3]{e}

1.3.2.8.

Answer.

For any constant

C,

F (x) + C

is an antiderivative of

f (x) .

So, for example,

F (x)

and

F (x) + 1

are both antiderivatives of

f (x) .

1.3.2.9.

Answer.

We differentiate with respect to $a .$ Recall $\frac{d}{d x} {\arccos x} = \frac{- 1}{\sqrt{1 - x^{2}}} .$ To differentiate $\frac{1}{2} a \sqrt{1 - a^{2}},$ we use the product and chain rules.
$\begin{aligned} \frac{d}{d a} {\frac{π}{4} - \frac{1}{2} \arccos (a) + \frac{1}{2} a \sqrt{1 - a^{2}}} \\ = 0 - \frac{1}{2} \cdot \frac{- 1}{\sqrt{1 - a^{2}}} + (\frac{1}{2} a) \cdot \frac{- 2 a}{2 \sqrt{1 - a^{2}}} + \frac{1}{2} \sqrt{1 - a^{2}} \\ = \frac{1}{2 \sqrt{1 - a^{2}}} - \frac{a^{2}}{2 \sqrt{1 - a^{2}}} + \frac{1 - a^{2}}{2 \sqrt{1 - a^{2}}} \\ = \frac{1 - a^{2} + 1 - a^{2}}{2 \sqrt{1 - a^{2}}} \\ = \frac{2 (1 - a^{2})}{2 \sqrt{1 - a^{2}}} \\ = \sqrt{1 - a^{2}} \end{aligned}$
$F (x) = \frac{5 π}{4} - \frac{1}{2} \arccos (x) + \frac{1}{2} x \sqrt{1 - x^{2}}$

1.3.2.10.

Answer.

(a) 0

(b),(c) The FTC does not apply, because the integrand is not continuous over the interval of integration.

1.3.2.11.

Answer.

1.3.2.12.

Answer.

(a) zero

(b) increasing when

0 < x < 1

and

3 < x < 4;

decreasing when

1 < x < 3

1.3.2.13.

Answer.

(a) zero

(b)

G (x)

is increasing when

1 < x < 3,

and it is decreasing when

0 < x < 1

and when

3 < x < 4 .

1.3.2.14.

Answer.

Using the definition of the derivative,

\begin{aligned} F^{'} (x) & = lim_{h \to 0} \frac{F (x + h) - F (x)}{h} \\ = lim_{h \to 0} \frac{\int_{a}^{x + h} t d t - \int_{a}^{x} t d t}{h} \\ = lim_{h \to 0} \frac{\int_{x}^{x + h} t d t}{h} \end{aligned}

The numerator describes the area of a trapezoid with base $h$ and heights $x$ and $x + h .$

\begin{aligned} = lim_{h \to 0} \frac{\frac{1}{2} h (x + x + h)}{h} \\ = lim_{h \to 0} (x + \frac{1}{2} h) \\ = x \end{aligned}

So,

F^{'} (x) = x .

1.3.2.15.

Answer.

f (t) = 0

1.3.2.16.

Answer.

\int \log (a x) d x = x \log (a x) - x + C,

where

a

is a given constant, and

C

is any constant.

1.3.2.17.

Answer.

\int x^{3} e^{x} d x = e^{x} (x^{3} - 3 x^{2} + 6 x - 6) + C

1.3.2.18.

Answer.

\int \frac{1}{\sqrt{x^{2} + a^{2}}} d x = \log | x + \sqrt{x^{2} + a^{2}} | + C

when

a

is a given constant. As usual,

C

is an arbitrary constant.

1.3.2.19.

Answer.

\int \frac{x}{\sqrt{x (a + x)}} d x = \sqrt{x (a + x)} - a \log (\sqrt{x} + \sqrt{a + x}) + C

1.3.2.20. (✳).

Answer.

5 - \cos 2

1.3.2.21. (✳).

Answer.

2

1.3.2.22.

Answer.

\frac{1}{5} \arctan (5 x) + C

1.3.2.23.

Answer.

\arcsin (\frac{x}{\sqrt{2}}) + C

1.3.2.24.

Answer.

\tan x - x + C

1.3.2.25.

Answer.

- \frac{3}{4} \cos (2 x) + C,

or equivalently,

\frac{3}{2} \sin^{2} x + C

1.3.2.26.

Answer.

\frac{1}{2} x + \frac{1}{4} \sin (2 x) + C

1.3.2.27. (✳).

Answer.

F^{'} (\frac{π}{2}) = \log (3)

G^{'} (\frac{π}{2}) = - \log (3)

1.3.2.28. (✳).

Answer.

f (x)

is increasing when

- \infty < x < 1

and when

2 < x < \infty .

1.3.2.29. (✳).

Answer.

F^{'} (x) = - \frac{\sin x}{\cos^{3} x + 6}

1.3.2.30. (✳).

Answer.

4 x^{3} e^{(1 + x^{4})^{2}}

1.3.2.31. (✳).

Answer.

(\sin^{6} x + 8) \cos x

1.3.2.32. (✳).

Answer.

F^{'} (1) = 3 e^{- 1}

1.3.2.33. (✳).

Answer.

\frac{\sin u}{1 + \cos^{3} u}

1.3.2.34. (✳).

Answer.

f (x) = 2 x

1.3.2.35. (✳).

Answer.

f (4) = 4 π

1.3.2.36. (✳).

Answer.

(a)

(2 x + 1) e^{- x^{2}}

(b)

x = - 1 / 2

1.3.2.37. (✳).

Answer.

e^{\sin x} - e^{\sin (x^{4} - x^{3})} (4 x^{3} - 3 x^{2})

1.3.2.38. (✳).

Answer.

- 2 x \cos (e^{- x^{2}}) - 5 x^{4} \cos (e^{x^{5}})

1.3.2.39. (✳).

Answer.

e^{x} \sqrt{\sin (e^{x})} - \sqrt{\sin (x)}

1.3.2.40. (✳).

Answer.

14

1.3.2.41. (✳).

Answer.

\frac{5}{2}

1.3.2.42. (✳).

Answer.

45 m

1.3.2.43. (✳).

Answer.

f^{'} (x) = (2 - 2 x) \log (1 + e^{2 x - x^{2}})

and

f (x)

achieves its absolute maximum at

x = 1,

because

f (x)

is increasing for

x < 1

and decreasing for

x > 1 .

1.3.2.44. (✳).

Answer.

The minimum is

\int_{0}^{- 1} \frac{d t}{1 + t^{4}} .

x

runs from

- \infty

\infty,

the function

f (x) = \int_{0}^{x^{2} - 2 x} \frac{d t}{1 + t^{4}}

decreases until

x

reaches 1 and then increases all

x > 1 .

So the minimum is achieved for

x = 1 .

x = 1,

x^{2} - 2 x = - 1 .

1.3.2.45. (✳).

Answer.

F

achieves its maximum value at

x = π .

1.3.2.46. (✳).

Answer.

2

1.3.2.47. (✳).

Answer.

\log 2

1.3.2.48.

Answer.

In the sketch below, open dots denote inflection points, and closed dots denote extrema.

1.3.2.49. (✳).

Answer.

(a)

3 x^{2} \int_{0}^{x^{3} + 1} e^{t^{3}} d t + 3 x^{5} e^{(x^{3} + 1)^{3}}

(b)

y = - 3 (x + 1)

1.3.2.50.

Answer.

Both students.

1.3.2.51.

Answer.

(a)

27 (1 - \cos 3)

(b)

x^{3} \sin (x) + 3 x^{2} [1 - \cos (x)]

1.3.2.52.

Answer.

f (x) = 0

for all

x,

then

F (x)

is even and possibly also odd.

f (x) \neq 0

for some

x,

then

F (x)

is not even. It might be odd, and it might be neither even nor odd.

(Perhaps surprisingly, every antiderivative of an odd function is even.)

1.4 Substitution
1.4.2 Exercises

1.4.2.1.

Answer.

(a) true

(b) false

1.4.2.2.

Answer.

The reasoning is not sound: when we do a substitution, we need to take care of the differential (

d x

). Remember the method of substitution comes from the chain rule: there should be a function and its derivative. Here’s the way to do it:

Problem: Evaluate $\int (2 x + 1)^{2} d x .$

Work: We use the substitution $u = 2 x + 1 .$ Then $d u = 2 d x,$ so $d x = \frac{1}{2} d u :$

$\begin{aligned} \int (2 x + 1)^{2} d x & = \int u^{2} \cdot \frac{1}{2} d u \\ = \frac{1}{6} u^{3} + C \\ = \frac{1}{6} {(2 x + 1)}^{3} + C \end{aligned}$

1.4.2.3.

Answer.

The problem is with the limits of integration, as in Question 1. Here’s how it ought to go:

Problem: Evaluate $\int_{1}^{π} \frac{\cos (\log t)}{t} d t .$

Work: We use the substitution $u = \log t,$ so $d u = \frac{1}{t} d t .$ When $t = 1,$ we have $u = \log 1 = 0$ and when $t = π,$ we have $u = \log (π) .$ Then:

$\begin{aligned} \int_{1}^{π} \frac{\cos (\log t)}{t} d t & = \int_{\log 1}^{\log (π)} \cos (u) d u \\ = \int_{0}^{\log (π)} \cos (u) d u \\ = \sin (\log (π)) - \sin (0) = \sin (\log (π)) . \end{aligned}$

1.4.2.4.

Answer.

This one is OK.

1.4.2.5. (✳).

Answer.

\int_{0}^{1} \frac{f (u)}{\sqrt{1 - u^{2}}} d u .

Because the denominator

\sqrt{1 - u^{2}}

vanishes when

u = 1,

this is what is known as an improper integral. Improper integrals will be discussed in Section 1.12.

1.4.2.6.

Answer.

some constant

C

1.4.2.7. (✳).

Answer.

\frac{1}{2} (\sin (e) - \sin (1))

1.4.2.8. (✳).

Answer.

\frac{1}{3}

1.4.2.9. (✳).

Answer.

- \frac{1}{300 {(x^{3} + 1)}^{100}} + C

1.4.2.10. (✳).

Answer.

\log 4

1.4.2.11. (✳).

Answer.

\log 2

1.4.2.12. (✳).

Answer.

\frac{4}{3}

1.4.2.13. (✳).

Answer.

e^{6} - 1

1.4.2.14. (✳).

Answer.

\frac{1}{3} (4 - x^{2})^{3 / 2} + C

1.4.2.15.

Answer.

e^{\sqrt{\log x}} + C

1.4.2.16. (✳).

Answer.

0

1.4.2.17. (✳).

Answer.

\frac{1}{2} [\cos 1 - \cos 2] \approx 0.478

1.4.2.18.

Answer.

\frac{1}{2} - \frac{1}{2} \log 2

1.4.2.19.

Answer.

\frac{1}{2} \tan^{2} θ - \log | \sec θ | + C

1.4.2.20.

Answer.

\arctan (e^{x}) + C

1.4.2.21.

Answer.

\frac{π}{4} - \frac{2}{3}

1.4.2.22.

Answer.

- \frac{1}{2} {(\log (\cos x))}^{2} + C

1.4.2.23. (✳).

Answer.

\frac{1}{2} \sin (1)

1.4.2.24. (✳).

Answer.

\frac{1}{3} [2 \sqrt{2} - 1] \approx 0.609

1.4.2.25.

Answer.

Using the definition of a definite integral with right Riemann sums:

\begin{aligned} \int_{a}^{b} 2 f (2 x) d x & = lim_{n \to \infty} \sum_{i = 1}^{n} Δ x \cdot 2 f (2 (a + i Δ x)) Δ x = \frac{b - a}{n} \\ = lim_{n \to \infty} \sum_{i = 1}^{n} (\frac{b - a}{n}) \cdot 2 f (2 (a + i (\frac{b - a}{n}))) \\ = lim_{n \to \infty} \sum_{i = 1}^{n} (\frac{2 b - 2 a}{n}) \cdot f (2 a + i (\frac{2 b - 2 a}{n})) \\ \int_{2 a}^{2 b} f (x) d x & = lim_{n \to \infty} \sum_{i = 1}^{n} Δ x \cdot f (2 a + i Δ x) Δ x = \frac{2 b - 2 a}{n} \\ = lim_{n \to \infty} \sum_{i = 1}^{n} (\frac{2 b - 2 a}{n}) \cdot f (2 a + i (\frac{2 b - 2 a}{n})) \end{aligned}

Since the Riemann sums are exactly the same,

\begin{aligned} \int_{a}^{b} 2 f (2 x) d x & = \int_{2 a}^{2 b} f (x) d x \end{aligned}

1.5 Area between curves
1.5.2 Exercises

1.5.2.1.

Answer.

Area between curves

\approx \frac{π}{4} (2 + \sqrt{2})

1.5.2.2.

Answer.

(a) Vertical rectangles:

(b) One possible answer:

1.5.2.3. (✳).

Answer.

\int_{0}^{\sqrt{2}} [2 x - x^{3}] d x

1.5.2.4. (✳).

Answer.

\int_{- 3 / 2}^{4} [\frac{4}{5} (6 - y^{2}) + 2 y] d y

1.5.2.5. (✳).

Answer.

\int_{0}^{4 a} [\sqrt{4 a x} - \frac{x^{2}}{4 a}] d x

1.5.2.6. (✳).

Answer.

\int_{1}^{25} [- \frac{1}{12} (x + 5) + \frac{1}{2} \sqrt{x}] d x

1.5.2.7. (✳).

Answer.

\frac{1}{8}

1.5.2.8. (✳).

Answer.

\frac{4}{3}

1.5.2.9. (✳).

Answer.

\frac{5}{3} - \frac{1}{\log 2}

1.5.2.10. (✳).

Answer.

\frac{8}{π} - 1

1.5.2.11. (✳).

Answer.

\frac{20}{9}

1.5.2.12. (✳).

Answer.

\frac{1}{6}

1.5.2.13.

Answer.

2 π

1.5.2.14. (✳).

Answer.

2 [π - \frac{1}{4} π^{2}]

1.5.2.15. (✳).

Answer.

\frac{31}{6}

1.5.2.16. (✳).

Answer.

\frac{26}{3}

1.5.2.17.

Answer.

\frac{7 π}{8} - \frac{1}{2}

1.5.2.18.

Answer.

12 \sqrt{2} - \frac{13}{4}

1.6 Volumes
1.6.2 Exercises

1.6.2.1.

Answer.

The horizontal cross-sections are circles, but the vertical cross-sections are not.

1.6.2.2.

Answer.

The columns have the same volume.

1.6.2.3.

Answer.

Washers when $1 < y \leq 6 :$ If $y > 1,$ then our washer has inner radius $2 + \frac{2}{3} y,$ outer radius $6 - \frac{2}{3} y,$ and height $d y .$
Washers when $0 \leq y < 1 :$ When $0 \leq y < 1,$ we have a “double washer,” two concentric rings. The inner washer has inner radius $r_{1} = y$ and outer radius $R_{1} = 2 - y .$ The outer washer has inner radius $r_{2} = 2 + \frac{2}{3} y$ and outer radius $R_{2} = 6 - \frac{2}{3} y .$ The thickness of the washers is $d y .$

1.6.2.4. (✳).

Answer.

(a)

π \int_{0}^{3} x e^{2 x^{2}} d x

(b)

\int_{0}^{1} π [(3 + \sqrt{y})^{2} - (3 - \sqrt{y})^{2}] d y + \int_{1}^{4} π [(5 - y)^{2} - (3 - \sqrt{y})^{2}] d y

1.6.2.5. (✳).

Answer.

(a)

\int_{- 1}^{1} π [{(5 - 4 x^{2})}^{2} - {(2 - x^{2})}^{2}] d x

(b)

\int_{- 1}^{0} π [(5 + \sqrt{y + 1})^{2} - (5 - \sqrt{y + 1})^{2}] d y

1.6.2.6. (✳).

Answer.

π \int_{- 2}^{2} [{(9 - x^{2})}^{2} - {(x^{2} + 1)}^{2}] d x

1.6.2.7.

Answer.

\frac{\sqrt{2}}{12} ℓ^{3}

1.6.2.8. (✳).

Answer.

\frac{π}{4} (e^{2 a^{2}} - 1)

1.6.2.9. (✳).

Answer.

π [\frac{38}{3} - \frac{514}{3^{4}}] = π \frac{512}{81}

1.6.2.10. (✳).

Answer.

(a)

8 π \int_{- 1}^{1} \sqrt{1 - x^{2}} d x

(b)

4 π^{2}

1.6.2.11. (✳).

Answer.

(a) The region

R

is the region between the blue and red curves, with

3 \leq x \leq 5,

in the figures below.

(b)

\frac{4}{3} π \approx 4.19

1.6.2.12. (✳).

Answer.

(a) The region

R

is sketched below.

(b)

π [4 \log 2 - \frac{3}{2}] \approx 3.998

1.6.2.13. (✳).

Answer.

π^{2} + 8 π^{3} + \frac{8 π^{6}}{5}

1.6.2.14. (✳).

Answer.

\frac{8}{3}

1.6.2.15. (✳).

Answer.

\frac{256 \times 8}{15} = 136.5 \dot{3}

1.6.2.16. (✳).

Answer.

\frac{28}{3} π h

1.6.2.17.

Answer.

(a) $\frac{4 π}{3 b^{2} a}$ cubic units,
(b) $a = \frac{1}{6356.752}$ and $b = \frac{1}{6378.137},$
(c) Approximately $1.08321 \times 10^{12} {km}^{3},$ or $1.08321 \times 10^{21} m^{3},$
(d) Absolute error is about $3.64 \times 10^{9} {km}^{3},$ and relative error is about $0.00336,$ or $0.336 % .$

1.6.2.18. (✳).

Answer.

(a)

\frac{9}{2}

(b)

π \int_{- 1}^{2} [{(4 - x)}^{2} - {(1 + (x - 1)^{2})}^{2}] d x

1.6.2.19. (✳).

Answer.

(a)

\frac{π}{2} - 1

(b)

\frac{π^{2}}{2} - π \approx 1.793

1.6.2.20. (✳).

Answer.

(a)

V_{1} = \frac{4}{3} π c^{2}

(b)

V_{2} = \frac{π c}{3} [4 \sqrt{2} - 2]

(c)

c = 0 or c = \sqrt{2} - \frac{1}{2}

1.6.2.21. (✳).

Answer.

\begin{aligned} \int_{π / 2}^{π} π [(5 + π \sin x)^{2} - (5 + 2 π - 2 x)^{2}] d x \\ + \int_{π}^{3 π / 2} π [(5 + 2 π - 2 x)^{2} - (5 + π \sin x)^{2}] d x \end{aligned}

1.6.2.22.

Answer.

(a)

\frac{6000 c π}{\log 2} (1 - \frac{1}{2^{10}}),

which is close to

\frac{6000 c π}{\log 2} .

(b) 6km: that is, there is roughly the same mass of air in the lowest 6 km of the column as there is in the remaining 54 km.

1.7 Integration by parts
1.7.2 Exercises

1.7.2.1.

Answer.

chain; product

1.7.2.2.

Answer.

The part chosen as

u

will be differentiated. The part chosen as

d v

will be antidifferentiated.

1.7.2.3.

Answer.

\int \frac{f^{'} (x)}{g (x)} d x = \frac{f (x)}{g (x)} + \int \frac{f (x) g^{'} (x)}{g^{2} (x)} d x

1.7.2.4.

Answer.

All the antiderivatives differ only by a constant, so we can write them all as

v (x) + C

for some

C .

Then, using the formula for integration by parts,

\begin{aligned} \int u (x) \cdot v^{'} (x) d x & = \underset{u}{\underset{⏟}{u (x)}} \underset{v}{\underset{⏟}{[v (x) + C]}} - \int \underset{v}{\underset{⏟}{[v (x) + C]}} \underset{d u}{\underset{⏟}{u^{'} (x) d x}} \\ = u (x) v (x) + C u (x) - \int v (x) u^{'} (x) d x - \int C u^{'} (x) d x \\ = u (x) v (x) + C u (x) - \int v (x) u^{'} (x) d x - C u (x) + D \\ = u (x) v (x) - \int v (x) u^{'} (x) d x + D \end{aligned}

where

D

is any constant.

Since the terms with

C

cancel out, it didn’t matter what we chose for

C

--all choices end up the same.

1.7.2.5.

Answer.

Suppose we choose

d v = f (x) d x,

u = 1 .

Then

v = \int f (x) d x,

and

d u = d x .

So, our integral becomes:

\begin{aligned} \int \underset{u}{\underset{⏟}{(1)}} \underset{d v}{\underset{⏟}{f (x) d x}} & = \underset{u}{\underset{⏟}{(1)}} \underset{v}{\underset{⏟}{\int f (x) d x}} - \int \underset{v}{\underset{⏟}{(\int f (x) d x)}} \underset{d u}{\underset{⏟}{d x}} \end{aligned}

In order to figure out the first product (and the second integrand), you need to know the antiderivative of

f (x)

--but that’s exactly what you’re trying to figure out!

1.7.2.6. (✳).

Answer.

\frac{x^{2} \log x}{2} - \frac{x^{2}}{4} + C

1.7.2.7. (✳).

Answer.

- \frac{\log x}{6 x^{6}} - \frac{1}{36 x^{6}} + C

1.7.2.8. (✳).

Answer.

π

1.7.2.9. (✳).

Answer.

\frac{π}{2} - 1

1.7.2.10.

Answer.

e^{x} (x^{3} - 3 x^{2} + 6 x - 6) + C

1.7.2.11.

Answer.

\frac{x^{2}}{2} \log^{3} x - \frac{3 x^{2}}{4} \log^{2} x + \frac{3 x^{2}}{4} \log x - \frac{3 x^{2}}{8} + C

1.7.2.12.

Answer.

(2 - x^{2}) \cos x + 2 x \sin x + C

1.7.2.13.

Answer.

(t^{3} - \frac{5}{2} t^{2} + 6 t) \log t - \frac{1}{3} t^{3} + \frac{5}{4} t^{2} - 6 t + C

1.7.2.14.

Answer.

e^{\sqrt{s}} (2 s - 4 \sqrt{s} + 4) + C

1.7.2.15.

Answer.

x \log^{2} x - 2 x \log x + 2 x + C

1.7.2.16.

Answer.

e^{x^{2} + 1} + C

1.7.2.17. (✳).

Answer.

y \arccos y - \sqrt{1 - y^{2}} + C

1.7.2.18. (✳).

Answer.

2 y^{2} \arctan (2 y) - y + \frac{1}{2} \arctan (2 y) + C

1.7.2.19.

Answer.

\frac{x^{3}}{3} \arctan x - \frac{1}{6} (1 + x^{2}) + \frac{1}{6} \log (1 + x^{2}) + C

1.7.2.20.

Answer.

\frac{2}{17} e^{x / 2} \cos (2 x) + \frac{8}{17} e^{x / 2} \sin (2 x) + C

1.7.2.21.

Answer.

\frac{x}{2} [\sin (\log x) - \cos (\log x)] + C

1.7.2.22.

Answer.

\frac{2^{x}}{\log 2} (x - \frac{1}{\log 2}) + C

1.7.2.23.

Answer.

2 e^{\cos x} [1 - \cos x] + C

1.7.2.24.

Answer.

\frac{x e^{- x}}{1 - x} + e^{- x} + C = \frac{e^{- x}}{1 - x} + C

1.7.2.25. (✳).

Answer.

(a) We integrate by parts with

u = \sin^{n - 1} x

and

d v = \sin x d x,

so that

d u = (n - 1) \sin^{n - 2} x \cos x

and

v = - \cos x .

\begin{aligned} \int \sin^{n} x d x & = \underset{u v}{\underset{⏟}{- \sin^{n - 1} x \cos x}} + \underset{- \int v d u}{\underset{⏟}{(n - 1) \int \cos^{2} x \sin^{n - 2} x d x}} \end{aligned}

Using the identity $\sin^{2} x + \cos^{2} x = 1,$

\begin{aligned} = - \sin^{n - 1} x \cos x + (n - 1) \int (1 - \sin^{2} x) \sin^{n - 2} x d x \\ = - \sin^{n - 1} x \cos x + (n - 1) \int \sin^{n - 2} x d x - (n - 1) \int \sin^{n} x d x \end{aligned}

Moving the last term on the right hand side to the left hand side gives

\begin{aligned} n \int \sin^{n} x d x & = - \sin^{n - 1} x \cos x + (n - 1) \int \sin^{n - 2} x d x \end{aligned}

Dividing across by

n

gives the desired reduction formula.

(b)

\frac{35}{256} π \approx 0.4295

1.7.2.26. (✳).

Answer.

(a) Area:

\frac{π}{4} - \frac{\log 2}{2}

(b) Volume:

\frac{π^{2}}{2} - π

1.7.2.27. (✳).

Answer.

π (\frac{17 e^{18} - 4373}{36})

1.7.2.28. (✳).

Answer.

12

1.7.2.29.

Answer.

\frac{2}{e}

1.8 Trigonometric Integrals
1.8.4 Exercises

1.8.4.1.

Answer.

(e)

1.8.4.2.

Answer.

\frac{1}{n} \sec^{n} x + C

1.8.4.3.

Answer.

We divide both sides by

\cos^{2} x,

and simplify.

\begin{aligned} \sin^{2} x + \cos^{2} x & = 1 \\ \frac{\sin^{2} x + \cos^{2} x}{\cos^{2} x} & = \frac{1}{\cos^{2} x} \\ \frac{\sin^{2} x}{\cos^{2} x} + 1 & = \sec^{2} x \\ \tan^{2} x + 1 & = \sec^{2} x \end{aligned}

1.8.4.4. (✳).

Answer.

\sin x - \frac{\sin^{3} x}{3} + C

1.8.4.5. (✳).

Answer.

\frac{π}{2}

1.8.4.6. (✳).

Answer.

\frac{\sin^{37} t}{37} - \frac{\sin^{39} t}{39} + C

1.8.4.7.

Answer.

\frac{1}{3 \cos^{3} x} - \frac{1}{\cos x} + C

1.8.4.8.

Answer.

\frac{π}{8} - \frac{9 \sqrt{3}}{64}

1.8.4.9.

Answer.

- \cos x + \frac{2}{3} \cos^{3} x - \frac{1}{5} \cos^{5} x + C

1.8.4.10.

Answer.

\frac{1}{2.2} \sin^{2.2} x + C

1.8.4.11.

Answer.

\frac{1}{2} \tan^{2} x + C,

or equivalently,

\frac{1}{2} \sec^{2} + C

1.8.4.12. (✳).

Answer.

\frac{1}{7} \sec^{7} x - \frac{1}{5} \sec^{5} x + C

1.8.4.13. (✳).

Answer.

\frac{\tan^{49} x}{49} + \frac{\tan^{47} x}{47} + C

1.8.4.14.

Answer.

\frac{1}{3.5} \sec^{3.5} x - \frac{1}{1.5} \sec^{1.5} x + C

1.8.4.15.

Answer.

\frac{1}{4} \sec^{4} x - \frac{1}{2} \sec^{2} x + C

\frac{1}{4} \tan^{4} x + C

1.8.4.16.

Answer.

\frac{1}{5} \tan^{5} x + C

1.8.4.17.

Answer.

\frac{1}{1.3} \sec^{1.3} x + \frac{1}{0.7} \cos^{0.7} x + C

1.8.4.18.

Answer.

= \frac{1}{4} \sec^{4} x - \sec^{2} x + \log | \sec x | + C

1.8.4.19.

Answer.

\frac{41}{45 \sqrt{3}} - \frac{π}{6}

1.8.4.20.

Answer.

\frac{1}{11} + \frac{1}{9}

1.8.4.21.

Answer.

2 \sqrt{\sec x} + C

1.8.4.22.

Answer.

\tan^{e + 1} θ (\frac{\tan^{6} θ}{7 + e} + \frac{3 \tan^{4} θ}{5 + e} + \frac{3 \tan^{2} θ}{3 + e} + \frac{1}{1 + e}) + C

1.8.4.23. (✳).

Answer.

(a) Using the trig identity

\tan^{2} x = \sec^{2} x - 1

and the substitution

y = \tan x,

d y = \sec^{2} x d x,

\begin{aligned} \int \tan^{n} x d x & = \int \tan^{n - 2} x \tan^{2} x d x \\ = \int \tan^{n - 2} x \sec^{2} x d x - \int \tan^{n - 2} x d x \\ = \int y^{n - 2} d y - \int \tan^{n - 2} x d x \\ = \frac{y^{n - 1}}{n - 1} - \int \tan^{n - 2} x d x \\ = \frac{\tan^{n - 1} x}{n - 1} - \int \tan^{n - 2} x d x \end{aligned}

(b)

\frac{13}{15} - \frac{π}{4} \approx 0.0813

1.8.4.24.

Answer.

\frac{1}{2 \cos^{2} x} + 2 \log | \cos x | - \frac{1}{2} \cos^{2} x + C

1.8.4.25.

Answer.

\tan θ + C

1.8.4.26.

Answer.

\log | \sin x | + C

1.8.4.27.

Answer.

\frac{1}{2} \sin^{2} (e^{x}) + C

1.8.4.28.

Answer.

(\sin^{2} x + 2) \cos (\cos x) + 2 \cos x \sin (\cos x) + C

1.8.4.29.

Answer.

\frac{x}{2} \sin^{2} x - \frac{x}{4} + \frac{1}{4} \sin x \cos x + C

1.9 Trigonometric Substitution
1.9.2 Exercises

1.9.2.1. (✳).

Answer.

(a)

x = \frac{4}{3} \sec θ

(b)

x = \frac{1}{2} \sin θ

(c)

x = 5 \tan θ

1.9.2.2.

Answer.

(a)

x - 2 = \sqrt{3} \sec u

(b)

x - 1 = \sqrt{5} \sin u

(c)

(2 x + \frac{3}{2}) = \frac{\sqrt{31}}{2} \tan u

(d)

x - \frac{1}{2} = \frac{1}{2} \sec u

1.9.2.3.

Answer.

(a)

\frac{\sqrt{399}}{20}

(b)

\frac{5 \sqrt{2}}{7}

(c)

\frac{\sqrt{x - 5}}{2}

1.9.2.4.

Answer.

(a)

\frac{\sqrt{4 - x^{2}}}{2}

(b)

\frac{1}{2}

(c)

\frac{1}{\sqrt{1 - x}}

1.9.2.5. (✳).

Answer.

\frac{1}{4} \cdot \frac{x}{\sqrt{x^{2} + 4}} + C

1.9.2.6. (✳).

Answer.

\frac{1}{2 \sqrt{5}}

1.9.2.7. (✳).

Answer.

\frac{π}{6}

1.9.2.8. (✳).

Answer.

\log | \sqrt{1 + \frac{x^{2}}{25}} + \frac{x}{5} | + C

1.9.2.9.

Answer.

\frac{1}{2} \sqrt{2 x^{2} + 4 x} + C

1.9.2.10. (✳).

Answer.

- \frac{1}{16} \frac{\sqrt{x^{2} + 16}}{x} + C

1.9.2.11. (✳).

Answer.

\frac{\sqrt{x^{2} - 9}}{9 x} + C

1.9.2.12. (✳).

Answer.

(a) We’ll use the trig identity

\cos 2 θ = 2 \cos^{2} θ - 1 .

It implies that

\begin{aligned} \cos^{2} θ = \frac{\cos 2 θ + 1}{2} ⟹ \cos^{4} θ & = \frac{1}{4} [\cos^{2} 2 θ + 2 \cos 2 θ + 1] \\ = \frac{1}{4} [\frac{\cos 4 θ + 1}{2} + 2 \cos 2 θ + 1] \\ = \frac{\cos 4 θ}{8} + \frac{\cos 2 θ}{2} + \frac{3}{8} \end{aligned}

So,

\begin{aligned} \int_{0}^{π / 4} \cos^{4} θ d θ & = \int_{0}^{π / 4} (\frac{\cos 4 θ}{8} + \frac{\cos 2 θ}{2} + \frac{3}{8}) d θ \\ = {[\frac{\sin 4 θ}{32} + \frac{\sin 2 θ}{4} + \frac{3}{8} θ]}_{0}^{π / 4} \\ = \frac{1}{4} + \frac{3}{8} \cdot \frac{π}{4} \\ = \frac{8 + 3 π}{32} \end{aligned}

as required.

(b)

\frac{8 + 3 π}{16}

1.9.2.13.

Answer.

1.9.2.14. (✳).

Answer.

2 \arcsin \frac{x}{2} + \frac{x}{2} \sqrt{4 - x^{2}} + C

1.9.2.15. (✳).

Answer.

\sqrt{25 x^{2} - 4} - 2 arcsec \frac{5 x}{2} + C

1.9.2.16.

Answer.

\frac{40}{3}

1.9.2.17. (✳).

Answer.

\arcsin \frac{x + 1}{2} + C

1.9.2.18.

Answer.

\frac{1}{4} (\arccos (\frac{1}{2 x - 3}) + \frac{\sqrt{4 x^{2} - 12 x + 8}}{(2 x - 3)^{2}}) + C,

or equivalently,

\frac{1}{4} (arcsec (2 x - 3) + \frac{\sqrt{4 x^{2} - 12 x + 8}}{(2 x - 3)^{2}}) + C

1.9.2.19.

Answer.

\log (1 + \sqrt{2}) - \frac{1}{\sqrt{2}}

1.9.2.20.

Answer.

\frac{1}{2} (\arctan x + \frac{x}{x^{2} + 1}) + C

1.9.2.21.

Answer.

\frac{3 + x}{2} \sqrt{x^{2} - 2 x + 2} + \frac{1}{2} \log | \sqrt{x^{2} - 2 x + 2} + x - 1 | + C

1.9.2.22.

Answer.

\frac{1}{\sqrt{3}} \log | (\frac{6}{5} x + 1) + \frac{2}{5} \sqrt{9 x^{2} + 15 x} | + C

1.9.2.23.

Answer.

\frac{1}{3} \sqrt{1 + x^{2}} (4 + x^{2}) + \log | \frac{1 - \sqrt{1 + x^{2}}}{x} | + C

1.9.2.24.

Answer.

\frac{8 π}{3} + 4 \sqrt{3}

1.9.2.25.

Answer.

Area:

\frac{4}{3} - \sqrt[4]{\frac{4}{3}}

Volume:

\frac{π^{2}}{6} - \frac{\sqrt{3} π}{4}

1.9.2.26.

Answer.

2 \sqrt{1 + e^{x}} + 2 \log | 1 - \sqrt{1 + e^{x}} | - x + C

1.9.2.27.

Answer.

$\frac{1}{1 - x^{2}}$
False
The work in the question is not correct. The most salient problem is that when we make the substitution $x = \sin θ,$ we restrict the possible values of $x$ to $[- 1, 1],$ since this is the range of the sine function. However, the original integral had no such restriction.

How can we be sure we avoid this problem in the future? In the introductory text to Section 1.9 (before Example 1.9.1), the notes tell us that we are allowed to write our old variable as a function of a new variable (say $x = s (u)$ ) as long as that function is invertible to recover our original variable $x .$ There is one very obvious reason why invertibility is necessary: after we antidifferentiate using our new variable $u,$ we need to get it back in terms of our original variable, so we need to be able to recover $x .$ Moreover, invertibility reconciles potential problems with domains: if an inverse function $u = s^{- 1} (x)$ exists, then for any $x,$ there exists a $u$ with $s (u) = x .$ (This was not the case in the work for the question, because we chose $x = \sin θ,$ but if $x = 2,$ there is no corresponding $θ .$ Note, however, that $x = \sin θ$ is invertible over $[- 1, 1],$ so the work is correct if we restrict $x$ to those values.)

1.9.2.28.

Answer.

(a), (b): None.

(c):

x < - a

1.10 Partial Fractions
1.10.4 Exercises

1.10.4.1.

Answer.

(a) (iii)

(b) (ii)

(d) (i)

1.10.4.2. (✳).

Answer.

\frac{A}{x - 1} + \frac{B}{(x - 1)^{2}} + \frac{C}{x + 1} + \frac{D}{(x + 1)^{2}} + \frac{E x + F}{x^{2} + 1}

1.10.4.3. (✳).

Answer.

3

1.10.4.4.

Answer.

(a)

\frac{x^{3} + 2 x + 2}{x^{2} + 1} = x + \frac{x + 2}{x^{2} + 1}

(b)

\frac{15 x^{4} + 6 x^{3} + 34 x^{2} + 4 x + 20}{5 x^{2} + 2 x + 8} = 3 x^{2} + 2 + \frac{4}{5 x^{2} + 2 x + 8}

(c)

\frac{2 x^{5} + 9 x^{3} + 12 x^{2} + 10 x + 30}{2 x^{2} + 5} = x^{3} + 2 x + 6

1.10.4.5.

Answer.

(a)

5 x^{3} - 3 x^{2} - 10 x + 6 = (x + \sqrt{2}) (x - \sqrt{2}) (5 x - 3)

(b)

\begin{aligned} x^{4} - 3 x^{2} - 5 \\ = (x + \sqrt{\frac{3 + \sqrt{29}}{2}}) (x - \sqrt{\frac{3 + \sqrt{29}}{2}}) (x^{2} + \frac{\sqrt{29} - 3}{2}) \end{aligned}

(c)

x^{4} - 4 x^{3} - 10 x^{2} - 11 x - 6 = (x + 1) (x - 6) (x^{2} + x + 1)

(d)

\begin{aligned} 2 x^{4} + 12 x^{3} - x^{2} - 52 x + 15 \\ = (x + 3) (x + 5) (x - (1 + \frac{\sqrt{2}}{2})) (x - (1 - \frac{\sqrt{2}}{2})) \end{aligned}

1.10.4.6.

Answer.

The goal of partial fraction decomposition is to write our integrand in a form that is easy to integrate. The antiderivative of (1) can be easily determined with the substitution

u = (a x + b) .

It’s less clear how to find the antiderivative of (2).

1.10.4.7. (✳).

Answer.

\log \frac{4}{3}

1.10.4.8. (✳).

Answer.

- \frac{1}{x} - \arctan x + C

1.10.4.9. (✳).

Answer.

4 \log | x - 3 | - 2 \log (x^{2} + 1) + C

1.10.4.10. (✳).

Answer.

F (x) = \log | x - 2 | + \log (x^{2} + 4) + 2 \arctan (x / 2) + D

1.10.4.11. (✳).

Answer.

- 2 \log | x - 3 | + 3 \log | x + 2 | + C

1.10.4.12. (✳).

Answer.

- 9 \log | x + 2 | + 14 \log | x + 3 | + C

1.10.4.13.

Answer.

5 x + \frac{1}{2} \log | x - 1 | - \frac{7}{2} \log | x + 1 | + C

1.10.4.14.

Answer.

x - \frac{2}{x} + \frac{5}{2} \arctan (2 x) + C

1.10.4.15.

Answer.

\frac{1}{x} - \frac{2}{x - 1} + C

1.10.4.16.

Answer.

- \frac{1}{2} \log | x - 2 | + \frac{1}{2} \log | x + 2 | + \frac{3}{2} \log | 2 x - 1 | + C

1.10.4.17.

Answer.

\log (\frac{4 \cdot 6^{3}}{5^{3}})

1.10.4.18.

Answer.

\frac{1}{2} \log | \frac{1 - \cos x}{1 + \cos x} | + C

1.10.4.19.

Answer.

\frac{- \cos x}{2 \sin^{2} x} + \frac{1}{4} \log | \frac{1 - \cos x}{1 + \cos x} | + C

1.10.4.20.

Answer.

3 \log 2 + \frac{1}{2} + \frac{2}{\sqrt{15}} (\arctan (\frac{7}{\sqrt{15}}) - \arctan (\frac{9}{\sqrt{15}}))

1.10.4.21.

Answer.

= \frac{9}{4 \sqrt{2}} \arctan (\frac{x}{\sqrt{2}}) - \frac{2 + 3 x}{4 (x^{2} + 2)} + C

1.10.4.22.

Answer.

\frac{3}{8} \arctan x + \frac{3 x^{3} + 5 x}{8 (1 + x^{2})^{2}} + C

1.10.4.23.

Answer.

\frac{3}{2} x^{2} + \frac{1}{\sqrt{5}} \arctan (\frac{x}{\sqrt{5}}) + \frac{3}{2} \log | x^{2} + 5 | - \frac{3}{2 x^{2} + 10} + C

1.10.4.24.

Answer.

\log | \frac{\sin θ - 1}{\sin θ - 2} | + C

1.10.4.25.

Answer.

t - \frac{1}{2} \log | e^{2 t} + e^{t} + 1 | - \frac{1}{\sqrt{3}} \arctan (\frac{2 e^{t} + 1}{\sqrt{3}}) + C

1.10.4.26.

Answer.

2 \sqrt{1 + e^{x}} + \log | \frac{\sqrt{1 + e^{x}} - 1}{\sqrt{1 + e^{x}} + 1} | + C

1.10.4.27. (✳).

Answer.

(a) The region

R

(b)

10 π \log \frac{9}{4} = 20 π \log \frac{3}{2}

(c)

20 π

1.10.4.28.

Answer.

2 \log \frac{5}{3} + \frac{4}{\sqrt{3}} \arctan \frac{1}{4 \sqrt{3}}

1.10.4.29.

Answer.

(a)

\frac{1}{6} (\log | 2 \cdot \frac{x - 3}{x + 3} |)

(b)

F^{'} (x) = \frac{1}{x^{2} - 9}

1.11 Numerical Integration
1.11.6 Exercises

1.11.6.1.

Answer.

Relative error:

\approx 0.08147;

absolute error:

0.113;

percent error:

\approx 8.147 % .

1.11.6.2.

Answer.

Midpoint rule:

Trapezoidal rule:

1.11.6.3.

Answer.

M = 6.25,

L = 2

1.11.6.4.

Answer.

One reasonable answer is

M = 3 .

1.11.6.5.

Answer.

(a)

\frac{π^{5}}{180 \cdot 8}

(b)

0

(c)

0

1.11.6.6.

Answer.

Possible answers:

f (x) = \frac{3}{2} x^{2} + C x + D

for any constants

C,

D .

1.11.6.7.

Answer.

my mother

1.11.6.8.

Answer.

(a) true

(b) false

1.11.6.9. (✳).

Answer.

True. Because

f (x)

is positive and concave up, the graph of

f (x)

is always below the top edges of the trapezoids used in the trapezoidal rule.

1.11.6.10.

Answer.

Any polynomial of degree at most 3 will do. For example,

f (x) = 5 x^{3} - 27,

f (x) = x^{2} .

1.11.6.11.

Answer.

Midpoint:

\begin{aligned} \int_{0}^{30} \frac{1}{x^{3} + 1} d x & \approx [\frac{1}{{(2.5)}^{3} + 1} + \frac{1}{{(7.5)}^{3} + 1} + \frac{1}{{(12.5)}^{3} + 1} + \frac{1}{{(17.5)}^{3} + 1} \\ + \frac{1}{{(22.5)}^{3} + 1} + \frac{1}{{(27.5)}^{3} + 1}] 5 \end{aligned}

Trapezoidal:

\begin{aligned} \int_{0}^{30} \frac{1}{x^{3} + 1} d x & \approx [\frac{1 / 2}{0^{3} + 1} + \frac{1}{5^{3} + 1} + \frac{1}{10^{3} + 1} + \frac{1}{15^{3} + 1} + \frac{1}{20^{3} + 1} \\ + \frac{1}{25^{3} + 1} + \frac{1 / 2}{30^{3} + 1}] 5 \end{aligned}

Simpson’s:

\begin{aligned} \int_{0}^{30} \frac{1}{x^{3} + 1} d x & \approx [\frac{1}{0^{3} + 1} + \frac{4}{5^{3} + 1} + \frac{2}{10^{3} + 1} + \frac{4}{15^{3} + 1} + \frac{2}{20^{3} + 1} \\ + \frac{4}{25^{3} + 1} + \frac{1}{30^{3} + 1}] \frac{5}{3} \end{aligned}

1.11.6.12. (✳).

Answer.

\frac{2 π}{3}

1.11.6.13. (✳).

Answer.

1720 π \approx 5403.5 {c m}^{3}

1.11.6.14. (✳).

Answer.

\frac{π}{12} (16.72) \approx 4.377 m^{3}

1.11.6.15. (✳).

Answer.

\frac{12.94}{6 π} \approx 0.6865 m^{3}

1.11.6.16. (✳).

Answer.

(a) 363,500

(b) 367,000

1.11.6.17. (✳).

Answer.

(a)

\frac{49}{2}

(b)

\frac{77}{3}

1.11.6.18. (✳).

Answer.

Let

f (x) = \sin (x^{2}) .

Then

f^{'} (x) = 2 x \cos (x^{2})

and

f^{″} (x) = 2 \cos (x^{2}) - 4 x^{2} \sin (x^{2}) .

Since

| x^{2} | \leq 1

when

| x | \leq 1,

and

| \sin θ | \leq 1

and

| \cos θ | \leq 1

for all

θ,

we have

\begin{aligned} | 2 \cos (x^{2}) - 4 x^{2} \sin (x^{2}) | & \leq 2 | \cos (x^{2}) | + 4 x^{2} | \sin (x^{2}) | \\ \leq 2 \times 1 + 4 \times 1 \times 1 = 2 + 4 = 6 \end{aligned}

We can therefore choose

M = 6,

and it follows that the error is at most

\begin{matrix} \frac{M [b - a]^{3}}{24 n^{2}} \leq \frac{6 \cdot [1 - (- 1)]^{3}}{24 \cdot 1000^{2}} = \frac{2}{10^{6}} = 2 \cdot 10^{- 6} \end{matrix}

1.11.6.19. (✳).

Answer.

\frac{3}{100}

1.11.6.20. (✳).

Answer.

(a)

\begin{aligned} \frac{1 / 3}{3} ((- 3)^{5} + 4 (\frac{1}{3} - 3)^{5} + 2 (\frac{2}{3} - 3)^{5} + 4 (- 2)^{5} + 2 (\frac{4}{3} - 3)^{5} \\ + 4 (\frac{5}{3} - 3)^{5} + (- 1)^{5}) \end{aligned}

(b) Simpson’s Rule results in a smaller error bound.

1.11.6.21. (✳).

Answer.

\frac{8}{15}

1.11.6.22. (✳).

Answer.

\frac{1}{180 \times 3^{4}} = \frac{1}{14580}

1.11.6.23. (✳).

Answer.

(a)

T_{4} = \frac{1}{4} [(\frac{1}{2} \times 1) + \frac{4}{5} + \frac{2}{3} + \frac{4}{7} + (\frac{1}{2} \times \frac{1}{2})],

(b)

S_{4} = \frac{1}{12} [1 + (4 \times \frac{4}{5}) + (2 \times \frac{2}{3}) + (4 \times \frac{4}{7}) + \frac{1}{2}]

(c)

| I - S_{4} | \leq \frac{24}{180 \times 4^{4}} = \frac{1}{1920}

1.11.6.24. (✳).

Answer.

(a)

T_{4} = 8.03515,

S_{4} \approx 8.03509

(b)

| \int_{a}^{b} f (x) d x - T_{n} | \leq \frac{2}{1000} \frac{8^{3}}{12 (4)^{2}} \leq 0.00533

| \int_{a}^{b} f (x) d x - S_{n} | \leq \frac{4}{1000} \frac{8^{5}}{180 (4)^{4}} \leq 0.00284

1.11.6.25. (✳).

Answer.

Any

n \geq 68

works.

1.11.6.26. (✳).

Answer.

\frac{472}{3} \approx 494 {f t}^{3}

1.11.6.27. (✳).

Answer.

(a)

0.025635

(b)

1.8 \times 10^{- 5}

1.11.6.28. (✳).

Answer.

(a)

\approx 0.6931698

(b)

n \geq 12

with

n

even

1.11.6.29. (✳).

Answer.

(a)

0.01345

(b)

n \geq 28

with

n

even

1.11.6.30. (✳).

Answer.

n \geq 259

1.11.6.31.

Answer.

(a) When

0 \leq x \leq 1,

then

x^{2} \leq 1

and

x + 1 \geq 1,

| f^{″} (x) | = \frac{x^{2}}{| x + 1 |} \leq \frac{1}{1} = 1

(b)

\frac{1}{2}

(c)

n \geq 65

(d)

n \geq 46

1.11.6.32.

Answer.

\frac{x - 1}{12} [1 + \frac{16}{x + 3} + \frac{4}{x + 1} + \frac{16}{3 x + 1} + \frac{1}{x}]

1.11.6.33.

Answer.

Note: for more detail, see the solutions.

First, we use Simpson’s rule with

n = 4

to approximate

\int_{1}^{2} \frac{1}{1 + x^{2}} d x .

The choice of this method (what we’re approximating, why

n = 4,

etc.) is explained in the solutions--here, we only show that it works.

\int_{1}^{2} \frac{1}{1 + x^{2}} d x \approx \frac{1}{12} [\frac{1}{2} + \frac{64}{41} + \frac{8}{13} + \frac{64}{65} + \frac{1}{5}] \approx 0.321748

For ease of notation, define

A = 0.321748 .

Now, we bound the error associated with this approximation. Define

N (x) = 24 (5 x^{4} - 10 x^{2} + 1)

and

D (x) = (x^{2} + 1)^{5},

N (x) / D (x)

gives the fourth derivative of

\frac{1}{1 + x^{2}} .

When

1 \leq x \leq 2,

| N (x) | \leq N (2) = 984

(because

N (x)

is increasing over that interval) and

| D (x) | \geq D (1) = 2^{5}

(because

D (x)

is also increasing over that interval), so

| \frac{d^{4}}{d x^{4}} {\frac{1}{1 + x^{2}}} | = | \frac{N (x)}{D (x)} | \leq \frac{984}{2^{5}} = 30.75 .

Now we find the error bound for Simpson’s rule with

L = 30.75,

b = 2,

a = 1,

and

n = 4 .

| \int_{1}^{2} \frac{1}{1 + x^{2}} d x - A | = | error | \leq \frac{L (b - a)^{5}}{180 \cdot n^{4}} = \frac{30.75}{180 \cdot 4^{4}} < 0.00067

So,

\begin{aligned} - 0.00067 & < \int_{1}^{2} \frac{1}{1 + x^{2}} d x - A < 0.00067 \\ A - 0.0067 & < \int_{1}^{2} \frac{1}{1 + x^{2}} d x < A + 0.00067 \\ A - 0.00067 & < \arctan (2) - \arctan (1) < A + 0.00067 \\ A - 0.00067 & < \arctan (2) - \frac{π}{4} < A + 0.00067 \\ \frac{π}{4} + A - 0.00067 & < \arctan (2) < \frac{π}{4} + A + 0.00067 \\ \frac{π}{4} + 0.321748 - 0.00067 & < \arctan (2) < \frac{π}{4} + 0.321748 + 0.00067 \\ \frac{π}{4} + 0.321078 & < \arctan (2) < \frac{π}{4} + 0.322418 \\ \frac{π}{4} + 0.321 & < \arctan (2) < \frac{π}{4} + 0.323 \end{aligned}

This was the desired bound.

1.12 Improper Integrals
1.12.4 Exercises

1.12.4.1.

Answer.

Any real number in

[1, \infty)

(- \infty, - 1],

and

b = \pm \infty .

1.12.4.2.

Answer.

b = \pm \infty

1.12.4.3.

Answer.

The red function is

f (x),

and the blue function is

g (x) .

1.12.4.4. (✳).

Answer.

False. For example, the functions

f (x) = e^{- x}

and

g (x) = 1

provide a counterexample.

1.12.4.5.

Answer.

Not enough information to decide. For example, consider $h (x) = 0$ versus $h (x) = - 1 .$
Not enough information to decide. For example, consider $h (x) = f (x)$ versus $h (x) = g (x) .$
$\int_{0}^{\infty} h (x) d x$ converges by the comparison test, since $| h (x) | \leq 2 f (x)$ and $\int_{0}^{\infty} 2 f (x) d x$ converges.

1.12.4.6. (✳).

Answer.

The integral diverges.

1.12.4.7. (✳).

Answer.

The integral diverges.

1.12.4.8. (✳).

Answer.

The integral does not converge.

1.12.4.9. (✳).

Answer.

The integral converges.

1.12.4.10.

Answer.

The integral diverges.

1.12.4.11.

Answer.

The integral diverges.

1.12.4.12.

Answer.

The integral diverges.

1.12.4.13.

Answer.

The integral diverges.

1.12.4.14. (✳).

Answer.

The integral diverges.

1.12.4.15. (✳).

Answer.

The integral converges.

1.12.4.16. (✳).

Answer.

The integral converges.

1.12.4.17.

Answer.

false

1.12.4.18. (✳).

Answer.

q = \frac{1}{5}

1.12.4.19.

Answer.

p > 1

1.12.4.20.

Answer.

\frac{\log 3 - π}{4} + \frac{1}{2} \arctan 2

1.12.4.21.

Answer.

The integral converges.

1.12.4.22.

Answer.

\frac{1}{2}

1.12.4.23. (✳).

Answer.

The integral converges.

1.12.4.24.

Answer.

The integral converges.

1.12.4.25. (✳).

Answer.

t = 10

and

n = 2042

will do the job. There are many other correct answers.

1.12.4.26.

Answer.

(a) The integral converges.

(b) The interval converges.

1.12.4.27.

Answer.

false

1.13 More Integration Examples

Exercises

1.13.1.

Answer.

(A)–(I), (B)–(IV), (C)–(II), (D)–(III)

1.13.2.

Answer.

\frac{1}{5} - \frac{2}{7} + \frac{1}{9} = \frac{8}{315}

1.13.3.

Answer.

\frac{3}{2 \sqrt{5}} \arcsin (x \sqrt{\frac{5}{3}}) + \frac{x}{2} \sqrt{3 - 5 x^{2}} + C

1.13.4.

Answer.

1.13.5.

Answer.

\log | \frac{x + 1}{3 x + 1} | + C

1.13.6.

Answer.

\frac{8}{3} \log 2 - \frac{7}{9}

1.13.7. (✳).

Answer.

\frac{1}{2} \log | x^{2} - 3 | + C

1.13.8. (✳).

Answer.

(a)

2

(b)

\frac{2}{15}

(c)

\frac{3 e^{4}}{16} + \frac{1}{16}

1.13.9. (✳).

Answer.

(a)

1

(b)

\frac{8}{15}

1.13.10. (✳).

Answer.

(a)

e^{2} + 1

(b)

\log (\sqrt{2} + 1)

(c)

\log \frac{15}{13} \approx 0.1431

1.13.11. (✳).

Answer.

(a)

\frac{9}{4} π

(b)

\log 2 - 2 + \frac{π}{2} \approx 0.264

(c)

2 \log 2 - \frac{1}{2} \approx 0.886

1.13.12.

Answer.

\frac{1}{3} \sin^{3} θ - 2 \sin θ + 12 \log | \frac{\sin θ - 3}{\sin θ - 2} | + C

1.13.13. (✳).

Answer.

(a)

\frac{1}{15}

(b)

\frac{1}{9} \cdot \frac{x}{\sqrt{x^{2} + 9}} + C

(c)

\frac{1}{2} \log | x - 1 | - \frac{1}{4} \log (x^{2} + 1) - \frac{1}{2} \arctan x + C

(d)

\frac{1}{2} [x^{2} \arctan x - x + \arctan x] + C

1.13.14. (✳).

Answer.

(a)

\frac{1}{12}

(b)

2 \sin^{- 1} \frac{x}{2} + x \sqrt{1 - \frac{x^{2}}{4}} + C

(c)

- 2 \log | x | + \frac{1}{x} + 2 \log | x - 1 | + C

1.13.15. (✳).

Answer.

(a)

\frac{2}{5}

(b)

\frac{1}{2 \sqrt{2}}

(c)

\log 2 - \frac{1}{2} \approx 0.193

(d)

\log 2 - \frac{1}{2} \approx 0.193

1.13.16. (✳).

Answer.

(a)

\frac{1}{2} x^{2} \log x - \frac{1}{4} x^{2} + C

(b)

\frac{1}{2} \log [x^{2} + 4 x + 5] - 3 \arctan (x + 2) + C

(c)

\frac{1}{2} \log | x - 3 | - \frac{1}{2} \log | x - 1 | + C

(d)

\frac{1}{3} \arctan x^{3} + C

1.13.17. (✳).

Answer.

(a)

\frac{π}{4} - \frac{1}{2} \log 2

(b)

\log | x^{2} - 2 x + 5 | + \frac{1}{2} \arctan \frac{x - 1}{2} + C

1.13.18. (✳).

Answer.

(a)

- \frac{1}{300 (x^{3} + 1)^{100}} + C

(b)

\frac{\sin^{5} x}{5} - \frac{\sin^{7} x}{7} + C

1.13.19.

Answer.

-2

1.13.20. (✳).

Answer.

(a)

- \frac{1}{4} \log | e^{x} + 1 | + \frac{1}{4} \log | e^{x} - 3 | + C

(b)

\frac{4 π}{3} - 2 \sqrt{3}

1.13.21. (✳).

Answer.

(a)

\frac{1}{2} \sec^{2} x + \log | \cos x | + C

(b)

\frac{1}{10} \arctan 8 \approx 0.1446

1.13.22.

Answer.

\frac{2}{5} (x - 1)^{5 / 2} + \frac{2}{3} (x - 1)^{3 / 2} + C

1.13.23.

Answer.

\log | x + \sqrt{x^{2} - 2} | - \frac{\sqrt{x^{2} - 2}}{x} + C

1.13.24.

Answer.

\frac{7}{24}

1.13.25.

Answer.

3 \log | x + 1 | + \frac{2}{x + 1} - \frac{5}{2 (x + 1)^{2}} + C

1.13.26.

Answer.

\frac{2}{\sqrt{3}} \arctan (\frac{2}{\sqrt{3}} x + \frac{1}{\sqrt{3}}) + C

1.13.27.

Answer.

\frac{1}{2} (x - \sin x \cos x) + C

1.13.28.

Answer.

\frac{1}{3} \log | x + 1 | - \frac{1}{6} \log | x^{2} + x + 1 | + \frac{1}{\sqrt{3}} \arctan (\frac{2 x - 1}{\sqrt{3}}) + C

1.13.29.

Answer.

3 x^{3} \arcsin x + 3 \sqrt{1 - x^{2}} - (1 - x^{2})^{3 / 2} + C

1.13.30.

Answer.

1.13.31.

Answer.

\frac{1}{4}

1.13.32.

Answer.

\log (\frac{\log (\cos (0.1))}{\log (\cos (0.2))})

1.13.33. (✳).

Answer.

(a)

\frac{1}{2} x [\sin (\log x) - \cos (\log x)] + C

(b)

2 \log 2 - \log 3 = \log \frac{4}{3}

1.13.34. (✳).

Answer.

(a)

\frac{9}{4} π + 9

(b)

2 \log | x - 2 | - \log (x^{2} + 4) + C

(c)

\frac{π}{2}

1.13.35.

Answer.

- \arcsin (\sqrt{1 - x}) - \sqrt{1 - x} \sqrt{x} + C

1.13.36.

Answer.

e^{e} (e - 1)

1.13.37.

Answer.

\frac{e^{x}}{x + 1} + C

1.13.38.

Answer.

x \sec x - \log | \sec x + \tan x | + C

1.13.39.

Answer.

\int x (x + a)^{n} d x = {\begin{cases} \frac{(x + a)^{(n + 2)}}{n + 2} - a \frac{(x + a)^{n + 1}}{n + 1} + C & if n \neq - 1, - 2 \\ (x + a) - a \log | x + a | + C & if n = - 1 \\ \log | x + a | + \frac{a}{x + a} + C & if n = - 2 \end{cases}

1.13.40.

Answer.

\begin{aligned} x \arctan (x^{2}) - \frac{1}{\sqrt{2}} (\frac{1}{2} \log | \frac{x^{2} - \sqrt{2} x + 1}{x^{2} + \sqrt{2} x + 1} | + \arctan (\sqrt{2} x + 1) \\ + \arctan (\sqrt{2} x - 1)) + C \end{aligned}

2 Applications of Integration
2.1 Work
2.1.2 Exercises

2.1.2.1.

Answer.

0.00294 J

2.1.2.2.

Answer.

The rock has mass

\frac{1}{9.8}

kg (about 102 grams); lifting it one metre takes 1 J of work.

2.1.2.3.

Answer.

(a) metres

(b) newtons

2.1.2.4.

Answer.

\frac{smoot \cdot barn}{megaFonzie}

(smoot-barns per megaFonzie)

2.1.2.5.

Answer.

10 cm below the bottom of the unloaded spring

2.1.2.6.

Answer.

x = 2

2.1.2.7. (✳).

Answer.

a = 3

2.1.2.8.

Answer.

(a) joules

(b)

c \log (\frac{ℓ - 1}{ℓ - 1.5}) J

2.1.2.9. (✳).

Answer.

\frac{1}{4} J

2.1.2.10. (✳).

Answer.

25

2.1.2.11. (✳).

Answer.

196 J

2.1.2.12.

Answer.

14700 J

2.1.2.13. (✳).

Answer.

\int_{0}^{3} (9.8) (8000) (2 + z) (3 - z)^{2} d z

joules

2.1.2.14.

Answer.

0.2352 J

2.1.2.15.

Answer.

\frac{20}{49}

kg, or about 408 grams

2.1.2.16.

Answer.

294 J

2.1.2.17.

Answer.

(a) 117.6 J

(b)

3.92 [30 - 2 \sqrt{3}] \approx 104 J

2.1.2.18.

Answer.

\frac{1}{2 \sqrt{5}}

m/sec, or about 22.36 cm/sec

2.1.2.19.

Answer.

yes (at least, the car won’t scrape the ground)

2.1.2.20.

Answer.

\approx 0.144 J

2.1.2.21. (✳).

Answer.

904,050 π J

2.1.2.22.

Answer.

1020 \frac{5}{6}

2.1.2.23.

Answer.

(a) 4900 N

(b)

\frac{44100}{x^{2}} N

(c)

29 400 J

2.1.2.24.

Answer.

220.5 J

2.1.2.25.

Answer.

About

7 \times 10^{28} J

2.1.2.26.

Answer.

true

2.1.2.27.

Answer.

9255 \frac{5}{9} J

2.1.2.28.

Answer.

\frac{7}{40} = 0.175 J

2.1.2.29.

Answer.

One possible answer:

\frac{1}{4} [\sqrt{1 - {(\frac{1}{8})}^{4}} + \sqrt{1 - {(\frac{3}{8})}^{4}}]

2.2 Averages
2.2.2 Exercises

2.2.2.1.

Answer.

The most straightforward of many possible answers is shown.

2.2.2.2.

Answer.

500 km

2.2.2.3.

Answer.

\frac{W}{b - a}

2.2.2.4.

Answer.

(a)

\frac{b - a}{n}

(b)

a + 3 \frac{b - a}{n}

(c)

f (a + 3 \frac{b - a}{n})

(d)

\frac{1}{n} \sum_{i = 1}^{n} f (a + (i - 1) \frac{b - a}{n})

2.2.2.5.

Answer.

(a) yes

(b) not enough information

2.2.2.6.

Answer.

2.2.2.7. (✳).

Answer.

1

2.2.2.8. (✳).

Answer.

\frac{1}{e - 1} [\frac{2}{9} e^{3} + \frac{1}{9}]

2.2.2.9. (✳).

Answer.

\frac{4}{π} + 1

2.2.2.10. (✳).

Answer.

\frac{2}{π}

2.2.2.11. (✳).

Answer.

\frac{10}{3} \log 7

degrees Celsius

2.2.2.12. (✳).

Answer.

\frac{1}{2 (e - 1)}

2.2.2.13. (✳).

Answer.

\frac{1}{2}

2.2.2.14.

Answer.

(a) 400 ppm

(b)

\approx 599.99

ppm

%

2.2.2.15.

Answer.

(a)

\frac{16 π}{5}

(b)

\frac{32 π}{5}

(c)

\frac{32 π}{5}

2.2.2.16.

Answer.

(a) 0

(b)

\sqrt{3}

2.2.2.17.

Answer.

\sqrt{\frac{4}{π} - 1} \approx 0.52

2.2.2.18.

Answer.

(a)

F (t) = 3 f (t) = 3 \sin (t π)

(b) 0

(c)

\frac{3}{\sqrt{2}} \approx 2.12

2.2.2.19. (✳).

Answer.

(a)

130 km

(b)

65 km/hr

2.2.2.20.

Answer.

(a)

A = e - 1

(b) 0

(c)

4 - 2 e + 2 (e - 1) \log (e - 1) \approx 0.42

2.2.2.21.

Answer.

(a) neither — both are zero

(b)

| f (x) - A |

has the larger average on

[0, 4]

2.2.2.22.

Answer.

(b - a) π R^{2}

2.2.2.23.

Answer.

2.2.2.24.

Answer.

Yes, but if

a \neq 0,

then

s = t .

2.2.2.25.

Answer.

A

2.2.2.26.

Answer.

(a)

\frac{b A (b) - a A (a)}{b - a}

(b)

f (t) = A (t) + t A^{'} (t)

2.2.2.27.

Answer.

One of many possible answers: $f (x) = {\begin{cases} - 1 & if x \leq 0 \\ 1 & if x > 0 \end{cases} .$
No such function exists.
- Note 1: Suppose $f (x) > 0$ for all $x$ in $[- 1, 1] .$ Then $\frac{1}{2} \int_{- 1}^{1} f (x) d x > \frac{1}{2} \int_{- 1}^{1} 0 d x = 0 .$ That is, the average value of $f (x)$ on the interval $[- 1, 1]$ is not zero — it’s something greater than zero.
- Note 2: Suppose $f (x) < 0$ for all $x$ in $[- 1, 1] .$ Then $\frac{1}{2} \int_{- 1}^{1} f (x) d x < \frac{1}{2} \int_{- 1}^{1} 0 d x = 0 .$ That is, the average value of $f (x)$ on the interval $[- 1, 1]$ is not zero — it’s something less than zero.
So, if the average value of $f (x)$ is zero, then $f (x) \geq 0$ for some $x$ in $[- 1, 1],$ and $f (y) \leq 0$ for some $y \in [- 1, 1] .$ Since $f$ is a continuous function, and 0 is between $f (x)$ and $f (y),$ by the intermediate value theorem (see the CLP-1 text) there is some value $c$ between $x$ and $y$ such that $f (c) = 0 .$ Since $x$ and $y$ are both in $[- 1, 1],$ then $c$ is as well. Therefore, no function exists as described in the question.

2.2.2.28.

Answer.

true

2.2.2.29.

Answer.

2.3 Centre of Mass and Torque
2.3.3 Exercises

2.3.3.1.

Answer.

(1, 1)

2.3.3.2.

Answer.

(0, 0)

2.3.3.3.

Answer.

In general, false.

2.3.3.4.

Answer.

3.5

metres from the left end

2.3.3.5.

Answer.

(a) to the left

(b) to the left

(d) along the line

x = a

(e) to the right

2.3.3.6.

Answer.

\frac{39200 π}{9} (12 - π) \approx 121, 212 J

2.3.3.7.

Answer.

(a), (b)

\frac{1}{x} d x

(c), (d)

\log 3

(e), (f)

\frac{2}{\log 3}

2.3.3.8.

Answer.

(a)

\frac{\sum_{i = 1}^{n} [\frac{b - a}{n} ρ (a + (i - \frac{1}{2}) (\frac{b - a}{n})) \times (a + (i - \frac{1}{2}) (\frac{b - a}{n}))]}{\sum_{i = 1}^{n} \frac{b - a}{n} ρ (a + (i - \frac{1}{2}) (\frac{b - a}{n}))}

(b)

\bar{x} = \frac{\int_{a}^{b} x ρ (x) d x}{\int_{a}^{b} ρ (x) d x}

2.3.3.9.

Answer.

(a)

(b)

(T (x) - B (x)) d x

(c)

T (x) - B (x)

(d)

\bar{x} = \frac{\int_{a}^{b} x (T (x) - B (x)) d x}{\int_{a}^{b} (T (x) - B (x)) d x}

2.3.3.10.

Answer.

(a) The strips between

x = a

and

x = a^{'}

at the left end of the figure all have the same centre of mass, which is the

y

-value where

T (x) = B (x),

x < 0 .

So, there should be multiple weights of different mass piled up at that

y

-value.

Similarly, the strips between

x = b^{'}

and

x = b

at the right end of the figure all have the same centre of mass, which is the

y

-value where

T (x) = B (x),

x > 0 .

So, there should be a second pile of weights of different mass, at that (higher)

y

-value.

Between these two piles, there are a collection of weights with identical mass distributed fairly evenly. The top and bottom ends of

R

(above the uppermost pile, and below the lowermost pile) have no weights.

One possible answer (using twelve slices):

(b) The area of the strip is

(T (x) - B (x)) d x,

and its centre of mass is at height

\frac{T (x) + B (x)}{2} .

(c)

\bar{y} = \frac{\int_{a}^{b} (T (x)^{2} - B (x)^{2}) d x}{2 \int_{a}^{b} (T (x) - B (x)) d x}

2.3.3.11. (✳).

Answer.

\bar{x} = - \frac{1}{3} \int_{- 1}^{0} 6 x^{2} d x

2.3.3.12.

Answer.

\bar{x} = \frac{14}{3}

2.3.3.13.

Answer.

\bar{x} = \frac{\log 10.1}{2 (\arctan 10 + \arctan (3))} \approx 0.43

2.3.3.14. (✳).

Answer.

\bar{y} = \frac{3}{4 e} - \frac{e}{4}

2.3.3.15. (✳).

Answer.

(a)

(b)

\frac{3 \log 3}{8 π}

2.3.3.16. (✳).

Answer.

\bar{x} = \frac{\frac{π}{4} \sqrt{2} - 1}{\sqrt{2} - 1}

and

\bar{y} = \frac{1}{4 (\sqrt{2} - 1)}

2.3.3.17. (✳).

Answer.

(a)

\bar{x} = \frac{k}{A} [\sqrt{2} - 1], \bar{y} = \frac{k^{2} π}{8 A}

(b)

k = \frac{8}{π} [\sqrt{2} - 1]

2.3.3.18. (✳).

Answer.

(a)

(b)

\frac{8}{3}

(c)

1

2.3.3.19. (✳).

Answer.

\frac{2}{π} \log 2 \approx 0.44127

2.3.3.20. (✳).

Answer.

\bar{x} = 0

and

\bar{y} = \frac{12}{24 + 9 π}

2.3.3.21. (✳).

Answer.

(a)

\frac{9}{4} π

(b)

\bar{x} = 0

and

\bar{y} = \frac{4}{π}

2.3.3.22.

Answer.

(\bar{x}, \bar{y}) = (1, - \frac{2}{π})

2.3.3.23.

Answer.

(\frac{e^{2} - 3 / 2}{e^{2} - 5 / 2}, \frac{e^{4} - 7}{4 e^{2} - 10}) \approx (1.2, 2.4)

2.3.3.24. (✳).

Answer.

\bar{y} = \frac{8}{5}

2.3.3.25. (✳).

Answer.

(a)

\bar{x} = \frac{8}{11},

\bar{y} = \frac{166}{55}

(b)

π \int_{0}^{4} y d y + π \int_{4}^{6} (6 - y)^{2} d y

2.3.3.26. (✳).

Answer.

(a)

\bar{y} = \frac{e}{4} - \frac{3}{4 e}

(b)

π (\frac{e^{2}}{2} + 2 e - \frac{3}{2})

2.3.3.27.

Answer.

(3, 1.5)

2.3.3.28.

Answer.

(0, 3.45)

2.3.3.29.

Answer.

(a)

\frac{h}{4}

(b)

\frac{\frac{1}{2} h^{2} k - \frac{2}{3} h k^{2} + \frac{1}{4} k^{3}}{h^{2} - h k + \frac{1}{3} k^{2}}

2.3.3.30.

Answer.

about 0.833 N

2.3.3.31.

Answer.

(a) 17,150

π

(b)

\frac{2450}{9} π (8 π - 9) \approx 13, 797 J

74 %

2.3.3.32.

Answer.

\bar{x} = \frac{π}{162} \sqrt{\frac{π}{2}} [\sin (\frac{π}{72}) + 2 \sin (\frac{π}{18}) + 9 \sin (\frac{π}{8}) + 8 \sin (\frac{2 π}{9}) + 25 \sin (\frac{25 π}{72}) + 9] \approx 0.976

2.4 Separable Differential Equations
2.4.7 Exercises

2.4.7.1.

Answer.

(a) yes

(b) yes

2.4.7.2.

Answer.

One possible answer: $f (x) = x,$ $g (y) = \frac{\sin y}{3 y} .$
One possible answer: $f (x) = e^{x},$ $g (y) = e^{y} .$
One possible answer: $f (x) = x - 1,$ $g (y) = 1 .$
The given equation is equivalent to the equation $\frac{d y}{d x} = x,$ which fits the form of a separable equation with $f (x) = x,$ $g (y) = 1 .$

2.4.7.3.

Answer.

The mnemonic allows us to skip from the separable differential equation we want to solve (very first line) to the equation

\int \frac{1}{g (y)} d y = \int f (x) d x

2.4.7.4.

Answer.

false

2.4.7.5.

Answer.

(a)

[0, \infty)

(b) No such function exists. If

| f (x) | = C x

and

f (x)

switches from

f (x) = C x

f (x) = - C x

at some point, then that point is a jump discontinuity. Where

f (x)

contains a discontinuity,

\frac{d y}{d x}

does not exist.

2.4.7.6.

Answer.

\frac{d Q}{d t} = - 0.003 Q (t)

2.4.7.7.

Answer.

\frac{d p}{d t} = α p (t) (1 - p (t)),

for some constant

α .

2.4.7.8.

Answer.

(a)

- 1

(b)

0

(c)

0.5

(d) Two possible answers are shown below:

Another possible answer is the constant function

y = 2 .

2.4.7.9.

Answer.

(a)

- \frac{1}{2}

(b)

\frac{3}{2}

(c)

- \frac{5}{2}

(d) Your sketch should look something like this:

(e) There are lots of possible answers. Several are shown below.

2.4.7.10. (✳).

Answer.

y = \log (x^{2} + 2)

2.4.7.11. (✳).

Answer.

y (x) = 3 \sqrt{1 + x^{2}}

2.4.7.12. (✳).

Answer.

y (t) = 3 \log (\frac{- 3}{C + \sin t})

2.4.7.13. (✳).

Answer.

y = \sqrt[3]{\frac{3}{2} e^{x^{2}} + C} .

2.4.7.14. (✳).

Answer.

y = - \log (C - \frac{x^{2}}{2})

The solution only exists for

C - \frac{x^{2}}{2} > 0,

i.e.

C > 0

and the function has domain

{x : | x | < \sqrt{2 C}} .

2.4.7.15. (✳).

Answer.

y = (3 e^{x} - 3 x^{2} + 24)^{1 / 3}

2.4.7.16. (✳).

Answer.

y = f (x) = - \frac{1}{\sqrt{x^{2} + 16}}

2.4.7.17. (✳).

Answer.

y = \sqrt{10 x^{3} + 4 x^{2} + 6 x - 4}

2.4.7.18. (✳).

Answer.

y (x) = e^{x^{4} / 4}

2.4.7.19. (✳).

Answer.

y = \frac{1}{1 - 2 x}

2.4.7.20. (✳).

Answer.

f (x) = e \cdot e^{x^{2} / 2}

2.4.7.21. (✳).

Answer.

y (x) = \sqrt{4 + 2 \log \frac{2 x}{x + 1}} .

Note that, to satisfy

y (1) = 2,

we need the positive square root.

2.4.7.22. (✳).

Answer.

y^{2} + \frac{2}{3} (y^{2} - 4)^{3 / 2} = 2 \sec x + 2

2.4.7.23. (✳).

Answer.

12 weeks

2.4.7.24. (✳).

Answer.

t = \sqrt{\frac{m}{k g}} \arctan (\sqrt{\frac{k}{m g}} v_{0})

2.4.7.25. (✳).

Answer.

(a)

k = \frac{1}{400}

(b)

t = 70 \sec

2.4.7.26. (✳).

Answer.

(a)

x (t) = \frac{3 - 4 e^{k t}}{1 - 2 e^{k t}}

(b) As

t \to \infty,

x \to 2 .

2.4.7.27. (✳).

Answer.

(a)

P = \frac{4}{1 + e^{- 4 t}}

(b) At

t = \frac{1}{2},

P \approx 3.523 .

t \to \infty,

P \to 4 .

2.4.7.28. (✳).

Answer.

(a)

\frac{d v}{d t} = - k v^{2}

(b)

v = \frac{400}{t + 1}

(c)

t = 7

2.4.7.29. (✳).

Answer.

(a)

B (t) = C e^{0.06 t - 0.02 \cos t}

with the arbitrary constant

C \geq 0 .

(b)

$ 1159.89

2.4.7.30. (✳).

Answer.

(a)

B (t) = {30000 - 50 m} e^{t / 50} + 50 m

(b)

$ 600

2.4.7.31. (✳).

Answer.

y (x) = \frac{4 - e^{1 - \cos x}}{2 - e^{1 - \cos x}} .

The largest allowed interval is

- \arccos (1 - \log 2) < x < \arccos (1 - \log 2)

or, roughly,

- 1.259 < x < 1.259 .

2.4.7.32. (✳).

Answer.

180, 000 \sqrt{\frac{3}{g}} \approx 99, 591 sec \approx 27.66 hr

2.4.7.33. (✳).

Answer.

t = \frac{4 \times 144}{15} \sqrt{\frac{12^{5}}{2 g}} \approx 2, 394 sec \approx 0.665 hr

2.4.7.34. (✳).

Answer.

(a)

3

(b)

y^{'} = (y - 1) (y - 2)

(c)

f (x) = \frac{4 - e^{x}}{2 - e^{x}}

2.4.7.35. (✳).

Answer.

p = \frac{1}{4}

2.4.7.36.

Answer.

One possible answer: $f (t) = 0$
$\frac{1}{\sqrt{x - a}} [f (x) - \frac{1}{2 (x - a)} \int_{a}^{x} f (t) d t] = \frac{f^{2} (x)}{2 \sqrt{\int_{a}^{x} f^{2} (t) d t}}$
$\frac{2}{x - a} \int_{a}^{x} f (t) d t [f (x) - \frac{1}{2 (x - a)} \int_{a}^{x} f (t) d t] = f^{2} (x)$
$Y (x) = D (x - a),$ where $D$ is any constant
$f (t) = D,$ for any nonnegative constant $D$

2.4.7.37.

Answer.

x = \frac{1}{4} (y - 1 + \frac{1}{4} \log | \frac{2 y - 1}{2 y + 1} |)

3 Sequences and series
3.1 Sequences
3.1.2 Exercises

3.1.2.1.

Answer.

(a)

- 2

(b)

0

3.1.2.2.

Answer.

true

3.1.2.3.

Answer.

(a)

\frac{A - B}{C}

(b)

0

(c)

\frac{A}{B}

3.1.2.4.

Answer.

Two possible answers, of many:

$a_{n} = {\begin{cases} 3000 - n & if n \leq 1000 \\ - 2 + \frac{1}{n} & if n > 1000 \end{cases}$
$a_{n} = \frac{1, 002, 001}{n} - 2$

3.1.2.5.

Answer.

One possible answer is

a_{n} = (- 1)^{n} = {- 1, 1, - 1, 1, - 1, 1, - 1, \dots} .

Another is

a_{n} = n (- 1)^{n} = {- 1, 2, - 3, 4, - 5, 6, - 7, \dots} .

3.1.2.6.

Answer.

One sequence of many possible is

a_{n} = \frac{(- 1)^{n}}{n} = {- 1, \frac{1}{2}, - \frac{1}{3}, \frac{1}{4}, - \frac{1}{5}, \frac{1}{6}, \dots} .

3.1.2.7.

Answer.

Some possible answers:

$\frac{- 1}{n} \leq \frac{\sin n}{n} \leq \frac{1}{n}$
$\frac{n^{2}}{13 e^{n}} \leq \frac{n^{2}}{e^{n} (7 + \sin n - 5 \cos n)} \leq \frac{n^{2}}{e^{n}}$ or $0 \leq \frac{n^{2}}{e^{n} (7 + \sin n - 5 \cos n)} \leq \frac{n^{2}}{e^{n}}$
$\frac{- 1}{n^{n}} \leq (- n)^{- n} \leq \frac{1}{n^{n}}$

3.1.2.8.

Answer.

(a)

a_{n} = b_{n} = h (n) = i (n),

c_{n} = j (n),

d_{n} = f (n),

e_{n} = g (n)

(b)

lim_{n \to \infty} a_{n} = lim_{n \to \infty} b_{n} = lim_{x \to \infty} h (x) = 1,

lim_{n \to \infty} c_{n} = lim_{n \to \infty} e_{n} = lim_{x \to \infty} g (x) = lim_{x \to \infty} j (x) = 0,

lim_{n \to \infty} d_{n},

lim_{x \to \infty} f (x)

and

lim_{x \to \infty} i (x)

do not exist.

3.1.2.9.

Answer.

(a) Some possible answers:

a_{22} \approx - 0.99996,

a_{66} \approx - 0.99965,

and

a_{110} \approx - 0.99902 .

(b) Some possible answers:

a_{11} \approx 0.0044,

a_{33} \approx - 0.0133,

and

a_{55} \approx 0.0221 .

The integers 11, 33, and 55 were found by approximating

π

\frac{22}{7}

and finding when an odd multiple of

\frac{11}{7}

(which is the corresponding approximation of

π / 2

) is an integer.

a_{44} \approx 0.9998,

a_{132} \approx 0.9986,

and

a_{220} \approx 0.09961 .

See the solution for how we found them.

3.1.2.10.

Answer.

(a)

\infty

(b)

\frac{3}{4}

3.1.2.11.

Answer.

\infty

3.1.2.12.

Answer.

3.1.2.13.

Answer.

3.1.2.14.

Answer.

3.1.2.15.

Answer.

3.1.2.16.

Answer.

3.1.2.17.

Answer.

\infty

3.1.2.18. (✳).

Answer.

lim_{k \to \infty} a_{k} = 0 .

3.1.2.19. (✳).

Answer.

The sequence converges to

0 .

3.1.2.20. (✳).

Answer.

9

3.1.2.21. (✳).

Answer.

\log 2

3.1.2.22.

Answer.

3.1.2.23.

Answer.

- \infty

3.1.2.24.

Answer.

100 \cdot 2^{99} .

3.1.2.25.

Answer.

Possible answers are

{a_{n}} = {n [f (a + \frac{1}{n}) - f (a)]}

{a_{n}} = {n [f (a) - f (a - \frac{1}{n})]} .

3.1.2.26.

Answer.

(a)

A_{n} = \frac{n}{2} \sin (\frac{2 π}{n})

(b)

π

3.1.2.27.

Answer.

$A_{n} = 1$ for all $n$
$lim_{n \to \infty} A_{n} = 1 .$
$g (x) = 0$
$\int_{0}^{\infty} g (x) d x = 0 .$

3.1.2.28.

Answer.

e^{3}

3.1.2.29.

Answer.

(a)

4

(b)

x = 4

3.1.2.30.

Answer.

(a) decreasing

(b)

f_{n} = \frac{1}{n} f_{1}

(c)

2 %

(d)

0.18 %

(e) “be”: 11,019,308; “and”: 7,346,205

3.2 Series
3.2.2 Exercises

3.2.2.1.

Answer.

\begin{array}{ll} N & S_{N} \\ 1 & 1 \\ 2 & 1 + \frac{1}{2} \\ 3 & 1 + \frac{1}{2} + \frac{1}{3} \\ 4 & 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} \\ 5 & 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} \end{array}

3.2.2.2.

Answer.

3.2.2.3.

Answer.

(a)

a_{n} = {\begin{cases} \frac{1}{2} & if n = 1 \\ \frac{1}{n (n + 1)} & else \end{cases}

(b)

0

3.2.2.4.

Answer.

a_{n} = {\begin{cases} 0 & if n = 1 \\ 2 (- 1)^{n} - \frac{1}{n (n - 1)} & else \end{cases}

3.2.2.5.

Answer.

a_{n} < 0

for all

n \geq 2

3.2.2.6.

Answer.

(a)

\sum_{n = 1}^{\infty} \frac{2}{4^{n}}

(b)

\frac{2}{3}

3.2.2.7.

Answer.

(a)

\sum_{n = 1}^{\infty} \frac{1}{9^{n}}

(b)

\frac{1}{8}

3.2.2.8.

Answer.

Two possible pictures:

3.2.2.9.

Answer.

\frac{5^{101} - 1}{4 \cdot 5^{100}}

3.2.2.10.

Answer.

All together, there were 36 cookies brought by Student 11 through Student 20.

3.2.2.11.

Answer.

\frac{5^{51} - 1}{4 \cdot 5^{100}}

3.2.2.12.

Answer.

(a) As time passes, your gains increase, approaching $1.

(b) 1

(d)

- \infty

3.2.2.13.

Answer.

A + B + C - c_{1}

3.2.2.14.

Answer.

in general, false

3.2.2.15. (✳).

Answer.

\frac{3}{2}

3.2.2.16. (✳).

Answer.

\frac{1}{7 \times 8^{6}}

3.2.2.17. (✳).

Answer.

6

3.2.2.18. (✳).

Answer.

\cos (\frac{π}{3}) - \cos (0) = - \frac{1}{2}

3.2.2.19. (✳).

Answer.

(a)

a_{n} = \frac{11}{16 n^{2} + 24 n + 5}

(b)

\frac{3}{4}

3.2.2.20. (✳).

Answer.

\frac{24}{5}

3.2.2.21. (✳).

Answer.

\frac{7}{30}

3.2.2.22. (✳).

Answer.

\frac{263}{99}

3.2.2.23. (✳).

Answer.

\frac{321}{999} = \frac{107}{333}

3.2.2.24. (✳).

Answer.

3

3.2.2.25. (✳).

Answer.

\frac{1}{2} + \frac{5}{7} = \frac{17}{14}

3.2.2.26. (✳).

Answer.

\frac{40}{3}

3.2.2.27.

Answer.

The series diverges to

- \infty .

3.2.2.28.

Answer.

- \frac{1}{2}

3.2.2.29.

Answer.

9.8 J

3.2.2.30.

Answer.

\frac{4 π}{3 (π^{3} - 1)}

3.2.2.31.

Answer.

\frac{\sin^{2} 3}{8} + 32 \approx 32.0025

3.2.2.32.

Answer.

a_{n} = {\begin{cases} \frac{2}{n (n - 1) (n - 2)} & if n \geq 3, \\ - \frac{5}{2} & if n = 2, \\ 2 & if n = 1 \end{cases}

3.2.2.33.

Answer.

\frac{5}{8}

3.3 Convergence Tests
3.3.11 Exercises

3.3.11.1.

Answer.

(B), (C)

3.3.11.2.

Answer.

(A)

3.3.11.3.

Answer.

(a) I am old

(b) not enough information to tell

(d) I am young

3.3.11.4.

Answer.

	if $\sum a_{n}$ converges	if $\sum a_{n}$ diverges
and if ${a_{n}}$ is the red series	then $\sum b_{n}$ CONVERGES	inconclusive
and if ${a_{n}}$ is the blue series	inconclusive	then $\sum b_{n}$ DIVERGES

3.3.11.5.

Answer.

(a) both direct comparison and limit comparison

(b) direct comparison

(d) neither

3.3.11.6.

Answer.

It diverges by the divergence test, because

lim_{n \to \infty} a_{n} \neq 0 .

3.3.11.7.

Answer.

We cannot use the divergence test to show that a series converges. It is inconclusive in this case.

3.3.11.8.

Answer.

The integral test does not apply because

f (x)

is not decreasing.

3.3.11.9.

Answer.

The inequality goes the wrong way, so the direct comparison test (with this comparison series) is inconclusive.

3.3.11.10.

Answer.

(B), (D)

3.3.11.11.

Answer.

One possible answer:

\sum_{n = 1}^{\infty} \frac{1}{n^{2}} .

3.3.11.12.

Answer.

By the divergence test, for a series

\sum a_{n}

to converge, we need

lim_{n \to \infty} a_{n} = 0 .

That is, the magnitude (absolute value) of the terms needs to be getting smaller. If

lim_{n \to \infty} | \frac{a_{n}}{a_{n + 1}} | < 1

or (equivalently)

lim_{n \to \infty} | \frac{a_{n + 1}}{a_{n}} | > 1,

then

| a_{n + 1} | > | a_{n} |

for sufficiently large

n,

so the terms are actually growing in magnitude. That means the series diverges, by the divergence test.

3.3.11.13.

Answer.

One possible answer:

f (x) = \sin (π x),

a_{n} = 0

for every

n .

By the integral test, any answer will use a function

f (x)

that is not both positive and decreasing.

3.3.11.14. (✳).

Answer.

One possible answer:

b_{n} = \frac{2^{n}}{3^{n}}

3.3.11.15. (✳).

Answer.

(a) In general false. The harmonic series

\sum_{n = 1}^{\infty} \frac{1}{n}

provides a counterexample.

(b) In general false. If

a_{n} = (- 1)^{n} \frac{1}{n},

then

\sum_{n = 1}^{\infty} (- 1)^{n} a_{n}

is again the harmonic series

\sum_{n = 1}^{\infty} \frac{1}{n},

which diverges.

a_{n} = 0

and

b_{n} = 1 .

3.3.11.16. (✳).

Answer.

No. It diverges.

3.3.11.17. (✳).

Answer.

It diverges.

3.3.11.18. (✳).

Answer.

The series diverges.

3.3.11.19.

Answer.

It diverges.

3.3.11.20.

Answer.

This is a geometric series with

r = 1.001 .

Since

| r | > 1,

it is divergent.

3.3.11.21.

Answer.

The series converges to

- \frac{1}{150} .

3.3.11.22.

Answer.

The series converges.

3.3.11.23.

Answer.

It diverges.

3.3.11.24.

Answer.

The series converges.

3.3.11.25.

Answer.

The series converges to

\frac{1}{3} .

3.3.11.26.

Answer.

The series converges.

3.3.11.27.

Answer.

It converges.

3.3.11.28. (✳).

Answer.

Let

f (x) = \frac{5}{x (\log x)^{3 / 2}} .

Then

f (x)

is positive and decreases as

x

increases. So the sum

\sum_{3}^{\infty} f (n)

and the integral

\int_{3}^{\infty} f (x) d x

either both converge or both diverge, by the integral test, which is Theorem 3.3.5. For the integral, we use the substitution

u = \log x,

d u = \frac{d x}{x}

to get

\begin{matrix} \int_{3}^{\infty} \frac{5 d x}{x (\log x)^{3 / 2}} = \int_{\log 3}^{\infty} \frac{5 d u}{u^{3 / 2}} \end{matrix}

which converges by the

p

-test (which is Example 1.12.8) with

p = \frac{3}{2} > 1 .

3.3.11.29. (✳).

Answer.

p > 1

3.3.11.30. (✳).

Answer.

It converges.

3.3.11.31. (✳).

Answer.

The series

\sum_{n = 2}^{\infty} \frac{\sqrt{3}}{n^{2}}

converges by the

p

-test with

p = 2 .

Note that

\begin{matrix} 0 < a_{n} = \frac{\sqrt{3 n^{2} - 7}}{n^{3}} < \frac{\sqrt{3 n^{2}}}{n^{3}} = \frac{\sqrt{3}}{n^{2}} \end{matrix}

for all

n \geq 2 .

As the series

\sum_{n = 2}^{\infty} \frac{\sqrt{3}}{n^{2}}

converges, the comparison test says that

\sum_{n = 2}^{\infty} \frac{\sqrt{3 n^{2} - 7}}{n^{3}}

converges too.

3.3.11.32. (✳).

Answer.

The series converges.

3.3.11.33. (✳).

Answer.

It diverges.

3.3.11.34. (✳).

Answer.

(a) diverges

(b) converges

3.3.11.35. (✳).

Answer.

The series diverges.

3.3.11.36. (✳).

Answer.

(a) converges

(b) diverges

3.3.11.37.

Answer.

\frac{1}{e^{5} - e^{4}}

3.3.11.38. (✳).

Answer.

\frac{1}{7}

3.3.11.39. (✳).

Answer.

(a) diverges by limit comparison with the harmonic series

(b) converges by the ratio test

3.3.11.40. (✳).

Answer.

(a) Converges by the limit comparison test with

b = \frac{1}{k^{5 / 3}} .

(b) Diverges by the ratio test.

3.3.11.41. (✳).

Answer.

It converges.

3.3.11.42. (✳).

Answer.

N = 5

3.3.11.43. (✳).

Answer.

N \geq 999

3.3.11.44. (✳).

Answer.

We need

N = 4

and then

S_{4} = \frac{1}{3^{2}} - \frac{1}{5^{2}} + \frac{1}{7^{2}} - \frac{1}{9^{2}}

3.3.11.45. (✳).

Answer.

(a) converges

(b) converges

3.3.11.46. (✳).

Answer.

(a) See the solution.

(b)

f (x) = \frac{x + \sin x}{1 + x^{2}}

is not a decreasing function.

3.3.11.47. (✳).

Answer.

The sum is between 0.9035 and 0.9535.

3.3.11.48. (✳).

Answer.

Since

lim_{n \to \infty} a_{n} = 0,

there must be some integer

N

such that

\frac{1}{2} > a_{n} \geq 0

for all

n > N .

Then, for

n > N,

\begin{aligned} \frac{a_{n}}{1 - a_{n}} & \leq \frac{a_{n}}{1 - 1 / 2} = 2 a_{n} \end{aligned}

From the information in the problem statement, we know

\begin{aligned} \sum_{n = N + 1}^{\infty} 2 a_{n} & = 2 \sum_{n = N + 1}^{\infty} a_{n} converges. \end{aligned}

So, by the direct comparison test,

\begin{aligned} \sum_{n = N + 1}^{\infty} \frac{a_{n}}{1 - a_{n}} & converges as well. \end{aligned}

Since the convergence of a series is not affected by its first $N$ terms, as long as $N$ is finite, we conclude

\begin{aligned} \sum_{n = 1}^{\infty} \frac{a_{n}}{1 - a_{n}} & converges. \end{aligned}

3.3.11.49. (✳).

Answer.

It diverges.

3.3.11.50. (✳).

Answer.

It converges to

- \log 2 = \log \frac{1}{2},

3.3.11.51. (✳).

Answer.

See the solution.

3.3.11.52.

Answer.

About 9% to 10%

3.3.11.53.

Answer.

The total population is between 29,820,091 and 30,631,021 people.

3.4 Absolute and Conditional Convergence
3.4.3 Exercises

3.4.3.1. (✳).

Answer.

False. For example,

b_{n} = \frac{1}{n}

provides a counterexample.

3.4.3.2.

Answer.

	$\sum a_{n}$ converges	$\sum a_{n}$ diverges
$\sum \| a_{n} \|$ converges	converges absolutely	not possible
$\sum \| a_{n} \|$ diverges	converges conditionally	diverges

3.4.3.3. (✳).

Answer.

Conditionally convergent

3.4.3.4. (✳).

Answer.

The series diverges.

3.4.3.5. (✳).

Answer.

It diverges.

3.4.3.6. (✳).

Answer.

It converges absolutely.

3.4.3.7. (✳).

Answer.

It converges absolutely.

3.4.3.8. (✳).

Answer.

It diverges.

3.4.3.9. (✳).

Answer.

It converges absolutely.

3.4.3.10.

Answer.

See solution.

3.4.3.11.

Answer.

See solution.

3.4.3.12.

Answer.

See solution.

3.4.3.13. (✳).

Answer.

(a) See the solution.

(b)

| S - S_{5} | \leq 24 \times 36 e^{- 6^{3}}

3.4.3.14.

Answer.

\cos 1 \approx \frac{389}{720};

the actual associated error (using a calculator) is about

0.000025 .

3.4.3.15.

Answer.

See solution.

3.5 Power Series
3.5.3 Exercises

3.5.3.1.

Answer.

3.5.3.2.

Answer.

f (x) = \sum_{n = 1}^{\infty} \frac{n (x - 5)^{n - 1}}{n! + 2}

3.5.3.3.

Answer.

Only

x = c

3.5.3.4.

Answer.

R = 6

3.5.3.5. (✳).

Answer.

(a)

R = \frac{1}{2}

(b)

\frac{2}{1 + 2 x}

for all

| x | < \frac{1}{2}

3.5.3.6. (✳).

Answer.

R = \infty

3.5.3.7. (✳).

Answer.

1

3.5.3.8. (✳).

Answer.

The interval of convergence is

- 1 < x + 2 \leq 1

(- 3, - 1] .

3.5.3.9. (✳).

Answer.

The radius of convergence is

3 .

The interval of convergence is

- 4 < x \leq 2,

or simply

(- 4, 2] .

3.5.3.10. (✳).

Answer.

- 3 \leq x < 7

[- 3, 7)

3.5.3.11. (✳).

Answer.

The given series converges if and only if

- 3 \leq x \leq - 1 .

Equivalently, the series has interval of convergence

[- 3, - 1] .

3.5.3.12. (✳).

Answer.

(a)

\frac{3}{4} \leq x < \frac{5}{4}

[\frac{3}{4}, \frac{5}{4})

(b)

- \frac{5}{8} \leq x < - \frac{3}{8}

[- \frac{5}{8}, - \frac{3}{8})

(c)

- \frac{3}{4} < x < - \frac{1}{4},

(- \frac{3}{4}, - \frac{1}{4})

3.5.3.13. (✳).

Answer.

The radius of convergence is

2 .

The interval of convergence is

- 1 < x \leq 3,

(- 1, 3] .

3.5.3.14. (✳).

Answer.

The interval of convergence is

a - 1 < x < a + 1,

(a - 1, a + 1) .

3.5.3.15. (✳).

Answer.

(a)

| x + 1 | \leq 9

- 10 \leq x \leq 8

[- 10, 8]

(b) This series converges only for

x = 1 .

3.5.3.16. (✳).

Answer.

\sum_{n = 0}^{\infty} x^{n + 3} = \sum_{n = 3}^{\infty} x^{n}

3.5.3.17.

Answer.

f (x) = 3 + \sum_{n = 1}^{\infty} \frac{(x - 1)^{n}}{n (n + 1)}

3.5.3.18. (✳).

Answer.

The series converges absolutely for

| x | < 9,

converges conditionally for

x = - 9

and diverges otherwise.

3.5.3.19. (✳).

Answer.

(a)

\sum_{n = 0}^{\infty} (- 1)^{n} \frac{x^{3 n + 1}}{3 n + 1} + C

(b) We need to keep two terms (the

n = 0

and

n = 1

terms).

3.5.3.20. (✳).

Answer.

(a) See the solution.

(b)

\sum_{n = 0}^{\infty} n^{2} x^{n} = \frac{x (1 + x)}{(1 - x)^{3}} .

The series converges for

- 1 < x < 1 .

3.5.3.21. (✳).

Answer.

See the solution.

3.5.3.22. (✳).

Answer.

(a)

1 .

(b) The series converges for

- 1 \leq x < 1,

i.e. for the interval

[- 1, 1)

3.5.3.23.

Answer.

\frac{5}{6}

3.5.3.24.

Answer.

The point

x = c

corresponds to a local maximum if

A_{2} < 0

and a local minimum if

A_{2} > 0 .

3.5.3.25.

Answer.

\frac{13}{80}

3.5.3.26.

Answer.

x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - \frac{x^{4}}{4}

3.5.3.27.

Answer.

x - \frac{x^{3}}{3} + \frac{x^{5}}{5}

3.6 Taylor Series
3.6.8 Exercises

3.6.8.1.

Answer.

A :

linear

B :

constant

C :

quadratic

3.6.8.2.

Answer.

T (5) = \arctan^{3} (e^{5} + 7)

3.6.8.3.

Answer.

A - V

radius

= 1

B - I

radius

= 1

C - IV

radius

= 1

D - VI

radius

= + \infty

E - II

radius

= + \infty

F - III

radius

= + \infty

3.6.8.4.

Answer.

(a)

f^{(20)} (3) = 20^{2} (\frac{20!}{20! + 1})

(b)

g^{(20)} (3) = 10^{2} (\frac{20!}{10! + 1})

(c)

h^{(20)} (0) = - \frac{20! \cdot 5^{11}}{11}; h^{(22)} (0) = 0

3.6.8.5.

Answer.

\sum_{n = 1}^{\infty} \frac{(- 1)^{n - 1}}{n} (x - 1)^{n}

3.6.8.6.

Answer.

\sum_{n = 0}^{\infty} \frac{(- 1)^{n + 1}}{(2 n + 1)!} (x - π)^{2 n + 1}

3.6.8.7.

Answer.

\frac{1}{10} \sum_{n = 0}^{\infty} {(\frac{10 - x}{10})}^{n}

with interval of convergence

(0, 20) .

3.6.8.8.

Answer.

\sum_{n = 0}^{\infty} \frac{3^{n} e^{3 a}}{n!} (x - a)^{n},

with infinite radius of convergence

3.6.8.9. (✳).

Answer.

- \sum_{n = 0}^{\infty} 2^{n} x^{n}

3.6.8.10. (✳).

Answer.

b_{n} = 3 (- 1)^{n} + 2^{n}

3.6.8.11. (✳).

Answer.

c_{5} = \frac{3^{5}}{5!}

3.6.8.12. (✳).

Answer.

\sum_{n = 0}^{\infty} (- 1)^{n} \frac{2^{n + 1} x^{n + 1}}{n + 1}

for all

| x | < \frac{1}{2}

3.6.8.13. (✳).

Answer.

a = 1,

b = - \frac{1}{3!} = - \frac{1}{6} .

3.6.8.14. (✳).

Answer.

\int \frac{e^{- x^{2}} - 1}{x} d x = C - \frac{x^{2}}{2} + \frac{x^{4}}{8} + \dots .

It is not clear from the wording of the question whether or not the arbitrary constant

C

is to be counted as one of the “first two nonzero terms”.

3.6.8.15. (✳).

Answer.

\sum_{n = 0}^{\infty} (- 1)^{n} \frac{2^{2 n + 1} x^{2 n + 6}}{(2 n + 1) (2 n + 6)} + C = \sum_{n = 0}^{\infty} (- 1)^{n} \frac{2^{2 n} x^{2 n + 6}}{(2 n + 1) (n + 3)} + C

3.6.8.16. (✳).

Answer.

f (x) = 1 + \sum_{n = 0}^{\infty} (- 1)^{n} \frac{3^{n}}{3 n + 2} x^{3 n + 2}

3.6.8.17. (✳).

Answer.

\frac{π}{2 \sqrt{3}}

3.6.8.18. (✳).

Answer.

\frac{1}{e}

3.6.8.19. (✳).

Answer.

e^{1 / e}

3.6.8.20. (✳).

Answer.

e^{1 / π} - 1

3.6.8.21. (✳).

Answer.

\log (3 / 2)

3.6.8.22. (✳).

Answer.

(e + 2) e^{e} - 2

3.6.8.23.

Answer.

The sum diverges — see the solution.

3.6.8.24.

Answer.

\frac{1 + \sqrt{2}}{\sqrt{2}}

3.6.8.25. (✳).

Answer.

(a) See the solution.

(b)

\frac{1}{2} (e + \frac{1}{e})

3.6.8.26.

Answer.

(a) 50,000

(b) three terms (

n = 0

n = 2

)

n = 0

n = 5

)

3.6.8.27.

Answer.

3.6.8.28.

Answer.

S_{13}

or higher

3.6.8.29.

Answer.

S_{9}

or higher

3.6.8.30.

Answer.

S_{18}

or higher

3.6.8.31.

Answer.

The error is in the interval

(\frac{- 5^{7}}{14 \cdot 3^{7}} [1 + \frac{1}{3^{7}}], \frac{- 5^{7}}{7 \cdot 6^{7}}) \approx (- 0.199, - 0.040)

3.6.8.32. (✳).

Answer.

- 1

3.6.8.33. (✳).

Answer.

\frac{1}{5!} = \frac{1}{120}

3.6.8.34.

Answer.

e^{2}

3.6.8.35.

Answer.

\sqrt{e}

3.6.8.36.

Answer.

\frac{2}{(6 / 7)^{3}} = \frac{343}{108}

3.6.8.37.

Answer.

\sum_{n = 0}^{\infty} (- 1)^{n} \frac{x^{2 n + 4}}{(2 n + 1) (2 n + 2)} = x^{3} \arctan x - \frac{x^{2}}{2} \log (1 + x^{2})

3.6.8.38.

Answer.

(a) the Maclaurin series for

f (x)

\sum_{n = 0}^{\infty} \frac{(2 n)!}{2^{2 n} (n!)^{2}} x^{n},

and its radius of convergence is

R = 1 .

(b) the Maclaurin series for

\arcsin x

\sum_{n = 0}^{\infty} \frac{(2 n)!}{2^{2 n} (n!)^{2} (2 n + 1)} x^{2 n + 1},

and its radius of convergence is

R = 1 .

3.6.8.39. (✳).

Answer.

\log (x) = \log 2 + \sum_{n = 1}^{\infty} \frac{(- 1)^{n - 1}}{n 2^{n}} (x - 2)^{n} .

It converges when

0 < x \leq 4 .

3.6.8.40. (✳).

Answer.

(a)

\sum_{n = 0}^{\infty} (- 1)^{n} \frac{x^{4 n + 1}}{4 n + 1}

(b)

0.493967

I (1 / 2)

3.6.8.41. (✳).

Answer.

\frac{1}{66}

3.6.8.42. (✳).

Answer.

Any interval of length 0.0002 that contains

0.03592

and

0.03600

is fine.

3.6.8.43. (✳).

Answer.

(a)

\sum_{n = 1}^{\infty} (- 1)^{n} \frac{x^{n}}{n n!}

(b)

- 0.80

3.6.8.44. (✳).

Answer.

(a)

Σ (x) = \sum_{n = 0}^{\infty} (- 1)^{n} \frac{x^{2 n + 1}}{(2 n + 1) (2 n + 1)!}

(b)

x = π

(c)

1.8525

3.6.8.45. (✳).

Answer.

(a)

I (x) = \sum_{n = 1}^{\infty} (- 1)^{n} \frac{x^{2 n - 1}}{(2 n)! (2 n - 1)}

(b)

I (1) = - \frac{1}{2} + \frac{1}{4! 3} \pm \frac{1}{6! 5} = - 0.486 \pm 0.001

(c)

I (1) < - \frac{1}{2} + \frac{1}{4! 3}

3.6.8.46. (✳).

Answer.

(a)

I (x) = \sum_{n = 1}^{\infty} (- 1)^{n + 1} \frac{x^{2 n - 1}}{(2 n)!} = \frac{1}{2!} x - \frac{1}{4!} x^{3} + \frac{1}{6!} x^{5} - \frac{1}{8!} x^{8} + \dots

(b) 0.460

(c)

I (1) < \frac{1}{2!} - \frac{1}{4!} + \frac{1}{6!} < 0.460

3.6.8.47. (✳).

Answer.

(a) See the solution.

(b) The series converges for all

x .

3.6.8.48. (✳).

Answer.

See the solution.

3.6.8.49. (✳).

Answer.

(a)

\cosh (x) = \sum_{\binom{n = 0}{n even}}^{\infty} \frac{x^{n}}{n!} = \sum_{n = 0}^{\infty} \frac{x^{2 n}}{(2 n)!}

for all

x .

(b), (c) See the solution.

3.6.8.50.

Answer.

(a)

\sqrt[3]{3} \approx 1.26

(b) 12 terms (

S_{11}

)

3.6.8.51.

Answer.

\frac{15!}{5! \cdot 5^{6}} - \frac{21!}{7! \cdot 11! \cdot 5^{11}} + \frac{27!}{9! \cdot 17! \cdot 5^{17}} - \frac{33!}{11! \cdot 23! \cdot 5^{23}}

3.6.8.52.

Answer.

(a)

(b) the constant function 0

(d) only at

x = 0

3.6.8.53.

Answer.

0

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Appendix F Answers to Exercises

1 Integration1.1 Definition of the Integral1.1.8 Exercises

1.1.8.1.

1.1.8.2.

1.1.8.3.

1.1.8.4.

1.1.8.5.

1.1.8.6.

1.1.8.7.

1.1.8.8.

1.1.8.9.

1.1.8.10.

1.1.8.11.

1.1.8.12. (✳).

1.1.8.13.

1.1.8.14.

1.1.8.15. (✳).

1.1.8.16.

1.1.8.17.

1.1.8.18. (✳).

1.1.8.19. (✳).

1.1.8.20. (✳).

1.1.8.21. (✳).

1.1.8.22. (✳).

1.1.8.23. (✳).

1.1.8.24. (✳).

1.1.8.25. (✳).

1.1.8.26.

1.1.8.27.

1.1.8.28. (✳).

1.1.8.29.

1.1.8.30.

1.1.8.31.

1.1.8.32.

1.1.8.33. (✳).

1.1.8.34. (✳).

1.1.8.35.

1.1.8.36.

1.1.8.37. (✳).

1.1.8.38. (✳).

1.1.8.39. (✳).

1.1.8.40. (✳).

1.1.8.41. (✳).

1.1.8.42.

1.1.8.43.

1.1.8.44.

1.1.8.45.

1.1.8.46.

1.1.8.47.

1.2 Basic properties of the definite integral1.2.3 Exercises

1.2.3.1.

1.2.3.2.

1.2.3.3. (✳).

1.2.3.4.

1.2.3.5. (✳).

1.2.3.6. (✳).

1.2.3.7. (✳).

1.2.3.8.

1.2.3.9. (✳).

1.2.3.10.

1.2.3.11.

1.2.3.12. (✳).

1.2.3.13. (✳).

1.2.3.14. (✳).

1.2.3.15.

1.2.3.16.

1.2.3.17.

1.2.3.18.

1.2.3.19.

1.2.3.20.

1.3 The Fundamental Theorem of Calculus1.3.2 Exercises

1.3.2.1. (✳).

1.3.2.2. (✳).

1.3.2.3. (✳).

1.3.2.4.

1.3.2.5.

1.3.2.6.

1.3.2.7.

1.3.2.8.

1.3.2.9.

1 Integration
1.1 Definition of the Integral
1.1.8 Exercises

1.2 Basic properties of the definite integral
1.2.3 Exercises

1.3 The Fundamental Theorem of Calculus
1.3.2 Exercises

1.4 Substitution
1.4.2 Exercises