Content check — Calculus Vol. 2, Chapter 1: Integration

Every problem in the chapter beside the book’s answer, a chapter at a time as you scroll.

Chapter 1: Integration

276 problems · 32 figures

Checkpoint 1.1 §1.1 source ↗

Write in sigma notation and evaluate the sum of terms 2ⁱ for

i = 3, 4, 5, 6 .

Answer

\sum_{i = 3}^{6} 2^{i} =

2^{3} +

2^{4} +

2^{5} +

2^{6} =

120

Checkpoint 1.2 §1.1 source ↗

Find the sum of the values of

4 + 3 i

for

i =

1,

2,…, 100 .

Answer

15,550

Checkpoint 1.3 §1.1 source ↗

Evaluate the sum indicated by the notation

\sum_{k = 1}^{20} (2 k + 1) .

Answer

440

Checkpoint 1.4 §1.1 source ↗

Sketch left-endpoint and right-endpoint approximations for

f (x) = \frac{1}{x}

[1, 2];

use

n = 4 .

Approximate the area using both methods.

Answer

The left-endpoint approximation is 0.7595. The right-endpoint approximation is 0.6345.

Two graphs side by side showing the left-endpoint approximation ad right-endpoint approximation of the area under the curve f(x) = 1/x from 1 to 2 with endpoints spaced evenly at .25 units. The heights of the left-endpoint approximation one are determined by the values of the function at the left endpoints, and the height of the right-endpoint approximation one are determined by the values of the function at the right endpoints.

Checkpoint 1.5 §1.1 source ↗

a. Find an upper sum for $f (x) = 10 - x^{2}$ on $[1, 2];$ let $n = 4 .$

b. Sketch the approximation.

Answer

a. $Upper sum =$ $8.0313 .$

A graph of the function f(x) = 10 − x^2 from 0 to 2. It is set up for a right endpoint approximation over the area [1,2], which is labeled a=x0 to x4. It is an upper sum.

Checkpoint 1.6 §1.1 source ↗

Using the function

f (x) = sin x

over the interval

[0, \frac{π}{2}],

find an upper sum; let

n = 6 .

Answer

A \approx 1.125

1-1 §1.1 source ↗

State whether the given sums are equal or unequal.

a. $\sum_{i = 1}^{10} i$ and $\sum_{k = 1}^{10} k$

b. $\sum_{i = 1}^{10} i$ and $\sum_{i = 6}^{15} (i - 5)$

c. $\sum_{i = 1}^{10} i (i - 1)$ and $\sum_{j = 0}^{9} (j + 1) j$

d. $\sum_{i = 1}^{10} i (i - 1)$ and $\sum_{k = 1}^{10} (k^{2} - k)$

Answer

a. They are equal; both represent the sum of the first 10 whole numbers. b. They are equal; both represent the sum of the first 10 whole numbers. c. They are equal by substituting

j = i - 1 .

d. They are equal; the first sum factors the terms of the second.

1-3 §1.1 source ↗

In the following exercises, use the rules for sums of powers of integers to compute the sums.

$\sum_{i = 5}^{10} i^{2}$

Answer

385 - 30 = 355

1-5 §1.1 source ↗

Suppose that $\sum_{i = 1}^{100} a_{i} = 15$ and $\sum_{i = 1}^{100} b_{i} = −12 .$ In the following exercises, compute the sums.

$\sum_{i = 1}^{100} (a_{i} - b_{i})$

Answer

15 - (−12) = 27

1-7 §1.1 source ↗

Suppose that $\sum_{i = 1}^{100} a_{i} = 15$ and $\sum_{i = 1}^{100} b_{i} = −12 .$ In the following exercises, compute the sums.

$\sum_{i = 1}^{100} (5 a_{i} + 4 b_{i})$

Answer

5 (15) +

4 (−12) =

27

1-9 §1.1 source ↗

In the following exercises, use summation properties and formulas to rewrite and evaluate the sums.

$\sum_{j = 1}^{50} (j^{2} - 2 j)$

Answer

\sum_{j = 1}^{50} j^{2} -

2 \sum_{j = 1}^{50} j =

\frac{(50) (51) (101)}{6} -

\frac{2 (50) (51)}{2} =

40,

​ 375

1-11 §1.1 source ↗

In the following exercises, use summation properties and formulas to rewrite and evaluate the sums.

$\sum_{k = 1}^{25} [{(2 k)}^{2} - 100 k]$

Answer

4 \sum_{k = 1}^{25} k^{2} -

100 \sum_{k = 1}^{25} k =

\frac{4 (25) (26) (51)}{6} -

50 (25) (26) =

−10,

​ 400

1-13 §1.1 source ↗

Let $L_{n}$ denote the left-endpoint sum using n subintervals and let $R_{n}$ denote the corresponding right-endpoint sum. In the following exercises, compute the indicated left and right sums for the given functions on the indicated interval.

R₄ for $g (x) =$ $cos (π x)$ on $[0, 1]$

Answer

R_{4} = -0.25

1-15 §1.1 source ↗

R₆ for $f (x) =$ $\frac{1}{x (x - 1)}$ on $[2, 5]$

Answer

R_{6} = 0.372

1-17 §1.1 source ↗

L₄ for $\frac{1}{x^{2} + 1}$ on $[−2, 2]$

Answer

L_{4} = 2.20

1-19 §1.1 source ↗

L₈ for $x^{2} - 2 x + 1$ on $[0, 2]$

Answer

L_{8} = 0.6875

1-21 §1.1 source ↗

Compute the left and right Riemann sums—L₆ and R₆, respectively—for $f (x) =$ $(3 - | 3 - x |)$ on $[0, 6] .$ Compute their average value and compare it with the area under the graph of f.

Answer

L_{6} =

9.000 =

R_{6} .

The graph of f is a triangle with area 9.

1-23 §1.1 source ↗

Compute the left and right Riemann sums—L₆ and R₆, respectively—for $f (x) =$ $\sqrt{9 - {(x - 3)}^{2}}$ on $[0, 6]$ and compare their values.

Answer

L_{6} =

13.12899 =

R_{6} .

They are equal.

1-25 §1.1 source ↗

Express the following endpoint sums in sigma notation but do not evaluate them.

L₁₀ for $f (x) = \sqrt{4 - x^{2}}$ on $[−2, 2]$

Answer

L_{10} =

\frac{4}{10} \sum_{i = 1}^{10} \sqrt{4 - (−2 + 4 \frac{(i - 1)}{10})^{2}}

1-27 §1.1 source ↗

Express the following endpoint sums in sigma notation but do not evaluate them.

R₁₀₀ for $ln x$ on $[1, e]$

Answer

R_{100} =

\frac{e - 1}{100} \sum_{i = 1}^{100} ln (1 + (e - 1) \frac{i}{100})

1-29 §1.1 source ↗

In the following exercises, graph the function then use a calculator or a computer program to evaluate the following left and right endpoint sums. If the two agree, say "neither."

[T] L₁₀₀ and R₁₀₀ for $y = x^{2}$ on the interval $[0, 1]$

Answer

A graph of the given function on the interval [0, 1]. It is set up for a left endpoint approximation and is an underestimate because the function is increasing. Ten rectangles are shown for visual clarity, but this behavior persists for more rectangles.

$R_{100} =$ $0.33835,$ $L_{100} =$ $0.32835 .$ The plot shows that the left Riemann sum is an underestimate because the function is increasing. Similarly, the right Riemann sum is an overestimate. The area lies between the left and right Riemann sums. Ten rectangles are shown for visual clarity. This behavior persists for more rectangles.

1-31 §1.1 source ↗

In the following exercises, graph the function then use a calculator or a computer program to evaluate the following left and right endpoint sums. If the two agree, say "neither."

[T] L₁₀₀ and R₁₀₀ for $y = x^{3} + 2$ on the interval $[−1, 1]$

Answer

A graph of the given function over [-1,1] set up for a left endpoint approximation. It is an underestimate since the function is increasing. Ten rectangles are shown for visual clarity, but this behavior persists for more rectangles.

$L_{100} =$ $3.980,$ $R_{100} =$ $4.020,$ $R_{100} >$ $L_{100} .$ The left endpoint sum is an underestimate because the function is increasing. Similarly, a right endpoint approximation is an overestimate. The area lies between the left and right endpoint estimates.

1-33 §1.1 source ↗

In the following exercises, graph the function then use a calculator or a computer program to evaluate the following left and right endpoint sums. If the two agree, say "neither."

[T] L₁₀₀ and R₁₀₀ for $y = e^{2 x}$ on the interval $[−1, 1]$

Answer

A graph of the given function over the interval -1 to 1 set up for a left endpoint approximation. It is an underestimate since the function is increasing. Ten rectangles are shown for isual clarity, but this behavior persists for more rectangles.

$L_{100} =$ $3.555,$ $R_{100} =$ $3.670$ The plot shows that the left Riemann sum is an underestimate because the function is increasing. Ten rectangles are shown for visual clarity. This behavior persists for more rectangles.

1-35 §1.1 source ↗

In the following exercises, graph the function then use a calculator or a computer program to evaluate the following left and right endpoint sums. If the two agree, say "neither."

Let $r_{j}$ denote the total rainfall in Portland on the jth day of the year in 2009. Interpret $\sum_{j = 1}^{31} r_{j} .$

Answer

The sum represents the cumulative rainfall in January 2009.

1-37 §1.1 source ↗

In the following exercises, graph the function then use a calculator or a computer program to evaluate the following left and right endpoint sums. If the two agree, say "neither."

To help get in shape, Joe gets a new pair of running shoes. If Joe runs 1 mi each day in week 1 and adds $\frac{1}{10}$ mi to his daily routine each week, what is the total mileage on Joe’s shoes after 25 weeks?

Answer

The total mileage is

7 \times \sum_{i = 1}^{25} (1 + \frac{(i - 1)}{10}) =

7 \times 25 +

\frac{7}{10} \times 12 \times 25 =

385 mi .

1-39 §1.1 source ↗

In the following exercises, graph the function then use a calculator or a computer program to evaluate the following left and right endpoint sums. If the two agree, say "neither."

The following table gives the approximate increase in sea level in inches over 20 years starting in the given year. Estimate the net change in mean sea level from 1870 to 2010.

Table 1.3 Approximate 20-Year Sea Level Increases, 1870–1990 in the book ↗

Starting Year	20-Year Change
1870	0.3
1890	1.5
1910	0.2
1930	2.8
1950	0.7
1970	1.1
1990	1.5

Source: http://link.springer.com/article/10.1007%2Fs10712-011-9119-1

Answer

Add the numbers to get 8.1-in. net increase.

1-41 §1.1 source ↗

In the following exercises, graph the function then use a calculator or a computer program to evaluate the following left and right endpoint sums. If the two agree, say "neither."

The following table gives the percent growth of the U.S. population beginning in July of the year indicated. If the U.S. population was 281,421,906 in July 2000, estimate the U.S. population in July 2010.

(Hint: To obtain the population in July 2001, multiply the population in July 2000 by 1.0112 to get 284,573,831.)

Table 1.5 Annual Percentage Growth of U.S. Population, 2000–2009 in the book ↗

Year	% Change/Year
2000	1.12
2001	0.99
2002	0.93
2003	0.86
2004	0.93
2005	0.93
2006	0.97
2007	0.96
2008	0.95
2009	0.88

Source: http://www.census.gov/popest/data.

Answer

309,389,957

1-43 §1.1 source ↗

In the following exercises, estimate the areas under the curves by computing the left Riemann sums, L₈.

The graph of a smooth curve going through the points (0,3), (1,2), (2,1), (3,2), (4,3), (5,4), (6,5), (7,4), and (8,3).

Answer

L_{8} =

3 +

2 +

1 +

2 +

3 +

4 +

5 +

4 =

24

1-45 §1.1 source ↗

In the following exercises, estimate the areas under the curves by computing the left Riemann sums, L₈.

The graph of a smooth curve going through the points (0, 3), (1, 5), (2, 7), (3, 6), (4, 8), (5, 6), (6, 5), (7, 4), and (8, 6).

Answer

L_{8} =

3 +

5 +

7 +

6 +

8 +

6 +

5 +

4 =

44

1-47 §1.1 source ↗

In the following exercises, estimate the areas under the curves by computing the left Riemann sums, L₈.

[T] Use a computer algebra system to compute the Riemann sum, L_N, for $N = 10, 30, 50$ for $f (x) = \frac{1}{\sqrt{1 + x^{2}}}$ on $[−1, 1] .$

Answer

L_{10} \approx

1.7604,

L_{30} \approx

1.7625,

L_{50} \approx

1.76265

1-49 §1.1 source ↗

In the following exercises, use a calculator or a computer program to evaluate the endpoint sums R_N and L_N for $N = 1,10,100 .$ How do these estimates compare with the exact answers, which you can find via geometry?

[T] $y = cos (π x)$ on the interval $[0, 1]$

Answer

R_{1} =

−1,

L_{1} =

1,

R_{10} =

−0.1,

L_{10} =

0.1,

L_{100} =

0.01,

and

R_{100} = −0.01 .

By symmetry of the graph, the exact area is zero.

1-51 §1.1 source ↗

In the following exercises, use a calculator or a computer program to evaluate the endpoint sums R_N and L_N for $N = 1,10,100 .$

[T] $y = x^{4} - 5 x^{2} + 4$ on the interval $[−2, 2],$ which has an exact area of $\frac{32}{15}$

Answer

R_{1} =

0,

L_{1} =

0,

R_{10} =

2.4499,

L_{10} =

2.4499,

R_{100} =

2.1365,

L_{100} =

2.1365

1-53 §1.1 source ↗

In the following exercises, use a calculator or a computer program to evaluate the endpoint sums R_N and L_N for $N = 1,10,100 .$

Explain why, if $f (a) \geq 0$ and f is increasing on $[a, b],$ that the left endpoint estimate is a lower bound for the area below the graph of f on $[a, b] .$

Answer

[c, d]

is a subinterval of

[a, b]

under one of the left-endpoint sum rectangles, then the area of the rectangle contributing to the left-endpoint estimate is

f (c) (d - c) .

But,

f (c) \leq f (x)

for

c \leq x \leq d,

so the area under the graph of f between c and d is

f (c) (d - c)

plus the area below the graph of f but above the horizontal line segment at height

f (c),

which is positive. As this is true for each left-endpoint sum interval, it follows that the left Riemann sum is less than or equal to the area below the graph of f on

[a, b] .

1-55 §1.1 source ↗

In the following exercises, use a calculator or a computer program to evaluate the endpoint sums R_N and L_N for $N = 1,10,100 .$

Show that, in general, $R_{N} -$ $L_{N} =$ $(b - a) \times \frac{f (b) - f (a)}{N} .$

Answer

L_{N} =

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

\frac{b - a}{N} \sum_{i = 0}^{N - 1} f (a + (b - a) \frac{i}{N})

and

R_{N} =

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

The left sum has a term corresponding to

i = 0

and the right sum has a term corresponding to

i = N .

R_{N} - L_{N},

any term corresponding to

i = 1, 2,…, N - 1

occurs once with a plus sign and once with a minus sign, so each such term cancels and one is left with

R_{N} -

L_{N} =

\frac{b - a}{N} (f (a + (b - a)) \frac{N}{N}) -

(f (a) + (b - a) \frac{0}{N}) =

\frac{b - a}{N} (f (b) - f (a)) .

1-57 §1.1 source ↗

In the following exercises, use a calculator or a computer program to evaluate the endpoint sums R_N and L_N for $N = 1,10,100 .$

For each of the three graphs:

a. Obtain a lower bound $L (A)$ for the area enclosed by the curve by adding the areas of the squares enclosed completely by the curve.

b. Obtain an upper bound $U (A)$ for the area by adding to $L (A)$ the areas $B (A)$ of the squares enclosed partially by the curve.

Answer

Graph 1: a.

L (A) =

0,

B (A) =

20;

U (A) = 20 .

Graph 2: a.

L (A) = 9;

B (A) =

11,

U (A) =

20 .

Graph 3: a.

L (A) = 11.0;

B (A) =

4.5,

U (A) =

15.5 .

1-59 §1.1 source ↗

In the following exercises, use a calculator or a computer program to evaluate the endpoint sums R_N and L_N for $N = 1,10,100 .$

A unit circle is made up of n wedges equivalent to the inner wedge in the figure. The base of the inner triangle is 1 unit and its height is $sin (\frac{2 π}{n}) .$ The base of the outer triangle is $B =$ $cos (\frac{π}{n}) +$ $sin (\frac{π}{n}) tan (\frac{π}{n})$ and the height is $H =$ $B sin (\frac{2 π}{n}) .$ Use this information to argue that the area of a unit circle is equal to π.

A wedge of a circle cut at an acute angle theta = 2pi / n. Several extra lines are drawn. The first is a line A connecting the ends of the two radii, creating a triangle. The second is another line B parallel to the A, connecting the radii a few units in from each endpoint. A concentric curve C connects the endpoints of B and is tangent to A near its midpoint. The area between this curve C and the edge of the circle is shaded in pink, and the rest of the wedge is purple. A final concentric curve is drawn very close to angle theta.

Answer

Let A be the area of the unit circle. The circle encloses n congruent triangles each of area

\frac{sin (\frac{2 π}{n})}{2},

\frac{n}{2} sin (\frac{2 π}{n}) \leq A .

Similarly, the circle is contained inside n congruent triangles each of area

\frac{B H}{2} =

\frac{1}{2} {\cos \frac{π}{n} + \sin \frac{π}{n} \tan \frac{π}{n}}^{2} \sin \frac{2 π}{n},

A \leq \frac{n}{2} sin (\frac{2 π}{n}) (cos (\frac{π}{n}) + sin (\frac{π}{n}) tan {(\frac{π}{n}))}^{2} .

n \to

\infty,

\frac{n}{2} sin (\frac{2 π}{n}) =

\frac{sin (\frac{2 π}{n})}{(\frac{2}{n})} Α \to

π,

so we conclude

π \leq A .

Also, as

n \to

\infty,

cos (\frac{π}{n}) +

sin (\frac{π}{n}) tan (\frac{π}{n}) \to

1,

so we also have

A \leq π .

By the squeeze theorem for limits, we conclude that

A = π .

Checkpoint 1.7 §1.2 source ↗

Use the definition of the definite integral to evaluate

\int_{0}^{3} (2 x - 1) d x .

Use a right-endpoint approximation to generate the Riemann sum.

Answer

Checkpoint 1.8 §1.2 source ↗

Use the formula for the area of a trapezoid to evaluate

\int_{2}^{4} (2 x + 3) d x .

Answer

18 square units

Checkpoint 1.9 §1.2 source ↗

Find the net signed area of $f (x) = x - 2$ over the interval $[0, 6],$ illustrated in the following image.

A graph of an increasing line going through (-2,-4), (0,-2), (2,0), (4,2) and (6,4). The area above the curve in quadrant four is shaded blue and labeled A2, and the area under the curve and to the left of x=6 in quadrant one is shaded and labeled A1.

Answer

Checkpoint 1.10 §1.2 source ↗

Find the total area between the function

f (x) = 2 x

and the x-axis over the interval

[−3, 3] .

Answer

Checkpoint 1.11 §1.2 source ↗

Use the properties of the definite integral to express the definite integral of

f (x) =

6 x^{3} -

4 x^{2} +

2 x -

3

over the interval

[1, 3]

as the sum of four definite integrals.

Answer

6 \int_{1}^{3} x^{3} d x -

4 \int_{1}^{3} x^{2} d x +

2 \int_{1}^{3} x d x -

\int_{1}^{3} 3 d x

Checkpoint 1.12 §1.2 source ↗

If it is known that

\int_{1}^{5} f (x) d x =

−3

and

\int_{2}^{5} f (x) d x =

4,

find the value of

\int_{1}^{2} f (x) d x .

Answer

−7

Checkpoint 1.13 §1.2 source ↗

Find the average value of

f (x) = 6 - 2 x

over the interval

[0, 3] .

Answer

1-61 §1.2 source ↗

In the following exercises, express the limits as integrals.

$lim_{n \to \infty} \sum_{i = 1}^{n} (5 {(x_{i}^{*})}^{2} - 3 {(x_{i}^{*})}^{3}) Δ x$ over $[0, 2]$

Answer

\int_{0}^{2} (5 x^{2} - 3 x^{3}) d x

1-63 §1.2 source ↗

In the following exercises, express the limits as integrals.

$lim_{n \to \infty} \sum_{i = 1}^{n} {cos}^{2} (2 π x_{i}^{*}) Δ x$ over $[0, 1]$

Answer

\int_{0}^{1} {cos}^{2} (2 π x) d x

1-65 §1.2 source ↗

In the following exercises, given L_n or R_n as indicated, express their limits as $n \to \infty$ as definite integrals, identifying the correct intervals.

$R_{n} =$ $\frac{1}{n} \sum_{i = 1}^{n} \frac{i}{n}$

Answer

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

1-67 §1.2 source ↗

In the following exercises, given L_n or R_n as indicated, express their limits as $n \to \infty$ as definite integrals, identifying the correct intervals.

$R_{n} =$ $\frac{3}{n} \sum_{i = 1}^{n} (3 + 3 \frac{i}{n})$

Answer

\int_{3}^{6} x d x

1-69 §1.2 source ↗

In the following exercises, given L_n or R_n as indicated, express their limits as $n \to \infty$ as definite integrals, identifying the correct intervals.

$R_{n} =$ $\frac{1}{n} \sum_{i = 1}^{n} (1 + \frac{i}{n}) log ({(1 + \frac{i}{n})}^{2})$

Answer

\int_{1}^{2} x log (x^{2}) d x

1-71 §1.2 source ↗

In the following exercises, evaluate the integrals of the functions graphed using the formulas for areas of triangles and circles, and subtracting the areas below the x-axis.

A graph of three isosceles triangles corresponding to the functions 1 - |x-1| over [0,2], 2 - |x-4| over [2,4], and 3 - |x-9| over [6,12]. The first triangle has endpoints at (0,0), (2,0), and (1,1). The second triangle has endpoints at (2,0), (6,0), and (4,2). The last has endpoints at (6,0), (12,0), and (9,3). All three are shaded.

Answer

1 +

2 \cdot 2 +

3 \cdot 3 =

14

1-73 §1.2 source ↗

In the following exercises, evaluate the integrals of the functions graphed using the formulas for areas of triangles and circles, and subtracting the areas below the x-axis.

A graph of three shaded triangles. The first has endpoints at (0, 0), (2, 0), and (1, 1) and corresponds to the function 1 - |x-1| over [0, 2]. The second has endpoints at (2, 0), (6, 0), and (4, -2) and corresponds to the function |x-4| - 2 over [2, 6]. The third has endpoints at (6, 0), (12, 0), and (9, 3) and corresponds to the function 3 - |x-9| over [6, 12].

Answer

1 - 4 + 9 = 6

1-75 §1.2 source ↗

In the following exercises, evaluate the integrals of the functions graphed using the formulas for areas of triangles and circles, and subtracting the areas below the x-axis.

A graph with three shaded parts. The first is a triangle with endpoints at (0, 0), (2, 0), and (1, 1), which corresponds to the function 1 - |x-1| over [0, 2] in quadrant 1. The second is the lower half of a circle with center at (4, 0) and radius two, which corresponds to the function –sqrt(-12 + 8x – x^2) over [2, 6]. The last is a triangle with endpoints at (6, 0), (12, 0), and (9, 3), which corresponds to the function 3 - |x-9| over [6, 12].

Answer

1 -

2 π +

9 =

10 -

2 π

1-77 §1.2 source ↗

In the following exercises, evaluate the integral using area formulas.

$\int_{2}^{3} (3 - x) d x$

Answer

The integral is the area of the triangle,

\frac{1}{2}

1-79 §1.2 source ↗

In the following exercises, evaluate the integral using area formulas.

$\int_{0}^{6} (3 - | x - 3 |) d x$

Answer

The integral is the area of the triangle, 9.

1-81 §1.2 source ↗

In the following exercises, evaluate the integral using area formulas.

$\int_{1}^{5} \sqrt{4 - {(x - 3)}^{2}} d x$

Answer

The integral is the area

\frac{1}{2} π r^{2} = 2 π .

1-83 §1.2 source ↗

In the following exercises, evaluate the integral using area formulas.

$\int_{−2}^{3} (3 - | x |) d x$

Answer

The integral is the area of the “big” triangle less the “missing” triangle,

9 - \frac{1}{2} .

1-85 §1.2 source ↗

In the following exercises, use averages of values at the left (L) and right (R) endpoints to compute the integrals of the piecewise linear functions with graphs that pass through the given list of points over the indicated intervals.

${(0, 2), (1, 0), (3, 5), (5, 5), (6, 2), (8, 0)}$ over $[0, 8]$

Answer

L =

2 +

0 +

10 +

5 +

4 =

21,

R =

0 +

10 +

10 +

2 +

0 =

22,

\frac{L + R}{2} =

21.5

1-87 §1.2 source ↗

${(−4, 0), (−2, 2), (0, 0), (1, 2), (3, 2), (4, 0)}$ over $[−4, 4]$

Answer

L =

0 +

4 +

0 +

4 +

2 =

10,

R =

4 +

0 +

2 +

4 +

0 =

10,

\frac{L + R}{2} =

10

1-89 §1.2 source ↗

Suppose that $\int_{0}^{4} f (x) d x = 5$ and $\int_{0}^{2} f (x) d x =$ $−3,$ and $\int_{0}^{4} g (x) d x =$ $−1$ and $\int_{0}^{2} g (x) d x =$ $2 .$ In the following exercises, compute the integrals.

$\int_{2}^{4} (f (x) + g (x)) d x$

Answer

\int_{2}^{4} f (x) d x +

\int_{2}^{4} g (x) d x =

8 -

3 =

5

1-91 §1.2 source ↗

$\int_{2}^{4} (f (x) - g (x)) d x$

Answer

\int_{2}^{4} f (x) d x -

\int_{2}^{4} g (x) d x =

8 +

3 =

11

1-93 §1.2 source ↗

$\int_{2}^{4} (4 f (x) - 3 g (x)) d x$

Answer

4 \int_{2}^{4} f (x) d x -

3 \int_{2}^{4} g (x) d x =

32 +

9 =

41

1-95 §1.2 source ↗

In the following exercises, use the identity $\int_{− A}^{A} f (x) d x =$ $\int_{− A}^{0} f (x) d x +$ $\int_{0}^{A} f (x) d x$ to compute the integrals.

$\int_{− \sqrt{π}}^{\sqrt{π}} \frac{t}{1 + cos t} d t$

Answer

The integrand is odd; the integral is zero.

1-97 §1.2 source ↗

In the following exercises, find the net signed area between $f (x)$ and the x-axis.

$\int_{2}^{4} {(x - 3)}^{3} d x$ (Hint: Look at the graph of f.)

Answer

The integrand is antisymmetric with respect to

x = 3 .

The integral is zero.

1-99 §1.2 source ↗

In the following exercises, given that $\int_{0}^{1} x d x =$ $\frac{1}{2},$ $\int_{0}^{1} x^{2} d x =$ $\frac{1}{3},$ and $\int_{0}^{1} x^{3} d x = \frac{1}{4},$ compute the integrals.

$\int_{0}^{1} (1 - x + x^{2} - x^{3}) d x$

Answer

1 -

\frac{1}{2} +

\frac{1}{3} -

\frac{1}{4} =

\frac{7}{12}

1-101 §1.2 source ↗

In the following exercises, given that $\int_{0}^{1} x d x =$ $\frac{1}{2},$ $\int_{0}^{1} x^{2} d x =$ $\frac{1}{3},$ and $\int_{0}^{1} x^{3} d x = \frac{1}{4},$ compute the integrals.

$\int_{0}^{1} {(1 - 2 x)}^{3} d x$

Answer

\int_{0}^{1} (1 - 6 x + 12 x^{2} - 8 x^{3}) d x =

(x - 3 x^{2} + 4 x^{3} - 2 x^{4}) =

(1 - 3 + 4 - 2) (0 - 0 + 0 - 0) =

0

1-103 §1.2 source ↗

In the following exercises, given that $\int_{0}^{1} x d x =$ $\frac{1}{2},$ $\int_{0}^{1} x^{2} d x =$ $\frac{1}{3},$ and $\int_{0}^{1} x^{3} d x = \frac{1}{4},$ compute the integrals.

$\int_{0}^{1} (7 - 5 x^{3}) d x$

Answer

7 - \frac{5}{4} = \frac{23}{4}

1-105 §1.2 source ↗

In the following exercises, use the comparison theorem.

Show that $\int_{−2}^{3} (x - 3) (x + 2) d x \leq 0 .$

Answer

The integrand is negative over

[−2, 3] .

1-107 §1.2 source ↗

In the following exercises, use the comparison theorem.

Show that $\int_{1}^{2} \sqrt{1 + x} d x \leq \int_{1}^{2} \sqrt{1 + x^{2}} d x .$

Answer

x \leq x^{2}

over

[1, 2],

\sqrt{1 + x} \leq \sqrt{1 + x^{2}}

over

[1, 2] .

1-109 §1.2 source ↗

In the following exercises, use the comparison theorem.

Show that $\int_{− π / 4}^{π / 4} cos t d t \geq π \sqrt{2} / 4 .$

Answer

cos (t) \geq \frac{\sqrt{2}}{2} .

Multiply by the length of the interval to get the inequality.

1-111 §1.2 source ↗

In the following exercises, find the average value f_ave of f between a and b, and find a point c, where $f (c) = f_{ave} .$

$f (x) =$ $x^{5},$ $a =$ $−1,$ $b =$ $1$

Answer

f_{ave} = 0; c = 0

1-113 §1.2 source ↗

In the following exercises, find the average value f_ave of f between a and b, and find a point c, where $f (c) = f_{ave} .$

$f (x) =$ $(3 - | x |),$ $a =$ $−3,$ $b =$ $3$

Answer

\frac{3}{2}

when

c = \pm \frac{3}{2}

1-115 §1.2 source ↗

In the following exercises, find the average value f_ave of f between a and b, and find a point c, where $f (c) = f_{ave} .$

$f (x) =$ $cos x,$ $a =$ $0,$ $b =$ $2 π$

Answer

f_{ave} =

0;

c =

\frac{π}{2},

\frac{3 π}{2}

1-117 §1.2 source ↗

In the following exercises, approximate the average value using Riemann sums L₁₀₀ and R₁₀₀. How does your answer compare with the exact given answer?

[T] $y = e^{x / 2}$ over the interval $[0, 1];$ the exact solution is $2 (\sqrt{e} - 1) .$

Answer

L_{100} =

1.294,

R_{100} =

1.301;

the exact average is between these values.

1-119 §1.2 source ↗

In the following exercises, approximate the average value using Riemann sums L₁₀₀ and R₁₀₀. How does your answer compare with the exact given answer?

[T] $y = \frac{x + 1}{\sqrt{4 - x^{2}}}$ over the interval $[−1, 1];$ the exact solution is $\frac{π}{6} .$

Answer

L_{100} \times (\frac{1}{2}) =

0.5178,

R_{100} \times (\frac{1}{2}) =

0.5294

1-121 §1.2 source ↗

In the following exercises, compute the average value using the left Riemann sums L_N for $N = 1, 10, 100 .$ How does the accuracy compare with the given exact value?

[T] $y = x e^{x^{2}}$ over the interval $[0, 2];$ the exact solution is $\frac{1}{4} (e^{4} - 1) .$

Answer

L_{1} =

0,

L_{10} \times (\frac{1}{2}) =

8.743493,

L_{100} \times (\frac{1}{2}) =

12.861728 .

The exact answer

\approx 26.799,

so L₁₀₀ is not accurate.

1-123 §1.2 source ↗

In the following exercises, compute the average value using the left Riemann sums L_N for $N = 1, 10, 100 .$ How does the accuracy compare with the given exact value?

[T] $y = x sin (x^{2})$ over the interval $[− π, 0];$ the exact solution is $\frac{cos (π^{2}) - 1}{2 π} .$

Answer

L_{1} \times (\frac{1}{π}) =

1.352,

L_{10} \times (\frac{1}{π}) =

−0.1837,

L_{100} \times (\frac{1}{π}) =

−0.2956 .

The exact answer

\approx - 0.303,

so L₁₀₀ is not accurate to first decimal.

1-125 §1.2 source ↗

In the following exercises, compute the average value using the left Riemann sums L_N for $N = 1, 10, 100 .$ How does the accuracy compare with the given exact value?

Suppose that $A =$ $\int_{− π / 4}^{π / 4} {sec}^{2} t d t =$ $π$ and $B =$ $\int_{− π / 4}^{π / 4} {tan}^{2} t d t .$ Show that $A - B = \frac{π}{2} .$

Answer

Use

{tan}^{2} θ +

1 =

{sec}^{2} θ .

Then,

B -

A =

\int_{− π / 4}^{π / 4} 1 d x =

\frac{π}{2} .

1-127 §1.2 source ↗

In the following exercises, compute the average value using the left Riemann sums L_N for $N = 1, 10, 100 .$ How does the accuracy compare with the given exact value?

Show that the average value of ${cos}^{2} t$ over $[0, 2 π]$ is equal to $1 / 2 .$ Without further calculation, determine whether the average value of ${cos}^{2} (t)$ over $[0, π]$ is also equal to $1 / 2 .$

Answer

\int_{0}^{2 π} {cos}^{2} t d t =

π,

so divide by the length 2π of the interval.

{cos}^{2} t

has period π, so yes, it is true.

1-129 §1.2 source ↗

In the following exercises, compute the average value using the left Riemann sums L_N for $N = 1, 10, 100 .$ How does the accuracy compare with the given exact value?

Suppose that parabola $p (x) =$ $a x^{2} +$ $b x +$ $c$ opens downward $(a < 0)$ and has a vertex of $y =$ $\frac{− b}{2 a} >$ $0 .$ For which interval $[A, B]$ is $\int_{A}^{B} (a x^{2} + b x + c) d x$ as large as possible?

Answer

The integral is maximized when one uses the largest interval on which p is nonnegative. Thus,

A =

\frac{− b - \sqrt{b^{2} - 4 a c}}{2 a}

and

B =

\frac{− b + \sqrt{b^{2} - 4 a c}}{2 a} .

1-131 §1.2 source ↗

In the following exercises, compute the average value using the left Riemann sums L_N for $N = 1, 10, 100 .$ How does the accuracy compare with the given exact value?

Suppose f and g are continuous functions such that $\int_{c}^{d} f (t) d t \leq \int_{c}^{d} g (t) d t$ for every subinterval $[c, d]$ of $[a, b] .$ Explain why $f (x) \leq g (x)$ for all values of x.

Answer

f (t_{0}) >

g (t_{0})

for some

t_{0} \in [a, b],

then since

f - g

is continuous, there is an interval containing t₀ such that

f (t) >

g (t)

over the interval

[c, d],

and then

\int_{d}^{d} f (t) d t >

\int_{c}^{d} g (t) d t

over this interval.

1-133 §1.2 source ↗

In the following exercises, compute the average value using the left Riemann sums L_N for $N = 1, 10, 100 .$ How does the accuracy compare with the given exact value?

Suppose that $[a, b]$ can be partitioned. taking $a =$ $a_{0} <$ $a_{1} <$ $⋯ <$ $a_{N} =$ $b$ such that the average value of f over each subinterval $[a_{i - 1}, a_{i}] = 1$ is equal to 1 for each $i = 1,…, N .$ Explain why the average value of f over $[a, b]$ is also equal to 1.

Answer

The integral of f over an interval is the same as the integral of the average of f over that interval. Thus,

\begin{matrix} \int_{a}^{b} f (t) d t = \int_{a_{0}}^{a_{1}} f (t) d t + \int_{a_{1}}^{a_{2}} f (t) d t + ⋯ + \int_{a_{N + 1}}^{a_{N}} f (t) d t = \int_{a_{0}}^{a_{1}} 1 d t + \int_{a_{1}}^{a_{2}} 1 d t + ⋯ + \int_{a_{N + 1}}^{a_{N}} 1 d t \\ = (a_{1} - a_{0}) + (a_{2} - a_{1}) + ⋯ + (a_{N} - a_{N - 1}) = a_{N} - a_{0} = b - a . \end{matrix}

Dividing through by

b - a

gives the desired identity.

1-135 §1.2 source ↗

In the following exercises, compute the average value using the left Riemann sums L_N for $N = 1, 10, 100 .$ How does the accuracy compare with the given exact value?

Suppose that for each i such that $1 \leq i \leq N$ one has $\int_{i - 1}^{i} f (t) d t =$ $i^{2} .$ Show that $\int_{0}^{N} f (t) d t =$ $\frac{N (N + 1) (2 N + 1)}{6} .$

Answer

\int_{0}^{N} f (t) d t =

\sum_{i = 1}^{N} \int_{i - 1}^{i} f (t) d t =

\sum_{i = 1}^{N} i^{2} =

\frac{N (N + 1) (2 N + 1)}{6}

1-137 §1.2 source ↗

In the following exercises, compute the average value using the left Riemann sums L_N for $N = 1, 10, 100 .$ How does the accuracy compare with the given exact value?

[T] Compute the left and right Riemann sums, L₁₀ and R₁₀, and their average $\frac{L_{10} + R_{10}}{2}$ for $f (t) = (4 - t^{2})$ over $[1, 2] .$ Given that $\int_{1}^{2} (4 - t^{2}) d t =$ $1. \bar{66},$ to how many decimal places is $\frac{L_{10} + R_{10}}{2}$ accurate?

Answer

L_{10} =

1.815,

R_{10} =

1.515,

\frac{L_{10} + R_{10}}{2} =

1.665,

so the estimate is accurate to two decimal places.

1-139 §1.2 source ↗

In the following exercises, compute the average value using the left Riemann sums L_N for $N = 1, 10, 100 .$ How does the accuracy compare with the given exact value?

Estimate $\int_{0}^{1} t d t$ using the left and right endpoint sums, each with a single rectangle. How does the average of these left and right endpoint sums compare with the actual value $\int_{0}^{1} t d t ?$

Answer

The average is

1 / 2,

which is equal to the integral in this case.

1-141 §1.2 source ↗

In the following exercises, compute the average value using the left Riemann sums L_N for $N = 1, 10, 100 .$ How does the accuracy compare with the given exact value?

From the graph of $sin (2 π x)$ shown:

a. Explain why $\int_{0}^{1} sin (2 π t) d t =$ $0 .$

b. Explain why, in general, $\int_{a}^{a + 1} sin (2 π t) d t =$ $0$ for any value of a.
A graph of the function f(x) = sin(2pi*x) over [0, 2]. The function is shaded over [.7, 1] above the curve and below to x axis, over [1,1.5] under the curve and above the x axis, and over [1.5, 1.7] above the curve and under the x axis. The graph is antisymmetric with respect o t = ½ over [0,1].

Answer

a. The graph is antisymmetric with respect to

t = \frac{1}{2}

over

[0, 1],

so the average value is zero. b. For any value of a, the graph between

[a, a + 1]

is a shift of the graph over

[0, 1],

so the net areas above and below the axis do not change and the average remains zero.

1-143 §1.2 source ↗

In the following exercises, compute the average value using the left Riemann sums L_N for $N = 1, 10, 100 .$ How does the accuracy compare with the given exact value?

If f is 1-periodic and $\int_{0}^{1} f (t) d t =$ $A,$ is it necessarily true that $\int_{a}^{1 + a} f (t) d t =$ $A$ for all A?

Answer

Yes, the integral over any interval of length 1 is the same.

Checkpoint 1.14 §1.3 source ↗

Find the average value of the function

f (x) = \frac{x}{2}

over the interval

[0, 6]

and find c such that

f (c)

equals the average value of the function over

[0, 6] .

Answer

Average value =

1.5;

c =

3

Checkpoint 1.15 §1.3 source ↗

Given

\int_{0}^{3} (2 x^{2} - 1) d x =

15,

find c such that

f (c)

equals the average value of

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

over

[0, 3] .

Answer

c = \sqrt{3}

Checkpoint 1.16 §1.3 source ↗

Use the Fundamental Theorem of Calculus, Part 1 to find the derivative of

g (r) =

\int_{0}^{r} \sqrt{x^{2} + 4} d x .

Answer

g^{'} (r) = \sqrt{r^{2} + 4}

Checkpoint 1.17 §1.3 source ↗

Let

F (x) =

\int_{1}^{x^{3}} cos t d t .

Find

F^{'} (x) .

Answer

F^{'} (x) =

3 x^{2} cos x^{3}

Checkpoint 1.18 §1.3 source ↗

Let

F (x) =

\int_{x}^{x^{2}} cos t d t .

Find

F^{'} (x) .

Answer

F^{'} (x) =

2 x cos x^{2} -

cos x

Checkpoint 1.19 §1.3 source ↗

Use The Fundamental Theorem of Calculus, Part 2 to evaluate

\int_{1}^{2} x^{−4} d x .

Answer

\frac{7}{24}

Checkpoint 1.20 §1.3 source ↗

Suppose James and Kathy have a rematch, but this time the official stops the contest after only 3 sec. Does this change the outcome?

Answer

Kathy still wins, but by a much larger margin: James skates 24 ft in 3 sec, but Kathy skates 29.3634 ft in 3 sec.

1-145 §1.3 source ↗

Two mountain climbers start their climb at base camp, taking two different routes, one steeper than the other, and arrive at the peak at exactly the same time. Is it necessarily true that, at some point, both climbers increased in altitude at the same rate?

Answer

Yes. It is implied by the Mean Value Theorem for Integrals.

1-147 §1.3 source ↗

Set

F (x) =

\int_{1}^{x} (1 - t) d t .

Find

F^{'} (2)

and the average value of

F^{′}

over

[1, 2] .

Answer

F^{'} (2) = −1;

average value of

F^{′}

over

[1, 2]

−1 / 2 .

1-149 §1.3 source ↗

In the following exercises, use the Fundamental Theorem of Calculus, Part 1, to find each derivative.

$\frac{d}{d x} \int_{1}^{x} e^{cos t} d t$

Answer

e^{cos x}

1-151 §1.3 source ↗

In the following exercises, use the Fundamental Theorem of Calculus, Part 1, to find each derivative.

$\frac{d}{d x} \int_{3}^{x} \frac{d s}{\sqrt{16 - s^{2}}}$

Answer

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

1-153 §1.3 source ↗

In the following exercises, use the Fundamental Theorem of Calculus, Part 1, to find each derivative.

$\frac{d}{d x} \int_{0}^{\sqrt{x}} t d t$

Answer

\sqrt{x} \frac{d}{d x} \sqrt{x} = \frac{1}{2}

1-155 §1.3 source ↗

In the following exercises, use the Fundamental Theorem of Calculus, Part 1, to find each derivative.

$\frac{d}{d x} \int_{cos x}^{1} \sqrt{1 - t^{2}} d t$

Answer

− \sqrt{1 - {cos}^{2} x} \frac{d}{d x} cos x =

| sin x | sin x

1-157 §1.3 source ↗

In the following exercises, use the Fundamental Theorem of Calculus, Part 1, to find each derivative.

$\frac{d}{d x} \int_{1}^{x^{2}} \frac{\sqrt{t}}{1 + t} d t$

Answer

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

1-159 §1.3 source ↗

In the following exercises, use the Fundamental Theorem of Calculus, Part 1, to find each derivative.

$\frac{d}{d x} \int_{1}^{e^{x}} ln u^{2} d u$

Answer

ln (e^{2 x}) \frac{d}{d x} e^{x} =

2 x e^{x}

1-161 §1.3 source ↗

In the following exercises, use the Fundamental Theorem of Calculus, Part 1, to find each derivative.

The graph of $y =$ $\int_{0}^{x} f (t) d t,$ where f is a piecewise constant function, is shown here.

A graph of a function with linear segments that goes through the points (0, 0), (1, -1), (2, 1), (3, 1), (4, -2), (5, -2), and (6, 0). The area over the function but under the x axis over the interval [0, 1.5] and [3.25, 6] is shaded. The area under the function but over the x axis over the interval [1.5, 3.25] is shaded.

a. Over which intervals is f positive? Over which intervals is it negative? Over which intervals, if any, is it equal to zero?

b. What are the maximum and minimum values of f?

c. What is the average value of f?

Answer

a. f is positive over

(1, 2)

and

(5, 6),

negative over

(0, 1)

and

(3, 4),

and zero over

(2, 3)

and

(4, 5) .

b. The maximum value is 2 and the minimum is −3. c. The average value is 0.

1-163 §1.3 source ↗

In the following exercises, use the Fundamental Theorem of Calculus, Part 1, to find each derivative.

The graph of $y =$ $\int_{0}^{x} ℓ (t) d t,$ where ℓ is a piecewise linear function, is shown here.

A graph of a function that goes through the points (0, 0), (1, 1), (2, 0), (3, -1), (4.5, 0), (5, 1), and (6, 2). The area under the function and over the x axis over the intervals [0, 2] and [4.5, 6] is shaded. The area over the function and under the x axis over the interval [2, 2.5] is shaded.

a. Over which intervals is ℓ positive? Over which intervals is it negative? Over which, if any, is it zero?

b. Over which intervals is ℓ increasing? Over which is it decreasing? Over which intervals, if any, is it constant?

c. What is the average value of ℓ?

Answer

a. ℓ is positive over

(0, 1)

and

(3, 6),

and negative over

(1, 3) .

b. It is increasing over

(0, 1)

and

(3, 5),

and it is constant over

(1, 3)

and

(5, 6) .

c. Its average value is

\frac{1}{3} .

1-165 §1.3 source ↗

In the following exercises, use a calculator to estimate the area under the curve by computing T₁₀, the average of the left- and right-endpoint Riemann sums using $N = 10$ rectangles. Then, using the Fundamental Theorem of Calculus, Part 2, determine the exact area.

[T] $y =$ $x^{3} +$ $6 x^{2} +$ $x -$ $5$ over $[−4, 2]$

Answer

T_{10} =

49.08,

\int_{−4}^{3} (x^{3} + 6 x^{2} + x - 5) d x = 48

1-167 §1.3 source ↗

[T] $y = \sqrt{x} + x^{2}$ over $[1, 9]$

Answer

T_{10} =

260.836,

\int_{1}^{9} (\sqrt{x} + x^{2}) d x = 260

1-169 §1.3 source ↗

[T] $\int \frac{4}{x^{2}} d x$ over $[1, 4]$

Answer

T_{10} =

3.058,

\int_{1}^{4} \frac{4}{x^{2}} d x = 3

1-171 §1.3 source ↗

In the following exercises, evaluate each definite integral using the Fundamental Theorem of Calculus, Part 2.

$\int_{−2}^{3} (x^{2} + 3 x - 5) d x$

Answer

F (x) =

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

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

5 x,

F (3) -

F (−2) =

-

\frac{35}{6}

1-173 §1.3 source ↗

In the following exercises, evaluate each definite integral using the Fundamental Theorem of Calculus, Part 2.

$\int_{2}^{3} (t^{2} - 9) (4 - t^{2}) d t$

Answer

F (x) =

-

\frac{t^{5}}{5} +

\frac{13 t^{3}}{3} -

36 t,

F (3) -

F (2) =

\frac{62}{15}

1-175 §1.3 source ↗

In the following exercises, evaluate each definite integral using the Fundamental Theorem of Calculus, Part 2.

$\int_{0}^{1} x^{99} d x$

Answer

F (x) =

\frac{x^{100}}{100},

F (1) -

F (0) =

\frac{1}{100}

1-177 §1.3 source ↗

In the following exercises, evaluate each definite integral using the Fundamental Theorem of Calculus, Part 2.

$\int_{1 / 4}^{4} (x^{2} - \frac{1}{x^{2}}) d x$

Answer

F (x) =

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

\frac{1}{x},

F (4) -

F (\frac{1}{4}) =

\frac{1125}{64}

1-179 §1.3 source ↗

In the following exercises, evaluate each definite integral using the Fundamental Theorem of Calculus, Part 2.

$\int_{1}^{4} \frac{1}{2 \sqrt{x}} d x$

Answer

F (x) =

\sqrt{x},

F (4) -

F (1) =

1

1-181 §1.3 source ↗

In the following exercises, evaluate each definite integral using the Fundamental Theorem of Calculus, Part 2.

$\int_{1}^{16} \frac{d t}{t^{1 / 4}}$

Answer

F (x) =

\frac{4}{3} t^{3 / 4},

F (16) -

F (1) =

\frac{28}{3}

1-183 §1.3 source ↗

In the following exercises, evaluate each definite integral using the Fundamental Theorem of Calculus, Part 2.

$\int_{0}^{π / 2} sin θ d θ$

Answer

F (x) =

− cos x,

F (\frac{π}{2}) -

F (0) =

1

1-185 §1.3 source ↗

In the following exercises, evaluate each definite integral using the Fundamental Theorem of Calculus, Part 2.

$\int_{0}^{π / 4} sec θ tan θ d θ$

Answer

F (x) =

sec x,

F (\frac{π}{4}) -

F (0) =

\sqrt{2} -

1

1-187 §1.3 source ↗

In the following exercises, evaluate each definite integral using the Fundamental Theorem of Calculus, Part 2.

$\int_{π / 4}^{π / 2} {csc}^{2} θ d θ$

Answer

F (x) =

− cot (x),

F (\frac{π}{2}) -

F (\frac{π}{4}) =

1

1-189 §1.3 source ↗

In the following exercises, evaluate each definite integral using the Fundamental Theorem of Calculus, Part 2.

$\int_{−2}^{−1} (\frac{1}{t^{2}} - \frac{1}{t^{3}}) d t$

Answer

F (t) =

-

\frac{1}{t} +

\frac{1}{2 t^{2}},

F (−1) -

F (−2) =

\frac{7}{8}

1-191 §1.3 source ↗

In the following exercises, use the evaluation theorem to express the integral as a function $F (x) .$

$\int_{1}^{x} e^{t} d t$

Answer

F (x) = e^{x} - e

1-193 §1.3 source ↗

In the following exercises, use the evaluation theorem to express the integral as a function $F (x) .$

$\int_{− x}^{x} sin t d t$

Answer

F (x) = 0

1-195 §1.3 source ↗

In the following exercises, identify the roots of the integrand to remove absolute values, then evaluate using the Fundamental Theorem of Calculus, Part 2.

$\int_{−2}^{4} | t^{2} - 2 t - 3 | d t$

Answer

\int_{−2}^{−1} (t^{2} - 2 t - 3) d t -

\int_{−1}^{3} (t^{2} - 2 t - 3) d t +

\int_{3}^{4} (t^{2} - 2 t - 3) d t =

\frac{46}{3}

1-197 §1.3 source ↗

In the following exercises, identify the roots of the integrand to remove absolute values, then evaluate using the Fundamental Theorem of Calculus, Part 2.

$\int_{− π / 2}^{π / 2} | sin t | d t$

Answer

− \int_{− π / 2}^{0} sin t d t + \int_{0}^{π / 2} sin t d t = 2

1-199 §1.3 source ↗

In the following exercises, identify the roots of the integrand to remove absolute values, then evaluate using the Fundamental Theorem of Calculus, Part 2.

Suppose the rate of gasoline consumption over the course of a year in the United States can be modeled by a sinusoidal function of the form $(11.21 - cos (\frac{π t}{6})) \times 10^{9}$ gal/mo.

a. What is the average monthly consumption, and for which values of t is the rate at time t equal to the average rate?

b. What is the number of gallons of gasoline consumed in the United States in a year?

c. Write an integral that expresses the average monthly U.S. gas consumption during the part of the year between the beginning of April $(t = 3)$ and the end of September $(t = 9).$

Answer

a. The average is

11.21 \times 10^{9}

since

cos (\frac{π t}{6})

has period 12 and integral 0 over any period. Consumption is equal to the average when

cos (\frac{π t}{6}) = 0,

when

t = 3,

and when

t = 9 .

b. Total consumption is the average rate times duration:

11.21 \times 12 \times 10^{9} =

1.35 \times 10^{11}

10^{9} (11.21 - \frac{1}{6} \int_{3}^{9} cos (\frac{π t}{6}) d t) =

10^{9} (11.21 + \frac{2}{π}) =

11.84 x 10^{9}

1-201 §1.3 source ↗

In the following exercises, identify the roots of the integrand to remove absolute values, then evaluate using the Fundamental Theorem of Calculus, Part 2.

Explain why, if f is continuous over $[a, b]$ and is not equal to a constant, there is at least one point $M \in [a, b]$ such that $f (M) >$ $\frac{1}{b - a} \int_{a}^{b} f (t) d t$ and at least one point $m \in [a, b]$ such that $f (m) <$ $\frac{1}{b - a} \int_{a}^{b} f (t) d t .$

Answer

If f is not constant, then its average is strictly smaller than the maximum and larger than the minimum, which are attained over

[a, b]

by the extreme value theorem.

1-203 §1.3 source ↗

In the following exercises, identify the roots of the integrand to remove absolute values, then evaluate using the Fundamental Theorem of Calculus, Part 2.

A point on an ellipse with major axis length 2a and minor axis length 2b has the coordinates $(a cos θ, b sin θ),$ $0 \leq θ \leq 2 π .$

a. Show that the distance from this point to the focus at $(− c, 0)$ is $d (θ) =$ $a +$ $c cos θ,$ where $c = \sqrt{a^{2} - b^{2}} .$

b. Use these coordinates to show that the average distance $\bar{d}$ from a point on the ellipse to the focus at $(− c, 0),$ with respect to angle θ, is a.

Answer

d^{2} θ =

{(a cos θ + c)}^{2} +

b^{2} {sin}^{2} θ =

a^{2} +

c^{2} {cos}^{2} θ +

2 a c cos θ =

{(a + c cos θ)}^{2};

\bar{d} =

\frac{1}{2 π} \int_{0}^{2 π} (a + c cos θ) d θ = a

1-205 §1.3 source ↗

In the following exercises, identify the roots of the integrand to remove absolute values, then evaluate using the Fundamental Theorem of Calculus, Part 2.

The force of gravitational attraction between the Sun and a planet is $F (θ) =$ $\frac{G m M}{r^{2} (θ)},$ where m is the mass of the planet, M is the mass of the Sun, G is a universal constant, and $r (θ)$ is the distance between the Sun and the planet when the planet is at an angle θ with the major axis of its orbit. Assuming that M, m, and the ellipse parameters a and b (half-lengths of the major and minor axes) are given, set up—but do not evaluate—an integral that expresses in terms of $G, m, M, a, b$ the average gravitational force between the Sun and the planet.

Answer

Mean gravitational force =

\frac{G m M}{2 π} \int_{0}^{2 π} \frac{1}{{(a + 2 \sqrt{a^{2} - b^{2}} cos θ)}^{2}} d θ .

Checkpoint 1.21 §1.4 source ↗

Find the definite integral of

f (x) = x^{2} - 3 x

over the interval

[1, 3] .

Answer

- \frac{10}{3}

Checkpoint 1.22 §1.4 source ↗

Find the net displacement and total distance traveled in meters given the velocity function

f (t) = \frac{1}{2} e^{t} - 2

over the interval

[0, 2] .

Answer

Net displacement:

\frac{e^{2} - 9}{2} \approx

-

0.8055 m;

total distance traveled:

4 ln 4 -

7.5 +

\frac{e^{2}}{2} \approx

1.740

Checkpoint 1.23 §1.4 source ↗

Suppose that, instead of remaining steady during the second half hour of Andrew’s outing, the wind starts to die down according to the function $v (t) =$ $−10 t +$ $20 .$ In other words, the wind speed is given by

v (t) = {\begin{cases} 20 t + 5 & for & 0 \leq t \leq \frac{1}{2} \\ - 10 t + 15 & for & \frac{1}{2} \leq t \leq 1 . \end{cases}

Under these conditions, how far from his starting point is Andrew after 1 hour?

Answer

17.5 mi

Checkpoint 1.24 §1.4 source ↗

Integrate the function

\int_{−2}^{2} x^{4} d x .

Answer

\frac{64}{5}

1-207 §1.4 source ↗

Use basic integration formulas to compute the following antiderivatives or definite integrals.

$\int (\sqrt{x} - \frac{1}{\sqrt{x}}) d x$

Answer

\int (\sqrt{x} - \frac{1}{\sqrt{x}}) d x =

\int x^{1 / 2} d x - \int x^{−1 / 2} d x =

\frac{2}{3} x^{3 / 2} +

C_{1} -

2 x^{1 / 2} +

C_{2} =

\frac{2}{3} x^{3 / 2} -

2 x^{1 / 2} +

C

1-209 §1.4 source ↗

Use basic integration formulas to compute the following antiderivatives or definite integrals.

$\int \frac{d x}{2 x}$

Answer

\int \frac{d x}{2 x} = \frac{1}{2} ln | x | + C

1-211 §1.4 source ↗

Use basic integration formulas to compute the following antiderivatives or definite integrals.

$\int_{0}^{π} (sin x - cos x) d x$

Answer

\int_{0}^{π} sin x d x - \int_{0}^{π} cos x d x =

− cos x |_{0}^{π} -

(sin x) |_{0}^{π} =

(− (−1) + 1) -

(0 - 0) =

2

1-213 §1.4 source ↗

Use basic integration formulas to compute the following antiderivatives or definite integrals.

Write an integral that expresses the increase in the perimeter $P (s)$ of a square when its side length s increases from 2 units to 4 units and evaluate the integral.

Answer

P (s) = 4 s,

\frac{d P}{d s} = 4

and

\int_{2}^{4} 4 d s = 8 .

1-215 §1.4 source ↗

Use basic integration formulas to compute the following antiderivatives or definite integrals.

A regular N-gon (an N-sided polygon with sides that have equal length s, such as a pentagon or hexagon) has perimeter Ns. Write an integral that expresses the increase in perimeter of a regular N-gon when the length of each side increases from 1 unit to 2 units and evaluate the integral.

Answer

\int_{1}^{2} N d s = N

1-217 §1.4 source ↗

Use basic integration formulas to compute the following antiderivatives or definite integrals.

A dodecahedron is a Platonic solid with a surface that consists of 12 pentagons, each of equal area. By how much does the surface area of a dodecahedron increase as the side length of each pentagon doubles from 1 unit to 2 units?

Answer

With p as in the previous exercise, each of the 12 pentagons increases in area from 2p to 4p units so the net increase in the area of the dodecahedron is 36p units.

1-219 §1.4 source ↗

Use basic integration formulas to compute the following antiderivatives or definite integrals.

Write an integral that quantifies the change in the area of the surface of a cube when its side length doubles from s unit to 2s units and evaluate the integral.

Answer

18 s^{2} =

6 \int_{s}^{2 s} 2 x d x

1-221 §1.4 source ↗

Use basic integration formulas to compute the following antiderivatives or definite integrals.

Write an integral that quantifies the increase in the surface area of a sphere as its radius doubles from R unit to 2R units and evaluate the integral.

Answer

12 π R^{2} =

8 π \int_{R}^{2 R} r d r

1-223 §1.4 source ↗

Use basic integration formulas to compute the following antiderivatives or definite integrals.

Suppose that a particle moves along a straight line with velocity $v (t) = 4 - 2 t,$ where $0 \leq t \leq 2$ (in meters per second). Find the displacement at time t and the total distance traveled up to $t = 2 .$

Answer

d (t) =

\int_{0}^{t} v (s) d s = 4 t - t^{2} .

The total distance is

d (2) = 4 m .

1-225 §1.4 source ↗

Use basic integration formulas to compute the following antiderivatives or definite integrals.

Suppose that a particle moves along a straight line with velocity defined by $v (t) =$ $| 2 t -$ $6 |,$ where $0 \leq t \leq 6$ (in meters per second). Find the displacement at time t and the total distance traveled up to $t = 6 .$

Answer

d (t) =

\int_{0}^{t} v (s) d s .

For

t <

3,

d (t) =

\int_{0}^{t} (6 - 2 t) d t = 6 t - t^{2} .

For

t >

3,

d (t) =

d (3) +

\int_{3}^{t} (2 t - 6) d t =

9 +

(t^{2} - 6 t) |_{3}^{6} .

The total distance is

d (6) = 18 m .

1-227 §1.4 source ↗

Use basic integration formulas to compute the following antiderivatives or definite integrals.

A ball is thrown upward from a height of 1.5 m at an initial speed of 40 m/sec. Acceleration resulting from gravity is −9.8 m/sec². Neglecting air resistance, solve for the velocity $v (t)$ and the height $h (t)$ of the ball t seconds after it is thrown and before it returns to the ground.

Answer

v (t) =

40 -

9.8 t m/sec;

h (t) =

1.5 +

40 t -

4.9 t^{2}

m/s

1-229 §1.4 source ↗

Use basic integration formulas to compute the following antiderivatives or definite integrals.

The area $A (t)$ of a circular shape is growing at a constant rate. If the area increases from 4π units to 9π units between times $t = 2$ and $t = 3,$ find the net change in the radius during that time.

Answer

The net increase is 1 unit.

1-231 §1.4 source ↗

Use basic integration formulas to compute the following antiderivatives or definite integrals.

Water flows into a conical tank with cross-sectional area πx² at height x and volume $\frac{π x^{3}}{3}$ up to height x. If water flows into the tank at a rate of 1 m³/min, find the height of water in the tank after 5 min. Find the change in height between 5 min and 10 min.

Answer

t = 5,

the height of water is

x =

{(\frac{15}{π})}^{1 / 3} m . .

The net change in height from

t = 5

t = 10

{(\frac{30}{π})}^{1 / 3} -

{(\frac{15}{π})}^{1 / 3}

1-233 §1.4 source ↗

Use basic integration formulas to compute the following antiderivatives or definite integrals.

The following table lists the electrical power in gigawatts—the rate at which energy is consumed—used in a certain city for different hours of the day, in a typical 24-hour period, with hour 1 corresponding to midnight to 1 a.m.

Hour	Power	Hour	Power
1	28	13	48
2	25	14	49
3	24	15	49
4	23	16	50
5	24	17	50
6	27	18	50
7	29	19	46
8	32	20	43
9	34	21	42
10	39	22	40
11	42	23	37
12	46	24	34

Find the total amount of energy in gigawatt-hours (gW-h) consumed by the city in a typical 24-hour period.

Answer

The total daily power consumption is estimated as the sum of the hourly power rates, or 911 gW-h.

1-235 §1.4 source ↗

Use basic integration formulas to compute the following antiderivatives or definite integrals.

The data in the following table are used to estimate the average power output produced by Peter Sagan for each of the last 18 sec of Stage 1 of the 2012 Tour de France.

Estimate the net energy used in kilojoules (kJ), noting that 1W = 1 J/s, and the average power output by Sagan during this time interval.

Table 1.6 Average Power Output in the book ↗

Second	Watts	Second	Watts
1	600	10	1200
2	500	11	1170
3	575	12	1125
4	1050	13	1100
5	925	14	1075
6	950	15	1000
7	1050	16	950
8	950	17	900
9	1100	18	780

Source: sportsexercisengineering.com

Answer

17 kJ

1-237 §1.4 source ↗

Use basic integration formulas to compute the following antiderivatives or definite integrals.

The distribution of incomes as of 2012 in the United States in $5000 increments is given in the following table. The kth row denotes the percentage of households with incomes between $$5000 x k$ and $5000 x k +$ $4999 .$ The row $k = 40$ contains all households with income between $200,000 and $250,000.

a. Estimate the percentage of U.S. households in 2012 with incomes less than $55,000.

b. What percentage of households had incomes exceeding $85,000?

c. Plot the data and try to fit its shape to that of a graph of the form $a (x + c) e^{− b (x + e)}$ for suitable $a, b, c .$

Table 1.8 Income Distributions in the book ↗

0	3.5	11	3.5	21	1.5	31	0.6
1	4.1	12	3.7	22	1.4	32	0.5
2	5.9	13	3.2	23	1.3	33	0.5
3	5.7	14	3.0	24	1.3	34	0.4
4	5.9	15	2.8	25	1.1	35	0.3
5	5.4	16	2.5	26	1.0	36	0.3
6	5.5	17	2.2	27	0.75	37	0.3
7	5.1	18	2.2	28	0.8	38	0.2
8	4.8	19	1.8	29	1.0	39	1.8
9	4.1	20	2.1	30	0.6	40	2.3
10	4.3

Source: http://www.census.gov/prod/2013pubs/p60-245.pdf

Answer

a. 54.3%; b. 27.00%; c. The curve in the following plot is $2.35 (t + 3) e^{−0.15 (t + 3)} .$

A graph of the data and a function approximating the data. The function is a very close approximation.

1-239 §1.4 source ↗

Use basic integration formulas to compute the following antiderivatives or definite integrals.

For a given motor vehicle, the maximum achievable deceleration from braking is approximately 7 m/sec² on dry concrete. On wet asphalt, it is approximately 2.5 m/sec². Given that 1 mph corresponds to 0.447 m/sec, find the total distance that a car travels in meters on dry concrete after the brakes are applied until it comes to a complete stop if the initial velocity is 67 mph (30 m/sec) or if the initial braking velocity is 56 mph (25 m/sec). Find the corresponding distances if the surface is slippery wet asphalt.

Answer

In dry conditions, with initial velocity

v_{0} = 30

m/s,

D = 64.3

and, if

v_{0} =

25,

D =

44.64 .

In wet conditions, if

v_{0} = 30,

and

D = 180

and if

v_{0} =

25,

D =

125 .

1-241 §1.4 source ↗

Use basic integration formulas to compute the following antiderivatives or definite integrals.

Sandra is a 25-year old woman who weighs 120 lb. She burns $300 - 50 t$ cal/hr while walking on her treadmill. Her caloric intake from drinking Gatorade is 100t calories/hour during the tth hour. What is her net decrease in calories after walking for 3 hours?

Answer

225 cal

1-243 §1.4 source ↗

Use basic integration formulas to compute the following antiderivatives or definite integrals.

Although some engines are more efficient at given a horsepower than others, on average, fuel efficiency decreases with horsepower at a rate of $1 / 25$ mpg/horsepower. If a typical 50-horsepower engine has an average fuel efficiency of 32 mpg, what is the average fuel efficiency of an engine with the following horsepower: 150, 300, 450?

Answer

E (150) =

28,

E (300) =

22,

E (450) =

16

1-245 §1.4 source ↗

Use basic integration formulas to compute the following antiderivatives or definite integrals.

[T] The following table provides hypothetical data regarding the level of service for a certain highway.

a. Plot vehicles per hour per lane on the x-axis and highway speed on the y-axis.

b. Compute the average decrease in speed (in miles per hour) per unit increase in congestion (vehicles per hour per lane) as the latter increases from 600 to 1000, from 1000 to 1500, and from 1500 to 2100. Does the decrease in miles per hour depend linearly on the increase in vehicles per hour per lane?

c. Plot minutes per mile (60 times the reciprocal of miles per hour) as a function of vehicles per hour per lane. Is this function linear?

Table 1.10 in the book ↗

Highway Speed Range (mph)	Vehicles per Hour per Lane	Density Range (vehicles/mi)
> 60	< 600	< 10
60–57	600–1000	10–20
57–54	1000–1500	20–30
54–46	1500–1900	30–45
46–30	1900–2100	45–70
<30	Unstable	70–200

Answer

A plot of the given data, which decreases in a roughly concave down manner from 600 to 2200.

b. Between 600 and 1000 the average decrease in vehicles per hour per lane is −0.0075. Between 1000 and 1500 it is −0.006 per vehicles per hour per lane, and between 1500 and 2100 it is −0.04 vehicles per hour per lane. c.

A graph of given data, showing that minutes per mile increases dramatically as wehicles per hour reaches 2000.

The graph is nonlinear, with minutes per mile increasing dramatically as vehicles per hour per lane reach 2000.

1-247 §1.4 source ↗

For the next two exercises use the data in the following table, which displays bald eagle populations from 1963 to 2000 in the continental United States.

[T] The graph below plots the cubic $p (t) =$ $0.07 t^{3} +$ $2.42 t^{2} -$ $25.63 t +$ $521.23$ against the data in the preceding table, normalized so that $t = 0$ corresponds to 1963. Estimate the average number of bald eagles per year present for the 37 years by computing the average value of p over $[0, 37] .$

A graph of the data and a cubic function that closely approximates it.

Table 1.11 Population of Breeding Bald Eagle Pairs in the book ↗

Year	Population of Breeding Pairs of Bald Eagles
1963	487
1974	791
1981	1188
1986	1875
1992	3749
1996	5094
2000	6471

Source: http://www.fws.gov/Midwest/eagle/population/chtofprs.html.

Answer

\frac{1}{37} \int_{0}^{37} p (t) d t = \frac{0.07 {(37)}^{3}}{4} +

\frac{2.42 {(37)}^{2}}{3} -

\frac{25.63 (37)}{2} +

521.23 \approx

2037

1-249 §1.4 source ↗

As a car accelerates, it does not accelerate at a constant rate; rather, the acceleration is variable. For the following exercises, use the following table, which contains the acceleration measured at every second as a driver merges onto a freeway.

Time (sec)	Acceleration (mph/sec)
1	11.2
2	10.6
3	8.1
4	5.4
5	0

As a car accelerates, it does not accelerate at a constant rate; rather, the acceleration is variable. For the next three exercises use the following table, which contains the acceleration measured at every second as a driver merges onto a freeway.

[T] The accompanying graph plots the best quadratic fit, $a (t) =$ $−0.70 t^{2} +$ $1.44 t +$ $10.44,$ to the data from the preceding table. Compute the average value of $a (t)$ to estimate the average acceleration between $t = 0$ and $t = 5 .$

A graph of the data and a curve that closely approximates the data.

Table 1.12 in the book ↗

Time (sec)	Acceleration (mph/sec)
1	11.2
2	10.6
3	8.1
4	5.4
5	0

Answer

Average acceleration is

-

\frac{0.7}{3} t^{3} +

\frac{1.44}{2} t^{2} +

10.44 t |_{0}^{5} =

-

\frac{0.7}{3} (5^{3}) +

10.44 (5) \approx

41

mph/s

1-251 §1.4 source ↗

Time (sec)	Acceleration (mph/sec)
1	11.2
2	10.6
3	8.1
4	5.4
5	0

[T] Using your velocity equation from the previous exercise, find the corresponding distance equation, assuming your initial distance is 0 mi. How far did you travel while you accelerated your car? (Hint: You will need to convert time units.)

Table 1.12 in the book ↗

Time (sec)	Acceleration (mph/sec)
1	11.2
2	10.6
3	8.1
4	5.4
5	0

Answer

s (t) =

v (t) d t =

\int (- \frac{7}{30} t^{3} + 0.72 t^{2} + 10.44 t + 23.967) d t =

-

\frac{7}{120} t^{4} +

\frac{0.72}{3} t^{3} +

\frac{10.44}{2} t^{2} +

23.967 t +

C =

174.24 feet

1-253 §1.4 source ↗

Table 1.12 in the book ↗

Time (sec)	Acceleration (mph/sec)
1	11.2
2	10.6
3	8.1
4	5.4
5	0

Time (sec)	Acceleration (mph/sec)
1	11.2
2	10.6
3	8.1
4	5.4
5	0

[T] An athlete runs by a motion detector, which records her speed, as displayed in the following table. The best linear fit to this data, $ℓ (t) =$ $−0.068 t +$ $5.14,$ is shown in the accompanying graph. Use the average value of $ℓ (t)$ between $t = 0$ and $t = 40$ to estimate the runner’s average speed.

Minutes	Speed (m/sec)
0	5
10	4.8
20	3.6
30	3.0
40	2.5

A graph of the data and a line to approximate the data.

Answer

\frac{1}{40} \int_{0}^{40} (−0.068 t + 5.14) d t =

-

\frac{0.068 (40)}{2} +

5.14 =

3.78 m/sec

Checkpoint 1.25 §1.5 source ↗

Use substitution to find the antiderivative

\int 3 x^{2} {(x^{3} - 3)}^{2} d x .

Answer

\int 3 x^{2} {(x^{3} - 3)}^{2} d x =

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

C

Checkpoint 1.26 §1.5 source ↗

Use substitution to find

\int x^{2} {(x^{3} + 5)}^{9} d x .

Answer

\frac{{(x^{3} + 5)}^{10}}{30} +

C

Checkpoint 1.27 §1.5 source ↗

Use substitution to evaluate the integral

\int \frac{cos t}{{sin}^{2} t} d t .

Answer

- \frac{1}{sin t} + C

Checkpoint 1.28 §1.5 source ↗

Use substitution to evaluate the indefinite integral

\int {cos}^{3} t sin t d t .

Answer

- \frac{{cos}^{4} t}{4} + C

Checkpoint 1.29 §1.5 source ↗

Use substitution to evaluate the definite integral

\int_{−1}^{0} y {(2 y^{2} - 3)}^{5} d y .

Answer

\frac{91}{3}

Checkpoint 1.30 §1.5 source ↗

Use substitution to evaluate

\int_{0}^{1} x^{2} cos (\frac{π}{2} x^{3}) d x .

Answer

\frac{2}{3 π} \approx 0.2122

1-255 §1.5 source ↗

2. If

f = g \circ h,

when reversing the chain rule,

\frac{d}{d x} (g \circ h) (x) =

g^{'} (h (x)) h^{'} (x),

should you take

u = g (x)

u = h (x) ?

Answer

u = h (x)

1-257 §1.5 source ↗

In the following exercises, verify each identity using differentiation. Then, using the indicated u-substitution, identify f such that the integral takes the form $\int f (u) d u .$

For $x >$ $1 : \int \frac{x^{2}}{\sqrt{x - 1}} d x =$ $\frac{2}{15} \sqrt{x - 1} (3 x^{2} + 4 x + 8) +$ $C;$ $u =$ $x -$ $1$

Answer

f (u) =

\frac{{(u + 1)}^{2}}{\sqrt{u}}

1-259 §1.5 source ↗

In the following exercises, verify each identity using differentiation. Then, using the indicated u-substitution, identify f such that the integral takes the form $\int f (u) d u .$

$\int \frac{x}{\sqrt{4 x^{2} + 9}} d x = \frac{1}{4} \sqrt{4 x^{2} + 9} + C;$ $u =$ $4 x^{2} +$ $9$

Answer

d u =

8 x d x;

f (u) =

\frac{1}{8 \sqrt{u}}

1-261 §1.5 source ↗

In the following exercises, find the antiderivative using the indicated substitution.

$\int {(x + 1)}^{4} d x;$ $u =$ $x +$ $1$

Answer

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

1-263 §1.5 source ↗

In the following exercises, find the antiderivative using the indicated substitution.

$\int {(2 x - 3)}^{−7} d x;$ $u =$ $2 x -$ $3$

Answer

-

\frac{1}{12 {(2 x - 3)}^{6}} +

C

1-265 §1.5 source ↗

In the following exercises, find the antiderivative using the indicated substitution.

$\int \frac{x}{\sqrt{x^{2} + 1}} d x;$ $u =$ $x^{2} +$ $1$

Answer

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

1-267 §1.5 source ↗

In the following exercises, find the antiderivative using the indicated substitution.

$\int (x - 1) {(x^{2} - 2 x)}^{3} d x;$ $u =$ $x^{2} -$ $2 x$

Answer

\frac{1}{8} {(x^{2} - 2 x)}^{4} +

C

1-269 §1.5 source ↗

In the following exercises, find the antiderivative using the indicated substitution.

$\int {cos}^{3} θ d θ;$ $u =$ $sin θ$ $(H i n t : {cos}^{2} θ = 1 - {sin}^{2} θ)$

Answer

sin θ -

\frac{{sin}^{3} θ}{3} +

C

1-271 §1.5 source ↗

In the following exercises, use a suitable change of variables to determine the indefinite integral.

$\int x {(1 - x)}^{99} d x$

Answer

\frac{{(1 - x)}^{101}}{101} -

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

C

1-273 §1.5 source ↗

In the following exercises, use a suitable change of variables to determine the indefinite integral.

$\int {(11 x - 7)}^{−3} d x$

Answer

\int {(11 x - 7)}^{−3} dx =

-

\frac{1}{22 {(11 x - 7)}^{2}} +

C

1-275 §1.5 source ↗

In the following exercises, use a suitable change of variables to determine the indefinite integral.

$\int {cos}^{3} θ sin θ d θ$

Answer

- \frac{{cos}^{4} θ}{4} + C

1-277 §1.5 source ↗

In the following exercises, use a suitable change of variables to determine the indefinite integral.

$\int {cos}^{2} (π t) sin (π t) d t$

Answer

-

\frac{{cos}^{3} (π t)}{3 π} +

C

1-279 §1.5 source ↗

In the following exercises, use a suitable change of variables to determine the indefinite integral.

$\int t sin (t^{2}) cos (t^{2}) d t$

Answer

-

\frac{1}{4} {cos}^{2} (t^{2}) +

C

1-281 §1.5 source ↗

In the following exercises, use a suitable change of variables to determine the indefinite integral.

$\int \frac{x^{2}}{{(x^{3} - 3)}^{2}} d x$

Answer

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

1-283 §1.5 source ↗

In the following exercises, use a suitable change of variables to determine the indefinite integral.

$\int \frac{y^{5}}{{(1 - y^{3})}^{3 / 2}} d y$

Answer

-

\frac{2 (y^{3} - 2)}{3 \sqrt{1 - y^{3}}} +

C

1-285 §1.5 source ↗

In the following exercises, use a suitable change of variables to determine the indefinite integral.

${\int (1 - {cos}^{3} θ)}^{10} {cos}^{2} θ sin θ d θ$

Answer

\frac{1}{33} {(1 - {cos}^{3} θ)}^{11} +

C

1-287 §1.5 source ↗

In the following exercises, use a suitable change of variables to determine the indefinite integral.

$\int ({sin}^{2} θ - 2 sin θ) {({sin}^{3} θ - 3 {sin}^{2} θ)}^{3} cos θ d θ$

Answer

\frac{1}{12} {({sin}^{3} θ - 3 {sin}^{2} θ)}^{4} +

C

1-289 §1.5 source ↗

In the following exercises, use a calculator to estimate the area under the curve using left Riemann sums with 50 terms, then use substitution to solve for the exact answer.

[T] $y = x {(1 - x^{2})}^{3}$ over $[−1, 2]$

Answer

L_{50} =

−8.5779 .

The exact area is

\frac{−81}{8}

1-291 §1.5 source ↗

In the following exercises, use a calculator to estimate the area under the curve using left Riemann sums with 50 terms, then use substitution to solve for the exact answer.

[T] $y = \frac{x}{{(x^{2} + 1)}^{2}}$ over $[−1, 1]$

Answer

L_{50} =

−0.006399

… The exact area is 0.

1-293 §1.5 source ↗

In the following exercises, use a change of variables to evaluate the definite integral.

$\int_{0}^{1} \frac{x}{\sqrt{1 + x^{2}}} d x$

Answer

u =

1 +

x^{2},

d u =

2 x d x,

\frac{1}{2} \int_{1}^{2} u^{−1 / 2} d u = \sqrt{2} - 1

1-295 §1.5 source ↗

In the following exercises, use a change of variables to evaluate the definite integral.

$\int_{0}^{1} \frac{t^{2}}{\sqrt{1 + t^{3}}} d t$

Answer

u =

1 +

t^{3},

d u =

3 t^{2} d t,

\frac{1}{3} \int_{1}^{2} u^{−1 / 2} d u = \frac{2}{3} (\sqrt{2} - 1)

1-297 §1.5 source ↗

In the following exercises, use a change of variables to evaluate the definite integral.

$\int_{0}^{π / 4} \frac{sin θ}{{cos}^{4} θ} d θ$

Answer

u =

cos θ,

d u =

− sin θ d θ,

\int_{1 / \sqrt{2}}^{1} u^{−4} d u = \frac{1}{3} (2 \sqrt{2} - 1)

1-299 §1.5 source ↗

In the following exercises, evaluate the indefinite integral $\int f (x) d x$ with constant $C = 0$ using u-substitution. Then, graph the function and the antiderivative over the indicated interval. If possible, estimate a value of C that would need to be added to the antiderivative to make it equal to the definite integral $F (x) =$ $\int_{a}^{x} f (t) d t,$ with a the left endpoint of the given interval.

[T] $\int \frac{cos (ln (2 x))}{x} d x$ on $[0, 2]$

Answer

Two graphs. The first shows the function f(x) = cos(ln(2x)) / x, which increases sharply over the approximate interval (0,.25) and then decreases gradually to the x axis. The second shows the function f(x) = sin(ln(2x)), which decreases sharply on the approximate interval (0, .25) and then increases in a gently curve into the first quadrant.

The antiderivative is $y =$ $sin (ln (2 x)) .$ Since the antiderivative is not continuous at $x = 0,$ one cannot find a value of C that would make $y =$ $sin (ln (2 x)) -$ $C$ work as a definite integral.

1-301 §1.5 source ↗

[T] $\int \frac{sin x}{{cos}^{3} x} d x$ over $[- \frac{π}{3}, \frac{π}{3}]$

Answer

Two graphs. The first is the function f(x) = sin(x) / cos(x)^3 over [-5pi/16, 5pi/16]. It is an increasing concave down function for values less than zero and an increasing concave up function for values greater than zero. The second is the fuction f(x) = ½ sec(x)^2 over the same interval. It is a wide, concave up curve which decreases for values less than zero and increases for values greater than zero.

The antiderivative is $y = \frac{1}{2} {sec}^{2} x .$ You should take $C = −2$ so that $F (- \frac{π}{3}) = 0 .$

1-303 §1.5 source ↗

[T] $\int 3 x^{2} \sqrt{2 x^{3} + 1} d x$ over $[0, 1]$

Answer

Two graphs. The first shows the function f(x) = 3x^2 * sqrt(2x^3 + 1). It is an increasing concave up curve starting at the origin. The second shows the function f(x) = 1/3 * (2x^3 + 1)^(1/3). It is an increasing concave up curve starting at about 0.3.

The antiderivative is $y =$ $\frac{1}{3} {(2 x^{3} + 1)}^{3 / 2} .$ One should take $C = - \frac{1}{3} .$

1-305 §1.5 source ↗

Is the substitution $u = 1 - x^{2}$ in the definite integral $\int_{0}^{2} \frac{x}{1 - x^{2}} d x$ okay? If not, why not?

Answer

No, because the integrand is discontinuous at

x = 1 .

1-307 §1.5 source ↗

In the following exercises, use a change of variables to show that each definite integral is equal to zero.

$\int_{0}^{\sqrt{π}} t cos (t^{2}) sin (t^{2}) d t$

Answer

u = sin (t^{2});

the integral becomes

\frac{1}{2} \int_{0}^{0} u d u .

1-309 §1.5 source ↗

In the following exercises, use a change of variables to show that each definite integral is equal to zero.

$\int_{0}^{1} \frac{1 - 2 t}{(1 + {(t - \frac{1}{2})}^{2})} d t$

Answer

u =

(1 + {(t - \frac{1}{2})}^{2});

the integral becomes

− \int_{5 / 4}^{5 / 4} \frac{1}{u} d u .

1-311 §1.5 source ↗

In the following exercises, use a change of variables to show that each definite integral is equal to zero.

$\int_{0}^{2} (1 - t) cos (π t) d t$

Answer

u = 1 - t;

the integral becomes

\begin{array}{l} \int_{1}^{−1} u cos (π (1 - u)) d u \\ = \int_{1}^{−1} u [cos π cos π u - sin π sin π u] d u \\ = − \int_{1}^{−1} u cos π u d u \\ = \int_{−1}^{1} u cos π u d u = 0 \end{array}

since the integrand is odd.

1-313 §1.5 source ↗

In the following exercises, use a change of variables to show that each definite integral is equal to zero.

Show that the average value of $f (x)$ over an interval $[a, b]$ is the same as the average value of $f (c x)$ over the interval $[\frac{a}{c}, \frac{b}{c}]$ for $c > 0 .$

Answer

Setting

u = c x

and

d u = c d x

gets you

\frac{1}{\frac{b}{c} - \frac{a}{c}} \int_{a / c}^{b / c} f (c x) d x = \frac{c}{b - a} \int_{u = a}^{u = b} f (u) \frac{d u}{c} = \frac{1}{b - a} \int_{a}^{b} f (u) d u .

1-315 §1.5 source ↗

In the following exercises, use a change of variables to show that each definite integral is equal to zero.

Find the area under the graph of $g (t) =$ $\frac{t}{{(1 - t^{2})}^{a}}$ between $t = 0$ and $t = x,$ where $0 < x < 1$ and $a > 0$ is fixed. Evaluate the limit as $x \to 1 .$

Answer

\int_{0}^{x} g (t) d t = \frac{1}{2} \int_{u = 1 - x^{2}}^{1} \frac{d u}{u^{a}} =

\frac{1}{2 (1 - a)} u^{1 - a} |_{u = 1 - x^{2}}^{1} =

\frac{1}{2 (1 - a)} (1 - {(1 - x^{2})}^{1 - a}) .

x \to 1

the limit is

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

a < 1,

and the limit diverges to +∞ if

a \geq 1 .

1-317 §1.5 source ↗

In the following exercises, use a change of variables to show that each definite integral is equal to zero.

The area of the top half of an ellipse with a major axis that is the x-axis from $x = - a$ to $x = a$ and with a minor axis that is the y-axis from $y = − b$ to $y = b$ can be written as $\int_{− a}^{a} b \sqrt{1 - \frac{x^{2}}{a^{2}}} d x .$ Use the substitution $x = a cos t$ to express this area in terms of an integral of a trigonometric function. You do not need to compute the integral.

Answer

\int_{t = π}^{0} b \sqrt{1 - {cos}^{2} t} \times (− a sin t) d t =

\int_{t = 0}^{π} a b {sin}^{2} t d t

1-319 §1.5 source ↗

In the following exercises, use a change of variables to show that each definite integral is equal to zero.

[T] The following graph is of a function of the form $f (x) =$ $a cos (n t) +$ $b cos (m t) .$ Estimate the coefficients a and b and the frequency parameters n and m. Use these estimates to approximate $\int_{0}^{\frac{π}{2}} f (t) d t .$

The graph of a function of the given form over [0, 2pi]. It begins at (0,1) and ends at (2pi, 1). It has five turning points, located just after pi/4, between pi/2 and 3pi/4, pi, between 5pi/4 and 3pi/2, and just before 7pi/4 at about -1.5, 2.5, -3, 2.5, and -1. It crosses the x axis between 0 and pi/4, just before pi/2, just after 3pi/4, just before 5pi/4, just after 3pi/2, and between 7pi/4 and 2pi.

Answer

f (t) =

2 cos (3 t) -

cos (2 t);

\int_{0}^{π} (2 cos (3 t) - cos (2 t)) d t =

0

Checkpoint 1.31 §1.6 source ↗

Find the antiderivative of the function using substitution:

x^{2} e^{−2 x^{3}} .

Answer

\int x^{2} e^{−2 x^{3}} d x = - \frac{1}{6} e^{−2 x^{3}} + C

Checkpoint 1.32 §1.6 source ↗

Find the antiderivative of

e^{x} {(3 e^{x} - 2)}^{2} .

Answer

\int e^{x} {(3 e^{x} - 2)}^{2} d x = \frac{1}{9} {(3 e^{x} - 2)}^{3}

Checkpoint 1.33 §1.6 source ↗

Evaluate the indefinite integral

\int 2 x^{3} e^{x^{4}} d x .

Answer

\int 2 x^{3} e^{x^{4}} d x =

\frac{1}{2} e^{x^{4}}

Checkpoint 1.34 §1.6 source ↗

Evaluate

\int_{0}^{2} e^{2 x} d x .

Answer

\frac{1}{2} \int_{0}^{4} e^{u} d u = \frac{1}{2} (e^{4} - 1)

Checkpoint 1.35 §1.6 source ↗

From Example 1.42, suppose the bacteria grow at a rate of

q (t) = 2^{t} .

Assume the culture still starts with 10,000 bacteria. Find

Q (t) .

How many bacteria are in the dish after 3 hours?

Answer

Q (t) =

\frac{2^{t}}{ln 2} +

8.557 .

There are 20,099 bacteria in the dish after 3 hours.

Checkpoint 1.36 §1.6 source ↗

Suppose the rate of growth of the fly population is given by

g (t) =

e^{0.01 t},

and the initial fly population is 100 flies. How many flies are in the population after 15 days?

Answer

There are 116 flies.

Checkpoint 1.37 §1.6 source ↗

Evaluate the definite integral using substitution:

\int_{1}^{2} \frac{1}{x^{3}} e^{4 x^{−2}} d x .

Answer

\int_{1}^{2} \frac{1}{x^{3}} e^{4 x^{−2}} d x =

\frac{1}{8} [e^{4} - e]

Checkpoint 1.38 §1.6 source ↗

Find the antiderivative of

\frac{1}{x + 2} .

Answer

ln | x + 2 | + C

Checkpoint 1.39 §1.6 source ↗

Find the antiderivative of

{log}_{3} x .

Answer

\frac{x}{ln 3} (ln x - 1) +

C

1-321 §1.6 source ↗

In the following exercises, compute each indefinite integral.

$\int e^{−3 x} d x$

Answer

\frac{−1}{3} e^{−3 x} + C

1-323 §1.6 source ↗

In the following exercises, compute each indefinite integral.

$\int 3^{− x} d x$

Answer

- \frac{3^{− x}}{ln 3} + C

1-325 §1.6 source ↗

In the following exercises, compute each indefinite integral.

$\int \frac{2}{x} d x$

Answer

2 \ln x +

C or \ln x^{2} +C

1-327 §1.6 source ↗

In the following exercises, compute each indefinite integral.

$\int \frac{1}{\sqrt{x}} d x$

Answer

2 \sqrt{x} + C

1-329 §1.6 source ↗

In the following exercises, find each indefinite integral by using appropriate substitutions.

$\int \frac{d x}{x {(ln x)}^{2}}$

Answer

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

1-331 §1.6 source ↗

In the following exercises, find each indefinite integral by using appropriate substitutions.

$\int \frac{d x}{x ln x ln (ln x)}$

Answer

ln (ln (ln x)) +

C

1-333 §1.6 source ↗

In the following exercises, find each indefinite integral by using appropriate substitutions.

$\int \frac{cos x - x sin x}{x cos x} d x$

Answer

ln (x cos x) + C

1-335 §1.6 source ↗

In the following exercises, find each indefinite integral by using appropriate substitutions.

$\int ln (cos x) tan x d x$

Answer

-

\frac{1}{2} {(ln (cos (x)))}^{2} +

C

1-337 §1.6 source ↗

In the following exercises, find each indefinite integral by using appropriate substitutions.

$\int x^{2} e^{− x^{3}} d x$

Answer

\frac{− e^{− x^{3}}}{3} + C

1-339 §1.6 source ↗

In the following exercises, find each indefinite integral by using appropriate substitutions.

$\int e^{tan x} {sec}^{2} x d x$

Answer

e^{tan x} + C

1-341 §1.6 source ↗

In the following exercises, find each indefinite integral by using appropriate substitutions.

$\int \frac{e^{ln (1 - t)}}{1 - t} d t$

Answer

t + C

1-343 §1.6 source ↗

In the following exercises, verify by differentiation that $\int ln x d x =$ $x (ln x - 1) +$ $C,$ then use appropriate changes of variables to compute the integral.

$\int x^{2} ln(x^{2}) d x$

Answer

\frac{2}{9} x^{3} (ln (x^{3}) - 1) +

C

1-345 §1.6 source ↗

In the following exercises, verify by differentiation that $\int ln x d x =$ $x (ln x - 1) +$ $C,$ then use appropriate changes of variables to compute the integral.

$\int \frac{ln x}{\sqrt{x}} d x$ $(H i n t : Set u = \sqrt{x} .)$

Answer

2 \sqrt{x} (ln x - 2) + C

1-347 §1.6 source ↗

In the following exercises, verify by differentiation that $\int ln x d x =$ $x (ln x - 1) +$ $C,$ then use appropriate changes of variables to compute the integral.

Write an integral to express the area under the graph of $y = e^{t}$ between $t = 0$ and $t = ln x,$ and evaluate the integral.

Answer

\int_{0}^{ln x} e^{t} d t = e^{t} |_{0}^{ln x} = e^{ln x} - e^{0} = x - 1

1-349 §1.6 source ↗

In the following exercises, use appropriate substitutions to express the trigonometric integrals in terms of compositions with logarithms.

$\int \frac{sin (3 x) - cos (3 x)}{sin (3 x) + cos (3 x)} d x$

Answer

- \frac{1}{3} ln sin (3 x) + cos (3 x) +C

1-351 §1.6 source ↗

In the following exercises, use appropriate substitutions to express the trigonometric integrals in terms of compositions with logarithms.

$\int x csc (x^{2}) d x$

Answer

-

\frac{1}{2} ln | csc (x^{2}) +

cot (x^{2}) | +

C

1-353 §1.6 source ↗

In the following exercises, use appropriate substitutions to express the trigonometric integrals in terms of compositions with logarithms.

$\int ln (csc x) cot x d x$

Answer

-

\frac{1}{2} {(ln (csc x))}^{2} +

C

1-355 §1.6 source ↗

In the following exercises, evaluate the definite integral.

$\int_{1}^{2} \frac{1 + 2 x + x^{2}}{3 x + 3 x^{2} + x^{3}} d x$

Answer

\frac{1}{3} ln (\frac{26}{7})

1-357 §1.6 source ↗

In the following exercises, evaluate the definite integral.

$\int_{0}^{π / 3} \frac{sin x - cos x}{sin x + cos x} d x$

Answer

ln (\sqrt{3} - 1)

1-359 §1.6 source ↗

In the following exercises, evaluate the definite integral.

$\int_{π / 4}^{π / 3} cot x d x$

Answer

\frac{1}{2} ln \frac{3}{2}

1-361 §1.6 source ↗

In the following exercises, integrate using the indicated substitution.

$\int \frac{y - 1}{y + 1} d y;$ $u =$ $y +$ $1$

Answer

y -

2 ln | y +

1 | +

C

1-363 §1.6 source ↗

In the following exercises, integrate using the indicated substitution.

$\int \frac{sin x + cos x}{sin x - cos x} d x;$ $u =$ $sin x -$ $cos x$

Answer

ln | sin x -

cos x | +

C

1-365 §1.6 source ↗

In the following exercises, integrate using the indicated substitution.

$\int ln (x) \frac{\sqrt{1 - {(ln x)}^{2}}}{x} d x;$ $u =$ $ln x$

Answer

-

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

C

1-367 §1.6 source ↗

In the following exercises, does the right-endpoint approximation overestimate or underestimate the exact area? Calculate the right endpoint estimate R₅₀ and solve for the exact area.

[T] $y = e^{− x}$ over $[0, 1]$

Answer

Exact solution:

\frac{e - 1}{e},

R_{50} =

0.6258 .

Since f is decreasing, the right endpoint estimate underestimates the area.

1-369 §1.6 source ↗

In the following exercises, does the right-endpoint approximation overestimate or underestimate the exact area? Calculate the right endpoint estimate R₅₀ and solve for the exact area.

[T] $y =$ $\frac{x + 1}{x^{2} + 2 x + 6}$ over $[0, 1]$

Answer

Exact solution:

\frac{2 ln (3) - ln (6)}{2},

R_{50} =

0.2033 .

Since f is increasing, the right endpoint estimate overestimates the area.

1-371 §1.6 source ↗

In the following exercises, does the right-endpoint approximation overestimate or underestimate the exact area? Calculate the right endpoint estimate R₅₀ and solve for the exact area.

[T] $y = − 2^{− x}$ over $[0, 1]$

Answer

Exact solution:

-

\frac{1}{ln (4)},

R_{50} =

−0.7164 .

Since f is negative and increasing on the interval, the right-endpoint estimate overestimates the value of the integral, but it underestimates the area of the region (absolute value of the integral).

1-373 §1.6 source ↗

In the following exercises, $f (x) \geq 0$ for $a \leq x \leq b .$ Find the area under the graph of $f (x)$ between the given values a and b by integrating.

$f (x) =$ $\frac{{log}_{2} (x)}{x};$ $a =$ $32,$ $b =$ $64$

Answer

\frac{11}{2} ln 2

1-375 §1.6 source ↗

In the following exercises, $f (x) \geq 0$ for $a \leq x \leq b .$ Find the area under the graph of $f (x)$ between the given values a and b by integrating.

$f (x) =$ $2^{− x};$ $a =$ $3,$ $b =$ $4$

Answer

\frac{1}{ln (65, 536)}

1-377 §1.6 source ↗

In the following exercises, $f (x) \geq 0$ for $a \leq x \leq b .$ Find the area under the graph of $f (x)$ between the given values a and b by integrating.

Compute the integral of $f (x) = x e^{− x^{2}}$ and find the smallest value of N such that the area under the graph $f (x) = x e^{− x^{2}}$ between $x = N$ and $x = N + 1$ is, at most, 0.01.

Answer

\int_{N}^{N + 1} x e^{− x^{2}} d x = \frac{1}{2} (e^{− N^{2}} - e^{− {(N + 1)}^{2}}) .

The quantity is less than 0.01 when

N = 2 .

1-379 §1.6 source ↗

In the following exercises, $f (x) \geq 0$ for $a \leq x \leq b .$ Find the area under the graph of $f (x)$ between the given values a and b by integrating.

Show that $\int_{a}^{b} \frac{d t}{t} =$ $\int_{1 / b}^{1 / a} \frac{d t}{t}$ when $0 < a \leq b .$

Answer

\int_{a}^{b} \frac{d x}{x} =

ln (b) -

ln (a) =

ln (\frac{1}{a}) -

ln (\frac{1}{b}) =

\int_{1 / b}^{1 / a} \frac{d x}{x}

1-381 §1.6 source ↗

In the following exercises, $f (x) \geq 0$ for $a \leq x \leq b .$ Find the area under the graph of $f (x)$ between the given values a and b by integrating.

Use the previous exercise to find the antiderivative of $h (x) =$ $x^{x} (1 + ln x)$ and evaluate $\int_{2}^{3} x^{x} (1 + ln x) d x .$

Answer

1-383 §1.6 source ↗

The following exercises are intended to derive the fundamental properties of the natural log starting from the definition $ln (x) =$ $\int_{1}^{x} \frac{d t}{t},$ using properties of the definite integral and making no further assumptions.

Use the identity $ln (x) =$ $\int_{1}^{x} \frac{d t}{t}$ to derive the identity $ln (\frac{1}{x}) =$ $− ln x .$

Answer

We may assume that

x >

1,

so \frac{1}{x} <

1 .

Then,

\int_{1}^{1 / x} \frac{d t}{t} .

Now make the substitution

u = \frac{1}{t},

d u = - \frac{d t}{t^{2}}

and

\frac{d u}{u} = - \frac{d t}{t},

and change endpoints:

\int_{1}^{1 / x} \frac{d t}{t} =

− \int_{1}^{x} \frac{d u}{u} =

− ln x .

1-385 §1.6 source ↗

Use the identity $ln x = \int_{1}^{x} \frac{d t}{t}$ to show that $ln (x)$ is an increasing function of x on $[0, \infty),$ and use the previous exercises to show that the range of $ln (x)$ is $(− \infty, \infty) .$ Without any further assumptions, conclude that $ln (x)$ has an inverse function defined on $(− \infty, \infty) .$

Answer

Answers will vary.

1-387 §1.6 source ↗

Pretend, for the moment, that we do not know that $e^{x}$ is the inverse function of $ln x,$ but keep in mind that $ln x$ has an inverse function defined on $(− \infty, \infty) .$ Call it E. Show that $E' (t) = E (t) .$

Answer

x = E (ln (x)) .

Then,

1 =

\frac{E' (ln x)}{x} or x =

E' (ln x) .

Since any number t can be written

t = ln x

for some x, and for such t we have

x = E (t),

it follows that for any

t,

E' (t) =

E (t) .

1-389 §1.6 source ↗

[T] The normal distribution in probability is given by $p (x) =$ $\frac{1}{σ \sqrt{2 π}} e^{− {(x - μ)}^{2} / 2 σ^{2}},$ where σ is the standard deviation and μ is the average. The standard normal distribution in probability, $p_{s},$ corresponds to $μ = 0 and σ = 1 .$ Compute the right endpoint estimates $R_{10} and R_{100}$ of $\int_{−1}^{1} \frac{1}{\sqrt{2 π}} e^{− \frac{x^{2}}{2}} d x .$

Answer

$R_{10} =$ $0.6811,$ $R_{100} =$ $0.6827$

A graph of the function f(x) = .5 * ( sqrt(2)*e^(-.5x^2)) / sqrt(pi). It is a downward opening curve that is symmetric across the y axis, crossing at about (0, .4). It approaches 0 as x goes to positive and negative infinity. Between 1 and -1, ten rectangles are drawn for a right endpoint estimate of the area under the curve.

Checkpoint 1.40 §1.7 source ↗

Evaluate the integral

\int \frac{d x}{\sqrt{1 - 16 x^{2}}} .

Answer

\frac{1}{4} {sin}^{−1} (4 x) +

C

Checkpoint 1.41 §1.7 source ↗

Find the indefinite integral using an inverse trigonometric function and substitution for

\int \frac{d x}{\sqrt{9 - x^{2}}} .

Answer

{sin}^{−1} (\frac{x}{3}) + C

Checkpoint 1.42 §1.7 source ↗

Use substitution to find the antiderivative

\int \frac{d x}{25 + 4 x^{2}} .

Answer

\frac{1}{10} {tan}^{−1} (\frac{2 x}{5}) +

C

Checkpoint 1.43 §1.7 source ↗

Evaluate the integral

\int \frac{d x}{16 + x^{2}} .

Answer

\frac{1}{4} {tan}^{−1} (\frac{x}{4}) +

C

Checkpoint 1.44 §1.7 source ↗

Evaluate the definite integral

\int_{0}^{2} \frac{d x}{4 + x^{2}} .

Answer

\frac{π}{8}

1-391 §1.7 source ↗

In the following exercises, evaluate each integral in terms of an inverse trigonometric function.

$\int_{0}^{\sqrt{3} / 2} \frac{d x}{\sqrt{1 - x^{2}}}$

Answer

{sin}^{−1} x |_{0}^{\sqrt{3} / 2} =

\frac{π}{3}

1-393 §1.7 source ↗

In the following exercises, evaluate each integral in terms of an inverse trigonometric function.

$\int_{\sqrt{3}}^{1} \frac{d x}{1 + x^{2}}$

Answer

{tan}^{−1} x |_{\sqrt{3}}^{1} =

-

\frac{π}{12}

1-395 §1.7 source ↗

In the following exercises, evaluate each integral in terms of an inverse trigonometric function.

$\int_{\frac{2}{\sqrt{3}}}^{\sqrt{2}} \frac{d x}{| x | \sqrt{x^{2} - 1}}$

Answer

{sec}^{−1} |_{\frac{2}{\sqrt{3}}}^{\sqrt{2}} =

\frac{π}{4} - \frac{π}{6} =

\frac{π}{12}

1-397 §1.7 source ↗

In the following exercises, find each indefinite integral, using appropriate substitutions.

$\int \frac{d x}{\sqrt{9 - x^{2}}}$

Answer

{sin}^{−1} (\frac{x}{3}) + C

1-399 §1.7 source ↗

In the following exercises, find each indefinite integral, using appropriate substitutions.

$\int \frac{d x}{9 + x^{2}}$

Answer

\frac{1}{3} {tan}^{−1} (\frac{x}{3}) +

C

1-401 §1.7 source ↗

In the following exercises, find each indefinite integral, using appropriate substitutions.

$\int \frac{d x}{| x | \sqrt{x^{2} - 9}}$

Answer

\frac{1}{3} {sec}^{−1} (\frac{x}{3}) +

C

1-403 §1.7 source ↗

In the following exercises, find each indefinite integral, using appropriate substitutions.

Explain the relationship $− {cos}^{−1} t +$ $C =$ $\int \frac{d t}{\sqrt{1 - t^{2}}} = {sin}^{−1} t + C .$ Is it true, in general, that ${cos}^{−1} t =$ $− {sin}^{−1} t ?$

Answer

cos (\frac{π}{2} - θ) =

sin θ .

So,

{sin}^{−1} t =

\frac{π}{2} -

{cos}^{−1} t .

They differ by a constant.

1-405 §1.7 source ↗

In the following exercises, find each indefinite integral, using appropriate substitutions.

Explain what is wrong with the following integral: $\int_{1}^{2} \frac{d t}{\sqrt{1 - t^{2}}} .$

Answer

\sqrt{1 - t^{2}}

is not defined as a real number when

t > 1 .

1-407 §1.7 source ↗

In the following exercises, solve for the antiderivative $\int f$ of f with $C = 0,$ then use a calculator to graph f and the antiderivative over the given interval $[a, b] .$ Identify a value of C such that adding C to the antiderivative recovers the definite integral $F (x) =$ $\int_{a}^{x} f (t) d t .$

[T] $\int \frac{1}{\sqrt{9 - x^{2}}} d x$ over $[−3, 3]$

Answer

Two graphs. The first shows the function f(x) = 1 / sqrt(9 – x^2). It is an upward opening curve symmetric about the y axis, crossing at (0, 1/3). The second shows the function F(x) = arcsin(1/3 x). It is an increasing curve going through the origin.

The antiderivative is ${sin}^{−1} (\frac{x}{3}) +$ $C .$ Taking $C = \frac{π}{2}$ recovers the definite integral.

1-409 §1.7 source ↗

[T] $\int \frac{cos x}{4 + {sin}^{2} x} d x$ over $[−6, 6]$

Answer

Two graphs. The first shows the function f(x) = cos(x) / (4 + sin(x)^2). It is an oscillating function over [-6, 6] with turning points at roughly (-3, -2.5), (0, .25), and (3, -2.5), where (0,.25) is a local max and the others are local mins. The second shows the function F(x) = .5 * arctan(.5*sin(x)), which also oscillates over [-6,6]. It has turning points at roughly (-4.5, .25), (-1.5, -.25), (1.5, .25), and (4.5, -.25).

The antiderivative is $\frac{1}{2} {tan}^{−1} (\frac{sin x}{2}) +$ $C .$ Taking $C =$ $\frac{1}{2} {tan}^{−1} (\frac{sin (6)}{2})$ recovers the definite integral.

1-411 §1.7 source ↗

In the following exercises, compute the antiderivative using appropriate substitutions.

$\int \frac{{sin}^{−1} t d t}{\sqrt{1 - t^{2}}}$

Answer

\frac{1}{2} {({sin}^{−1} t)}^{2} +

C

1-413 §1.7 source ↗

In the following exercises, compute the antiderivative using appropriate substitutions.

$\int \frac{{tan}^{−1} (2 t)}{1 + 4 t^{2}} d t$

Answer

\frac{1}{4} {({tan}^{−1} (2 t))}^{2} +

C

1-415 §1.7 source ↗

In the following exercises, compute the antiderivative using appropriate substitutions.

$\int \frac{{sec}^{−1} (\frac{t}{2})}{| t | \sqrt{t^{2} - 4}} d t$

Answer

\frac{1}{4} ({sec}^{−1} {(\frac{t}{2})}^{2}) +

C

1-417 §1.7 source ↗

In the following exercises, use a calculator to graph the antiderivative $\int f$ with $C = 0$ over the given interval $[a, b] .$ Approximate a value of C, if possible, such that adding C to the antiderivative gives the same value as the definite integral $F (x) =$ $\int_{a}^{x} f (t) d t .$

[T] $\int \frac{1}{x \sqrt{x^{2} - 4}} d x$ over $[2, 6]$

Answer

A graph of the function f(x) = -.5 * arctan(2 / ( sqrt(x^2 – 4) ) ) in quadrant four. It is an increasing concave down curve with a vertical asymptote at x=2.

The antiderivative is $\frac{1}{2} {sec}^{−1} (\frac{x}{2}) +$ $C .$ Taking $C = 0$ recovers the definite integral over $[2, 6] .$

1-419 §1.7 source ↗

[T] $\int \frac{(sin x + x cos x)}{1 + x^{2} {sin}^{2} x} d x$ over $[−6, 6]$

Answer

The graph of f(x) = arctan(x sin(x)) over [-6,6]. It has five turning points at roughly (-5, -1.5), (-2,1), (0,0), (2,1), and (5,-1.5).

The general antiderivative is ${tan}^{−1} (x sin x) +$ $C .$ Taking $C =$ $− {tan}^{−1} (6 sin (6))$ recovers the definite integral.

1-421 §1.7 source ↗

[T] $\int \frac{1}{x + x {ln}^{2} x} d x$ over $[0, 2]$

Answer

A graph of the function f(x) = arctan(ln(x)) over (0, 2]. It is an increasing curve with x-intercept at (1,0).

The general antiderivative is ${tan}^{−1} (ln x) +$ $C .$ Taking $C = \frac{π}{2} = {tan}^{−1} \infty$ recovers the definite integral.

1-423 §1.7 source ↗

In the following exercises, compute each integral using appropriate substitutions.

$\int \frac{e^{t}}{\sqrt{1 - e^{2 t}}} d t$

Answer

{sin}^{−1} (e^{t}) + C

1-425 §1.7 source ↗

In the following exercises, compute each integral using appropriate substitutions.

$\int \frac{d t}{t \sqrt{1 - {ln}^{2} t}}$

Answer

{sin}^{−1} (ln t) +

C

1-427 §1.7 source ↗

In the following exercises, compute each integral using appropriate substitutions.

$\int \frac{{cos}^{−1} (2 t)}{\sqrt{1 - 4 t^{2}}} d t$

Answer

-

\frac{1}{4} {({cos}^{−1} (2 t))}^{2} +

C

1-429 §1.7 source ↗

In the following exercises, compute each definite integral.

$\int_{0}^{1 / 2} \frac{tan ({sin}^{−1} t)}{\sqrt{1 - t^{2}}} d t$

Answer

\frac{1}{2} ln (\frac{4}{3})

1-431 §1.7 source ↗

In the following exercises, compute each definite integral.

$\int_{0}^{1 / 2} \frac{sin ({tan}^{−1} t)}{1 + t^{2}} d t$

Answer

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

1-433 §1.7 source ↗

In the following exercises, compute each definite integral.

For $A > 0,$ compute $I (A) =$ $\int_{− A}^{A} \frac{d t}{1 + t^{2}}$ and evaluate $lim_{A \to \infty} I (A),$ the area under the graph of $\frac{1}{1 + t^{2}}$ on $[− \infty, \infty] .$

Answer

2 {tan}^{−1} (A) \to π

A \to \infty

1-435 §1.7 source ↗

In the following exercises, compute each definite integral.

Use the substitution $u = \sqrt{2} cot x$ and the identity $1 +$ ${cot}^{2} x =$ ${csc}^{2} x$ to evaluate $\int \frac{d x}{1 + {cos}^{2} x} .$ (Hint: Multiply the top and bottom of the integrand by ${csc}^{2} x .)$

Answer

Using the hint, one has

\int \frac{{csc}^{2} x}{{csc}^{2} x + {cot}^{2} x} d x =

\int \frac{{csc}^{2} x}{1 + 2 {cot}^{2} x} d x .

Set

u = \sqrt{2} cot x .

Then,

d u = − \sqrt{2} {csc}^{2} x

and the integral is

-

\frac{1}{\sqrt{2}} \int \frac{d u}{1 + u^{2}} = - \frac{1}{\sqrt{2}} {tan}^{−1} u +

C =

\frac{1}{\sqrt{2}} {tan}^{−1} (\sqrt{2} cot x) +

C .

If one uses the identity

{tan}^{−1} s +

{tan}^{−1} (\frac{1}{s}) =

\frac{π}{2},

then this can also be written

\frac{1}{\sqrt{2}} {tan}^{−1} (\frac{tan x}{\sqrt{2}}) +

C .

1-437 §1.7 source ↗

In the following exercises, compute each definite integral.

[T] Approximate the points at which the graphs of $f (x) = x^{2} - 1$ and ${g (x) = (x^{2} + 1)}^{\frac{1}{2}}$ intersect to four decimal places, and approximate the area between their graphs to three decimal places.

Answer

x \approx \pm 1.7321 .

The left endpoint estimate with

N = 100

is 4.781 and these decimals persist for

N = 500 .

1-439 §1.R source ↗

True or False. Justify your answer with a proof or a counterexample. Assume all functions $f$ and $g$ are continuous over their domains.

If $f (x) >$ $0,$ $f^{'} (x) >$ $0$ for all $x,$ then the right-hand rule underestimates the integral $\int_{a}^{b} f (x) .$ Use a graph to justify your answer.

Answer

False

1-441 §1.R source ↗

True or False. Justify your answer with a proof or a counterexample. Assume all functions $f$ and $g$ are continuous over their domains.

If $f (x) \leq g (x)$ for all $x \in [a, b],$ then $\int_{a}^{b} f (x) \leq \int_{a}^{b} g (x) .$

Answer

True

1-443 §1.R source ↗

Evaluate the Riemann sums $L_{4} and R_{4}$ for the following functions over the specified interval. Compare your answer with the exact answer, when possible, or use a calculator to determine the answer.

$y = 3 x^{2} - 2 x + 1$ over $[−1, 1]$

Answer

L_{4} =

5.25,

R_{4} =

3.25,

exact answer: 4

1-445 §1.R source ↗

$y = x^{2} sin x$ over $[0, π]$

Answer

L_{4} =

5.364,

R_{4} =

5.364,

exact answer: 5.870

1-447 §1.R source ↗

Evaluate the following integrals.

$\int_{−1}^{1} (x^{3} - 2 x^{2} + 4 x) d x$

Answer

- \frac{4}{3}

1-449 §1.R source ↗

Evaluate the following integrals.

$\int_{π / 3}^{π / 2} 2 sec (2 θ) tan (2 θ) d θ$

Answer

1-451 §1.R source ↗

Find the antiderivative.

$\int \frac{d x}{{(x + 4)}^{3}}$

Answer

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

1-453 §1.R source ↗

Find the antiderivative.

$\int \frac{4 x^{2}}{\sqrt{1 - x^{6}}} d x$

Answer

\frac{4}{3} {sin}^{−1} (x^{3}) +

C

1-455 §1.R source ↗

Find the derivative.

$\frac{d}{d t} \int_{0}^{t} \frac{sin x}{\sqrt{1 + x^{2}}} d x$

Answer

\frac{sin t}{\sqrt{1 + t^{2}}}

1-457 §1.R source ↗

Find the derivative.

$\frac{d}{d x} \int_{1}^{ln (x)} (4 t + e^{t}) d t$

Answer

4 \frac{ln x}{x} + 1

1-459 §1.R source ↗

The following problems consider the historic average cost per gigabyte of RAM on a computer.

Year	5-Year Change ($)
1980	0
1985	−5,468,750
1990	−755,495
1995	−73,005
2000	−29,768
2005	−918
2010	−177

If the average cost per gigabyte of RAM in 2010 is $12, find the average cost per gigabyte of RAM in 1980.

Answer

$6,328,113

1-461 §1.R source ↗

The following problems consider the historic average cost per gigabyte of RAM on a computer.

Year	5-Year Change ($)
1980	0
1985	−5,468,750
1990	−755,495
1995	−73,005
2000	−29,768
2005	−918
2010	−177

Find the average cost of 1GB RAM for 2005 to 2010.

Answer

$73.36

1-463 §1.R source ↗

The following problems consider the historic average cost per gigabyte of RAM on a computer.

Year	5-Year Change ($)
1980	0
1985	−5,468,750
1990	−755,495
1995	−73,005
2000	−29,768
2005	−918
2010	−177

What is the average velocity of the bullet for the first half-second?

Answer

\frac{19117}{12} ft/sec,

or 1593 ft/sec

Loading chapter 2…

Problems and answers are adapted from Calculus Volume 2 by Gilbert Strang and Edwin “Jed” Herman, OpenStax, and are used under the CC BY-NC-SA 4.0 licence. Access for free at https://openstax.org/books/calculus-volume-2/pages/1-introduction. Illustrations: Copyright Rice University, OpenStax, under CC BY-NC-SA 4.0 license. This adaptation is released under the same licence, is non-commercial, and is neither affiliated with nor endorsed by OpenStax or Rice University.

Hour	Power	Hour	Power
1	28	13	48
2	25	14	49
3	24	15	49
4	23	16	50
5	24	17	50
6	27	18	50
7	29	19	46
8	32	20	43
9	34	21	42
10	39	22	40
11	42	23	37
12	46	24	34

Hour	Power	Hour	Power
1	28	13	48
2	25	14	49
3	24	15	49
4	23	16	50
5	24	17	50
6	27	18	50
7	29	19	46
8	32	20	43
9	34	21	42
10	39	22	40
11	42	23	37
12	46	24	34

Hour	Power	Hour	Power
1	28	13	48
2	25	14	49
3	24	15	49
4	23	16	50
5	24	17	50
6	27	18	50
7	29	19	46
8	32	20	43
9	34	21	42
10	39	22	40
11	42	23	37
12	46	24	34