Unit 6

Applications of Integration

Area between curves, volumes by disks, washers and cross sections, average value and arc length, plus integration by parts and improper integrals.

Topics in this unit

Average Value of a Function · Area Between Curves · Area & Volume from a Graph · Volumes: Known Cross Sections · Volumes: Rectangle Cross Sections · Volumes: Disk & Washer · Volumes: Revolving About Other Axes · Volumes: Shell Method · Area & Volume (Calculator-Active) · Estimating with Tables (Riemann & Trapezoid) · Arc Length · Integration by Parts · Partial Fractions · Improper Integrals

Average Value of a Function

Question 1

no calculatormedium

Let

f

be a differentiable function with

f (0) = 3

and

f (2) = 3

. Which of the following must be true on

0 \leq x \leq 2

? I. The average rate of change of

f

0

.
II. The average value of

f^{'}

0

.
III.

f^{'} (c) = 0

for some

c

(0, 2)

Show the answer

The answer is C.

Since

f (0) = f (2)

: I, average rate of change

= \frac{3 - 3}{2} = 0

. II, average value of

f^{'} = \frac{f ( 2 ) - f ( 0 )}{2} = 0

(FTC). III, Rolle's Theorem (a special case of the MVT) guarantees an interior

c

with

f^{'} (c) = 0

. All three hold.

Why not A:III is also guaranteed: since $f (0) = f (2)$ and $f$ is differentiable, Rolle's Theorem gives some $c$ with $f^{'} (c) = 0$ .
Why not B:II is also true: the average value of $f^{'}$ is $\frac{1}{2} \int_{0}^{2} f^{'} d x = \frac{f ( 2 ) - f ( 0 )}{2} = 0$ .
Why not D:I is also true: the average rate of change is $\frac{f ( 2 ) - f ( 0 )}{2 - 0} = 0$ .

Question 2

no calculatorhard

The graph of

f

is shown below. It consists of two line segments and a semicircle of radius

1

. What is the average value of

f

on the interval

[0, 7]

Show the answer

The answer is D.

Average value

= \frac{1}{7} \int_{0}^{7} f (x) d x

. Read the integral as signed area: a

2 \times 3

rectangle on

[0, 2]

(area

6

), a triangle on

[2, 5]

with base

3

and height

3

(area

\frac{9}{2}

), and a semicircle of radius

1

below the axis on

[5, 7]

(area

\frac{π}{2}

, counted negative). So

\int_{0}^{7} f = 6 + \frac{9}{2} - \frac{π}{2} = \frac{21 - π}{2}

, and the average value is

\frac{21 - π}{14}

Why not A:That is $\int_{0}^{7} f (x) d x$ itself. The average value divides by the length of the interval, $7$ .
Why not B:The semicircle on $[5, 7]$ is BELOW the axis, so its area $\frac{π}{2}$ enters with a minus sign.
Why not C:A semicircle of radius $1$ has area $\frac{1}{2} π (1)^{2} = \frac{π}{2}$ , not $π$ .

Area Between Curves

Question 3

no calculatormedium

The shaded region in the figure is bounded above by the line

y = 4

, below by the curve

y = x^{2}

, and on the left by the

y

-axis. Which of the following gives the area of the region?

Show the answer

The answer is D.

The region runs from

x = 0

to where

x^{2} = 4

(at

x = 2

). Its area is

\int_{0}^{2} (top - bottom) d x = \int_{0}^{2} (4 - x^{2}) d x

Why not A:Subtracted bottom minus top; the line $y = 4$ is the upper boundary, so the integrand is $4 - x^{2}$ .
Why not B:This is the area under the curve, not the region above it up to $y = 4$ .
Why not C:Used $y = 4$ as the upper $x$ -limit; the region runs only to $x = 2$ , where the curve meets $y = 4$ .

Volumes: Known Cross Sections

Question 4

no calculatormedium

The base of a solid is the region in the first quadrant enclosed by the graph of

y = 2 - x

and the coordinate axes. If every cross section of the solid perpendicular to the

y

-axis is a square, the volume of the solid is given by

Show the answer

The answer is B.

For a cross section at height

y

, the side runs from

x = 0

x = 2 - y

, so the side length is

2 - y

and the area is

(2 - y)^{2}

. Stacking in

y

V = \int_{0}^{2} (2 - y)^{2} d y

Why not A:The $π$ belongs to circular cross sections (disks); squares have area $s^{2}$ with no $π$ .
Why not C:Cross sections perpendicular to the $y$ -axis are stacked in $y$ , so integrate in $y$ with side $x = 2 - y$ .
Why not D:Forgot to square: a square's area is (side) $^{2} = (2 - y)^{2}$ , not $2 - y$ .

Volumes: Disk & Washer

Question 5

no calculatorhard

The region enclosed by

y = x

and

y = x^{2}

is revolved about the

x

-axis. What is the volume?

Show the answer

The answer is A.

Washers: outer

R = x

, inner

r = x^{2}

V = π \int_{0}^{1} (x^{2} - x^{4}) d x = π (\frac{1}{3} - \frac{1}{5}) = \frac{2 π}{15}

Why not B:That's $π \int_{0}^{1} (x - x^{2}) d x$ , with nothing squared. Washers square each radius: $π \int (R^{2} - r^{2}) d x$ .
Why not C:That's $π \int_{0}^{1} (x - x^{2})^{2} d x$ , which squares the gap between the curves. Square each radius first: $π \int (x^{2} - x^{4}) d x$ .
Why not D:That's $π \int_{0}^{1} (x^{2} + x^{4}) d x$ . The inner disk is a hole, so subtract it: $π (\frac{1}{3} - \frac{1}{5}) = \frac{2 π}{15}$ .

Question 6

no calculatorhard

The region bounded by

y = x^{2}

and

y = 2 x

is revolved about the

x

-axis. What is the volume of the solid generated?

Show the answer

The answer is B.

The curves meet at

x = 0, 2

, with

y = 2 x

outside and

y = x^{2}

inside. Washers give

V = π \int_{0}^{2} [(2 x)^{2} - (x^{2})^{2}] d x = π \int_{0}^{2} (4 x^{2} - x^{4}) d x = π (\frac{32}{3} - \frac{32}{5}) = \frac{64}{15} π .

Why not A: $\frac{32}{15} π$ is exactly half; you likely lost a factor of $2$ somewhere in $4 x^{2} - x^{4}$ .
Why not C: $\frac{32}{5} π$ is only $π \int_{0}^{2} x^{4} d x$ (the inner part). You must subtract this from the outer $π \int_{0}^{2} 4 x^{2} d x$ .
Why not D: $\frac{64}{3} π$ is $π \int_{0}^{2} 4 x^{2} d x$ , the outer radius alone, without subtracting the inner radius squared.

Volumes: Revolving About Other Axes

Question 7

no calculatormedium

The region in the first quadrant bounded by the graph of

y = x^{3}

, the line

y = 8

, and the

y

-axis is revolved about the line

x = - 1

. Which of the following gives the volume of the resulting solid?

Show the answer

The answer is D.

Slice perpendicular to the axis, so integrate in

y

from

0

8

. The region runs from

x = 0

x = y^{1/3}

. Measured from

x = - 1

, the outer radius is

y^{1/3} + 1

and the inner radius is

1

V = π \int_{0}^{8} [(y^{1/3} + 1)^{2} - 1^{2}] d y

Why not A:The axis is to the left of the region, so distances are $x - (- 1) = x + 1$ , not $x - 1$ .
Why not B: $π \int_{0}^{8} y^{2/3} d y$ revolves about the $y$ -axis. The axis $x = - 1$ is one unit left, so the radii grow by $1$ and there is a hole.
Why not C:A vertical axis calls for horizontal slices in $y$ . This integrand treats $x^{3} + 1$ as a radius, which mixes up the variables.

Arc Length

Question 8

BCno calculatorhard

What is the length of the curve

y = \frac{2}{3} x^{3/2}

from

x = 0

x = 3

Show the answer

The answer is C.

y^{'} = x^{1/2}

, so

1 + (y^{'})^{2} = 1 + x

(a perfect setup).

L = \int_{0}^{3} 1 + x d x = [\frac{2}{3} (1 + x)^{3/2}]_{0}^{3} = \frac{2}{3} (8 - 1) = \frac{14}{3} .

Why not A: $\frac{7}{3}$ is half the result; $(1 + x)^{3/2}$ goes from $1$ to $8$ , a difference of $7$ , then times $\frac{2}{3}$ gives $\frac{14}{3}$ .
Why not B: $\frac{16}{3}$ uses $8$ without subtracting the lower-bound value $1$ in $(1 + x)^{3/2}$ .
Why not D: $\frac{26}{3}$ is too big. At $x = 3$ , $(1 + x)^{3/2} = 4^{3/2} = 8$ , and at $x = 0$ it is $1$ , so $L = \frac{2}{3} (8 - 1) = \frac{14}{3}$ .

Integration by Parts

Question 9

BCno calculatorhard

\int arcsin x d x =

Show the answer

The answer is A.

Integrate by parts with

u = arcsin x

d v = d x

(so

d u = \frac{1}{1 - x ^{2}} d x

v = x

\int arcsin x d x = x arcsin x - \int \frac{x}{1 - x ^{2}} d x

. The remaining integral is

- 1 - x^{2}

, so

\int arcsin x d x = x arcsin x + 1 - x^{2} + C

Why not B:Sign slip. With $u = arcsin x, d v = d x$ : $\int arcsin x d x = x arcsin x - \int \frac{x}{1 - x ^{2}} d x$ , and $\int \frac{x}{1 - x ^{2}} d x = - 1 - x^{2}$ , so the result is $x arcsin x + 1 - x^{2} + C$ .
Why not C: $\frac{1}{1 - x ^{2}}$ is the DERIVATIVE of $arcsin x$ , not the integral of $\frac{x}{1 - x ^{2}}$ . That integral is $- 1 - x^{2}$ , giving $x arcsin x + 1 - x^{2} + C$ .
Why not D:That is $\frac{d}{d x} arcsin x$ , not its antiderivative. Use integration by parts: $x arcsin x + 1 - x^{2} + C$ .

Improper Integrals

Question 10

BCno calculatorhard

Which of the following improper integrals converge? I.

\int_{1}^{\infty} \frac{1}{x} d x

II.

\int_{1}^{\infty} \frac{1}{x ^{2}} d x

III.

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

Show the answer

The answer is B.

I diverges (

\int_{1}^{\infty} \frac{d x}{x} = \infty

). II converges to

1

. III converges to

2

(the singularity at

0

is mild enough since the exponent

\frac{1}{2} < 1

). So II and III only.

Why not A:I diverges ( $\int_{1}^{\infty} \frac{1}{x} = lim ln b = \infty$ ); III converges ( $\int_{0}^{1} x^{- 1/2} = 2$ ).
Why not C:II also converges: $\int_{1}^{\infty} x^{- 2} = 1$ . Only I diverges.
Why not D:I diverges, so not all three converge.

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