AP Calculus AB Unit 8: Applications of Integration

Explore AP Calculus Unit 8 questions on average value, motion, net change, areas, cross sections, and solids of revolution.

Covered syllabus points

8.1 Finding the Average Value of a Function on an Interval 8.2 Connecting Position, Velocity, and Acceleration of Functions Using Integrals 8.3 Using Accumulation Functions and Definite Integrals in Applied Contexts 8.4 Finding the Area Between Curves Expressed as Functions of x 8.5 Finding the Area Between Curves Expressed as Functions of y 8.6 Finding the Area Between Curves That Intersect at More Than Two Points 8.7 Volumes with Cross Sections: Squares and Rectangles 8.8 Volumes with Cross Sections: Triangles and Semicircles 8.9 Volume with Disc Method: Revolving Around the x- or y-Axis 8.10 Volume with Disc Method: Revolving Around Other Axes 8.11 Volume with Washer Method: Revolving Around the x- or y-Axis 8.12 Volume with Washer Method: Revolving Around Other Axes

[Maximum number: 2]

An invasive species of plant appears in a fruit grove at time t=0 and begins to spread. The

function C defined by $C(t)=7.6 \arctan (0.2 t)$ models the number of acres in the fruit grove

affected by the species t weeks after the species appears. It can be shown that $C^{\prime}(t)=\frac{38}{25+t^{2}}$ .

(Note: Your calculator should be in radian mode.)

Find the average number of acres affected by the invasive species from time t=0 to time

t=4 weeks. Show the setup for your calculations.


$\frac{1}{4-0} \int_{0}^{4} C(t) d t$	Average value formula	Point 1 (P1)
$=\frac{1}{4}(11.112896)=2.778224$	Answer	Point 2 (P2)
From time t=0 to t=4 weeks, the average number of acres affected by the invasive species was 2.778 acres.

Scoring Notes for Part A

- P1 is earned for the correct integral, with or without the differential, along with evidence of division

by 4. In the presence of the correct integral, the correct answer will suffice as evidence of division

by 4. These may appear all in one step, as in the model solution, or in multiple steps.

- P2 is earned for the correct answer, with or without supporting work. A reported answer should be

accurate to three places after the decimal point, rounded or truncated. An inappropriately rounded

answer does not earn the point.

- Incorrect or unclear communication between the correct integral and the correct answer is treated as

scratch work and is not considered in scoring. For example:

○ $\int_{0}^{4} C(t) d t=11.112896$ so the average velocity is 2.778224.

Note: This response earns P1 for the correct integral with the correct answer as evidence of

division by 4. It also earns P2 for the correct answer.

○ $\int_{0}^{4} C(t) d t=\frac{11.112896}{4}=2.778224$

Note: This response earns P1 for the correct integral with the correct answer as evidence of

division by 4. It also earns P2 for the correct answer. (In this instance, incorrect linkage is not

considered in scoring.)

○ $\int_{0}^{4} C(t) d t=2.778224$

Note: This response earns P1 for the correct integral with the correct answer as evidence of

division by 4. It also earns P2 for the correct answer. (In this instance, incorrect linkage is not

considered in scoring.)

- Note that the values $\frac{1}{4}(11.112)$ and $\frac{1}{4}(11.113)$ are accurate to three digits after the decimal and

therefore earn P2.

[Maximum number: 3]

Two particles, H and J, are moving along the x-axis. For $0 \leq t \leq 5$ , the position of particle H at

time t is given by $x_{H}(t)=e^{t^{2}-4 t}$ and the velocity of particle J at time t is given by

$v_{J}(t)=2 t\left(t^{2}-1\right)^{3}$ .

Particle J is at position x=7 at time t=0. Find the position of particle J at time t=2. Show

the work that leads to your answer.

D Particle J is at position x=7 at time t=0. Find the position of particle J at time t=2. Show the

work that leads to your answer.

$\begin{aligned} & x_{J}(2)=x_{J}(0)+\int_{0}^{2} v_{J}(t) d t=7+\int_{0}^{2} 2 t\left(t^{2}-1\right)^{3} d t & =7+\left[\frac{1}{4}\left(t^{2}-1\right)^{4}\right]_{0}^{2} & =7+\frac{1}{4}\left((3)^{4}-(-1)^{4}\right)=7+\frac{1}{4}(80)=27 \end{aligned}$

Integrand	Point 7 (P7)
Antiderivative	Point 8 (P8)
Answer	Point 9 (P9)

Scoring Notes for Part D

- To earn P7, a response must present an indefinite or definite integral with an integrand of $v_{J}(t)$ or

$2 t\left(t^{2}-1\right)^{3}$ . (See below for notes on how to handle a missing differential d t.)

- P8 is earned for an antiderivative of the form $k\left(t^{2}-1\right)^{4}$ or equivalent, for k>0. If $k \neq \frac{1}{4}$ , then

the response is not eligible to earn P9.

- A response of $7+\frac{1}{4}\left((3)^{4}-(-1)^{4}\right)$ or equivalent banks P9 (i.e., subsequent errors in simplification

will not be considered in scoring for P9).

Note: An ambiguous response, such as $7+\frac{1}{4}\left((3)^{4}-(-1)^{4}\right.$ , does not bank P9 and therefore must go

on to resolve the ambiguity with a correct final answer (e.g., $7+\frac{1}{4}(80)$ or 27) to earn P9.

- If the differential d t is missing:

○ Writing $\int_{0}^{2} v_{J}(t)$ earns P7 and is eligible to earn P8 and P9.

○ Writing $7+\int_{0}^{2} v_{J}(t)$ earns P7 and is eligible to earn P8 and P9.

○ Writing $\int_{0}^{2} v_{J}(t)+7$ introduces an ambiguity for the intended integrand.

- $\int_{0}^{2} v_{J}(t)+7=\left[\frac{1}{4}\left(t^{2}-1\right)^{4}\right]_{0}^{2}+7$ resolves the ambiguity.

Therefore, this earns P7 and P8 and is eligible for P9.

- $\int_{0}^{2} v_{J}(t)+7=\left[\frac{1}{4}\left(t^{2}-1\right)^{4}+7 t\right]_{0}^{2}$ confirms that an incorrect integrand was used.

Therefore, this does not earn P7, earns P8, and is not eligible for P9.

- If the ambiguity is not resolved, this does not earn P7, P8, or P9.

- Alternate solution using u-substitution:

$\begin{aligned} & \text { Let } u=t^{2}-1, \text { then } d u=2 t d t. \\ & t=0 \Rightarrow u=-1 \\ & t=2 \Rightarrow u=3 \\ & x_{J}(2)=x_{J}(0)+\int_{0}^{2} v_{J}(t) d t=7+\int_{0}^{2} 2 t\left(t^{2}-1\right)^{3} d t \\ & =7+\int_{-1}^{3} u^{3} d u=7+\left[\frac{1}{4} u^{4}\right]_{-1}^{3} \\ & =7+\frac{1}{4}\left((3)^{4}-(-1)^{4}\right)=7+\frac{1}{4}(80)=27 \end{aligned}$

- Alternate solution using indefinite integral:

$\begin{aligned} & \int 2 t\left(t^{2}-1\right)^{3} d t=\frac{1}{4}\left(t^{2}-1\right)^{4}+C \\ & x_{J}(0)=7=\frac{1}{4}\left(0^{2}-1\right)^{4}+C \Rightarrow C=\frac{27}{4} \\ & x_{J}(t)=\frac{1}{4}\left(t^{2}-1\right)^{4}+\frac{27}{4} \\ & x_{J}(2)=\frac{1}{4}\left(2^{2}-1\right)^{4}+\frac{27}{4}=\frac{108}{4}=27 \end{aligned}$

Part B (AB): Graphing calculator not allowed

[Maximum number: 1]

A function f(t) gives the rate of evaporation of water, in liters per hour, from a pond, where t is measured in hours since 12 noon. Which of the following gives the meaning of $\int_{4}^{10} f(t) d t$ in the context described?

The total volume of water, in liters, that evaporated from the pond during the first 10 hours after 12 noon

The total volume of water, in liters, that evaporated from the pond between 4 P.M. and 10 P.M.

The net change in the rate of evaporation, in liters per hour, from the pond between 4 P.M. and 10 P.M.

The average rate of evaporation, in liters per hour, from the pond between 4 P.M. and 10 P.M.

The average rate of change in the rate of evaporation, in liters per hour per hour, from the pond between 4 P.M. and 10 P.M.

[Maximum number: 7]

The shaded region R is bounded by the graphs of the functions f and g, where $f(x)=x^{2}-2 x$

and $g(x)=x+\sin (\pi x)$ , as shown in the figure.

Figure for Question Unit 8: Applications of Integration question 4 — AP Calculus AB

(Note: Your calculator should be in radian mode.)

(a)

Find the area of R. Show the setup for your calculations.

[ 2 ]

A	Find the area of R. Show the setup for your calculations.
		$\int_{0}^{3}(g(x)-f(x)) d x$	Form of integrand Point 1 (P1)
		= 5.136620	Answer Point 2 (P2)
		Scoring Notes for Part A
		- P1 is earned for a response that presents an integrand of g(x)-f(x),\|g(x)-f(x)\|, f(x)-g(x), or \|f(x)-g(x)\| in a definite integral, with or without the differential d x. - P2 is earned for the correct answer, with or without supporting work. A reported answer should be accurate to three places after the decimal point, rounded or truncated. An inappropriately rounded answer does not earn the point. - Incorrect communication between the integral and the correct answer is treated as scratch work and is not considered in scoring. Note: This response earns P1 for the integral. It also earns P2 for the correct answer. Note: This response earns P1 for the integral. It also earns P2 for the correct answer. (In this instance, incorrect linkage is not considered in scoring.)

(b)

Region R is the base of a solid. For this solid, at each x the cross section perpendicular to the

x-axis is a rectangle with height x and base in region R. Find the volume of the solid. Show

the setup for your calculations.

[ 2 ]

B Region R is the base of a solid. For this solid, at each x the cross section perpendicular to the x-axis is a

rectangle with height x and base in region R. Find the volume of the solid. Show the setup for your

calculations.

$\int_{0}^{3} x(g(x)-f(x)) d x$	Form of integrand	Point 3 (P3)
=7.704930<br>The volume of the solid is 7.705 (or 7.704).	Answer	Point 4 (P4)
Scoring Notes for Part B

- P3 is earned for a definite integral with an integrand presented as a product of two nonconstant

factors, with one of the factors equal to x, g(x)-f(x), or f(x)-g(x).

- The presence or absence of the differential d x will not be considered in scoring P3 or P4.

- P4 is earned for the correct answer, with or without supporting work. A reported answer should be

accurate to three places after the decimal point, rounded or truncated. An inappropriately rounded

answer does not earn the point, unless an earlier point was not earned due to inappropriate rounding.

- Incorrect or unclear communication between the integral and the correct answer is treated as scratch

work and is not considered in scoring. For example:

○ $\int_{0}^{3} x(f(x)-g(x)) d x=-7.705$ so the volume is 7.705.

Note: This response earns P3 for the integral. It also earns P4 for the correct answer.

○ $\int_{0}^{3} x(f(x)-g(x)) d x=7.705$

Note: This response earns P3 for the integral. It also earns P4 for the correct answer. (In this

instance, incorrect linkage is not considered in scoring.)

- The exact answer is $\frac{12+27 \pi}{4 \pi}$ .

(c)

Write, but do not evaluate, an integral expression for the volume of the solid generated when

the region R is rotated about the horizontal line y=-2.

[ 3 ]

C Write, but do not evaluate, an integral expression for the volume of the solid generated when the region

R is rotated about the horizontal line y=-2.

Volume $=\pi \int_{0}^{3}\left((g(x)-(-2))^{2}-(f(x)-(-2))^{2}\right) d x$	Point 5 (P5)
	Integrand	Point 6 (P6)
	Limits, constant, and differential	Point 7 (P7)
Scoring Notes for Part C

- P5 is earned for a definite integral with an integrand of $R^{2}-r^{2}$ or $\left|R^{2}-r^{2}\right|$ , where one of $\{R, r\}$

is correct or a difference between g and a nonzero constant, and the other is correct or a difference

between f and a nonzero constant.

- P6 is earned for the integral $\int_{0}^{3}\left((g(x)+2)^{2}-(f(x)+2)^{2}\right) d x, \int_{0}^{3}\left((f(x)+2)^{2}-(g(x)+2)^{2}\right) d x$ ,

or a mathematically equivalent expression.

- Note P5 and P6 could be earned for a difference of definite integrals.

- A response that presents an integral expression that does not include the constant $\pi$ is eligible for

P5 and P6 but does not earn P7.

- To be eligible for P7, a response must have earned P5.

- A response that reverses the difference of squares must resolve the reversal with either the constant

or the limits of integration AND include the differential to earn P7. For example:

○ A response of $-\pi \int_{0}^{3}\left((f(x)+2)^{2}-(g(x)+2)^{2}\right) d x$ or $\pi \int_{3}^{0}\left((f(x)+2)^{2}-(g(x)+2)^{2}\right) d x$

earns P5, P6, and P7.

○ A response of $\pi \int_{0}^{3}\left((f(x)+2)^{2}-(g(x)+2)^{2}\right) d x$ earns P5 and P6 but does not earn P7.

- A response of only $\pi \int_{0}^{3}\left((g(x)-(-2))^{2}-(f(x)-(-2))^{2}\right) d x$ earns P5, P6, and P7.

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Covered syllabus points

Unit 8: Applications of Integration question 1

Unit 8: Applications of Integration question 2

Unit 8: Applications of Integration question 3

Unit 8: Applications of Integration question 4

Question (a)

Question (b)

Question (c)