AP Calculus BC Unit 8: Applications of Integration

Practice AP Calculus BC Unit 8 questions on applying definite integrals to average value, motion, accumulation, area, volume, and length.

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 8.13 The Arc Length of a Smooth, Planar Curve and Distance Traveled

[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: 6]

Johanna jogs along a straight path. For $0 \leq t \leq 40$ , Johanna's velocity is given by a differentiable function v. Selected values of v(t), where t is measured in minutes and v(t) is measured in meters per minute, are given in the table above.

(a)

Using correct units, explain the meaning of the definite integral $\int_{0}^{40}|v(t)| d t$ in the context of the problem. Approximate the value of $\int_{0}^{40}|v(t)| d t$ using a right Riemann sum with the four subintervals indicated in the table.

[ 3 ]

$\int_{0}^{40}|v(t)| d t$ is the total distance Johanna jogs, in meters, over the time interval $0 \leq t \leq 40$ minutes.

\begin{aligned} \int_{0}^{40}|v(t)| d t & \approx 12 \cdot|v(12)|+8 \cdot|v(20)|+4 \cdot|v(24)|+16 \cdot|v(40)| \\ & =12 \cdot 200+8 \cdot 240+4 \cdot 220+16 \cdot 150 \\ & =2400+1920+880+2400 \\ & =7600 \text { meters } \end{aligned}

(b)

Based on the model B from part (c), find Bob's average velocity during the interval $0 \leq t \leq 10$ .

[ 3 ]

\begin{aligned} \text { Avg vel } & =\frac{1}{10} \int_{0}^{10}\left(t^{3}-6 t^{2}+300\right) d t \\ & =\frac{1}{10}\left[\frac{t^{4}}{4}-2 t^{3}+300 t\right]_{0}^{10} \\ & =\frac{1}{10}\left[\frac{10000}{4}-2000+3000\right]=350 \text { meters } / \mathrm{min} \end{aligned}

: approximation

3:\left\{\begin{array}{l} 1: \text { explanation } \\ 1: \text { right Riemann sum } \\ 1: \text { approximation } \end{array}\right.

2:\left\{\begin{array}{l} 1: \text { uses } B^{\prime}(t) \\ 1: \text { answer } \end{array}\right.

3:\left\{\begin{array}{l} 1: \text { integral } \\ 1: \text { antiderivative } \\ 1: \text { answer } \end{array}\right.

[Maximum number: 2]

The temperature of water in a tub at time t is modeled by a strictly increasing, twice-differentiable function W, where W(t) is measured in degrees Fahrenheit and t is measured in minutes. At time t=0, the temperature of the water is 55°F. The water is heated for 30 minutes, beginning at time t=0. Values of W(t) at selected times t for the first 20 minutes are given in the table above.

Use the data in the table to evaluate $\int_{0}^{20} W^{\prime}(t) d t$ . Using correct units, interpret the meaning of $\int_{0}^{20} W^{\prime}(t) d t$ in the context of this problem.

$\int_{0}^{20} W^{\prime}(t) d t=W(20)-W(0)=71.0-55.0=16$
The water has warmed by 16 °F over the interval from t=0 to t=20 minutes.

[Maximum number: 3]

The graphs of the functions f and g are shown in the figure for $0 \leq x \leq 3$ . It is known that $g(x)=\frac{12}{3+x}$ for $x \geq 0$ . The twice-differentiable function f, which is not explicitly given, satisfies f(3)=2 and $\int_{0}^{3} f(x) d x=10$ .

Find the area of the shaded region enclosed by the graphs of f and g.

Area = \int_{0}^{3}(f(x)-g(x)) d x=\int_{0}^{3} f(x) d x-\int_{0}^{3} g(x) d x\)

Integrand

1 point

$=10-\int_{0}^{3} \frac{12}{3+x} d x=10-12[\ln |3+x|]_{0}^{3}$

Antiderivative of
g(x)

1 point

$=10-12(\ln 6-\ln 3)=10-12(\ln 2)$

Answer

1 point

Scoring notes:

- The first point is earned for any of the integrands f(x)-g(x), g(x)-f(x),|f(x)-g(x)|, or

|g(x)-f(x)| in any definite integral. If the limits are incorrect, the response does not earn the

third point.

- The first point is earned with an implied integrand for f and explicit integrand for g, such as

$10-\int_{0}^{3} g(x) d x$ .

- The second point is earned for finding $a \int \frac{d x}{3+x}=a \cdot \ln |3+x|$ or $a \cdot \ln (3+x)$ .

- A response is eligible for the third point only if it has earned the first 2 points. The third point is

earned only for the correct answer. The answer does not need to be simplified; however, if

simplification is attempted, it must be correct.

- A response is not eligible for the third point with incorrect limits of integration for u-substitution, for

example, $\int_{0}^{3} \frac{12}{3+x} d x=\int_{0}^{3} \frac{12}{u} d u=12[\ln (x+3)]_{0}^{3}$ .

- A response with incorrect communication, such as "Area $=\int_{0}^{3}(g(x)-f(x)) d x=10-12(\ln 2)$ ,"

does not earn the third point. However, a response of " $\int_{0}^{3}(g(x)-f(x)) d x=12(\ln 2)-10$ , so the

area is $10-12(\ln 2)$ " earns all 3 points.

Total for part (a) 3 points

(b) Evaluate the improper integral $\int_{0}^{\infty}(g(x))^{2} d x$ , or show that the integral diverges.

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Covered syllabus points

Unit 8: Applications of Integration question 1

Unit 8: Applications of Integration question 2

Question (a)

Question (b)

Unit 8: Applications of Integration question 3

Unit 8: Applications of Integration question 4