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Calculus AB : Find Average Value

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Example Questions

Example Question #1 : Find Average Value

Which of the following theorems is related to finding the Average Value of a Function?

Possible Answers:

Mean Value Theorem for Integrals

Extreme Value Theorem

Fundamental Theorem of Calculus

Intermediate Value Theorem

Correct answer:

Mean Value Theorem for Integrals

Explanation:

The following equation is used for finding the Average Value of a Function: $f_{avg}=\frac{1}{b-a} \int_{a}^{b} f(x) dx$ . A rearrangement of this equation could be multiplying (b-a) to both sides. Making this rearrangement, and substituting $f_{avg}=f(c)$ with a<c<b , results in the following: $\int_{a}^{b}f(x) dx=f(c)(b-a)$ . Assuming f(x) is continuous, this is the correct equation for the Mean Value Theorem for Integrals.

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Example Question #2 : Find Average Value

Find the average value of the function f(x)=2x^7+5x^2+4 over the interval [1,3] . Round to the nearest hundredth.

Possible Answers:

1691.33

846.50

103.45

845.67

Correct answer:

845.67

Explanation:

When finding the average value of a function, it is useful to keep the following formula in mind: $f_{avg}=\frac{1}{b-a}\int_{a}^{b}f(x) dx$ . This equation is very helpful because it provides a simple way to determine the average value by substituting in values of the bounds and the function itself:

$f_{avg}=\frac{1}{3-1}\int_{1}^{3}2x^7+5x^2+4 dx=(12)[\frac{1}{4}x^8+\frac{5}{3}x^3+4x] |_1^3$ $=(\frac{1}{2})(\frac{6561}{4}+45+12-1\frac{}{4}-\frac{5}{3}-4)=845.67$

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Example Question #3 : Find Average Value

Find the average value of the function $f(x)=4e^{2x}$ over the interval [0,5]

Possible Answers:

$\frac{2}{5}(e^{10}-1)$

2e^10

e^5

$\frac{2}{5}(e^5-1)$

Correct answer:

$\frac{2}{5}(e^{10}-1)$

Explanation:

$f_{avg}=\frac{1}{5-0}\int_{0}^{5}4e^{2x} dx=(\frac{1}{5})[4e^{2x} * \frac{1}{2}] |_0^5 =\frac{2}{5}e^{2(5)}-\frac{2}{5}e^{2(0)}=\frac{2}{5}(e^{10}-1)$

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Example Question #4 : Find Average Value

Find the average value of the function f(t)=4+2cos(2t) over the interval $[0,\frac{\pi}{4}]$ .

Possible Answers:

$4-\frac{4}{\pi}$

$\frac{\pi}{4} + 2$

$1+\frac{4}{\pi}$

$4+\frac{4}{\pi}$

Correct answer:

$4+\frac{4}{\pi}$

Explanation:

When finding the average value of a function, it is useful to keep the following formula in mind: $f_{avg}=\frac{1}{b-a}\int_{a}^{b}f(x) dx$ . This equation allows the substitution of the function and interval to solve for the average value. While the function in this problem contains a trigonometric function, the same approach can be applied. Remember that the function is in terms of t, so the definite integral expression should likewise be in terms of .

$f_{avg}=\frac{1}{\frac{\pi}{4}-0}\int_{0}^{\frac{\pi}{4}}4+2cos(2t) dt=(\frac{4}{\pi})[4t+sin(2t)] |_0^{\frac{\pi}{4}}$ $=(\frac{4}{\pi})[\pi +1-0-0]=4+\frac{4}{\pi}$

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Example Question #5 : Find Average Value

Identify the correct integral expression for the average value of the function f(t)=6-3tsin(t^2) over the interval $[0,\pi]$ .

Possible Answers:

$\frac{1}{\pi}\int_{0}^{\pi}6-3tsin(t^2) dt$

$\int_{0}^{\pi}6-3tsin(t^2) dx$

$\frac{1}{\pi}\int_{0}^{\pi}6-3xsin(x^2) dt$

$\int_{0}^{\pi}6-3tsin(t^2) dt$

Correct answer:

$\frac{1}{\pi}\int_{0}^{\pi}6-3tsin(t^2) dt$

Explanation:

When finding the average value of a function, it is useful to keep the following formula in mind: $f_{avg}=\frac{1}{b-a}\int_{a}^{b}f(x) dx$ . This equation allows the substitution of the function and interval to solve for the average value. While the function in this problem contains a trigonometric function, the same approach can be applied. Remember that the function is in terms of , so the definite integral expression should likewise be in terms of .

$f_{avg}=\frac{1}{\pi-0}\int_{0}^{\pi}6-(3t)sin(t^2) dt =\frac{1}{\pi}\int_{0}^{\pi}6-3tsin(t^2) dt$

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Example Question #6 : Find Average Value

Find the average value of the function f(x)=2x^5-6x^3+7 over the interval [1,2] .

Possible Answers:

5.5

7.5

Correct answer:

5.5

Explanation:

Because the objective of this problem is to find the average value of the function, the formula f $f_{avg}=\frac{1}{b-a}\int_{a}^{b}f(x) dx$ will be useful. Since the interval and function are provided, this problem consists of recognizing the base components and making the appropriate substitutions:

$f_{avg}=\frac{1}{b-a}\int_{a}^{b}f(x) dx=\frac{1}{2-1}\int_{1}^{2}2x^5-6x^3+7 dx =[\frac{1}{3}x^6-\frac{3}{2}x^4+7x] |_1^2$ $=\frac{64}{3}-24+14-\frac{1}{3}+\frac{3}{2}-7=5.5$

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Example Question #7 : Find Average Value

Identify the correct integral expression for the average value of the function f(x)=1x+2x over the interval [1,4] .

Possible Answers:

$\frac{1}{3}\int_{1}^{4}\frac{1}{x}+\sqrt{2x} dx$

$\frac{1}{3}\int_{1}^{4}\frac{1}{t}+\sqrt{2t} dx$

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

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

Correct answer:

$\frac{1}{3}\int_{1}^{4}\frac{1}{x}+\sqrt{2x} dx$

Explanation:

When finding the average value of a function, it is useful to keep the following formula in mind: $f_{avg}=\frac{1}{b-a}\int_{a}^{b}f(x) dx$ . This equation allows the substitution of the function and interval to solve for the average value. Because the function indicated in the problem is in terms of , the definite integral expression should also be in terms of .

$f_{avg}=\frac{1}{4-1}\int_{1}^{4}\frac{1}{x}+\sqrt{2x} dx =\frac{1}{3}\int_{1}^{4}\frac{1}{x}+\sqrt{2x} dx$

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Example Question #8 : Find Average Value

Find the average value of the function f(x)=20x+x^3 over the interval [-1,1] .

Possible Answers:

Correct answer:

Explanation:

When finding the average value of a function, it is useful to keep the following formula in mind: $f_{avg}=\frac{1}{b-a}\int_{a}^{b}f(x) dx$ . This equation allows the substitution of the function and interval to solve for the average value. Because the function indicated in the problem is in terms of , the definite integral expression should also be in terms of .

$f_{avg}=\frac{1}{1-(-1)}\int_{-1}^{1}20x+x^3 dx$ $=\frac{1}{2}[10x^2+(\frac{1}{4})(x)^4] |_-1^1=\frac{1}{2}(10+\frac{1}{4}-10-\frac{1}{4})=0$

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Example Question #9 : Find Average Value

Let f(x)=4x+11 . What value of c allows the average value of f(x) over the interval [0,c] to be ?

Possible Answers:

Correct answer:

Explanation:

$27=\frac{1}{c-0}\int_{c}^{0}4x +11 dx$

Next, the definite integral can be taken to continue solving for .

$27=\frac{1}{c}[2x^2+11x] |_0^c$

$27=(\frac{1}{c})(2c^2+11c)$

27=2c+11

c=8

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Example Question #10 : Find Average Value

Let f(x)=3x2+4x-3 . What value of allows the average value of f(x) over the interval [0,c] to be ?

Possible Answers:

Correct answer:

Explanation:

$12=\frac{1}{c-0}\int_{0}^{c}3x^2+4x-3 dx$

Next, the definite integral can be taken to continue solving for .

$12=\frac{1}{c}[x^3+2x^2-3x] |_0^c$

$12=(\frac{1}{c})(c^3+2c^2-3c)$

12=c^2+2c-3

0=c^2+2c-15

0=(c+5)(c-3)

c={-5,3}

Because the problem states that c>0 , the answer c=-5 can be eliminated. Therefore, the correct answer is c=3 .

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Calculus AB : Find Average Value

Study concepts, example questions & explanations for Calculus AB

All Calculus AB Resources

Example Questions

Example Question #1 : Find Average Value

Example Question #2 : Find Average Value

Example Question #3 : Find Average Value

Example Question #4 : Find Average Value

Example Question #5 : Find Average Value

Example Question #6 : Find Average Value

Example Question #7 : Find Average Value

Example Question #8 : Find Average Value

Example Question #9 : Find Average Value

Example Question #10 : Find Average Value

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