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Calculus 2 : Average Values and Lengths of Functions

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Example Question #11 : Average Values And Lengths Of Functions

What is the average value of the function $f(x)=e^{2x}$ on the interval [0,5] ?

Possible Answers:

$\frac{1}{10}e^{10}$

$e^{5}$

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

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

Correct answer:

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

Explanation:

The average value of a function is found by taking the integral of the function over the interval and dividing by the length of the interval.

The general rule for this type of integration is as follows.

$\frac{1}{b-a}\int_a^b e^u dx=\frac{1}{b-a}\left[\frac{1}{\frac{du}{dx}}e^u\right]\left |_a^b$

So we have:

$\frac{1}{5-0}\int_{0}^5 e^{2x} dx=\frac{1}{5}\left(\frac{1}{2}e^{2x}\right)|_{0}^5=\frac{1}{10}(e^{10}-1)$

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Example Question #12 : Average Values And Lengths Of Functions

What is the average value of the function f(x)=2x+3x^2 on the interval [0,2] ?

Possible Answers:

Correct answer:

Explanation:

The average value of a function is found by taking the integral of the function over the interval and dividing by the length of the interval.

The general rule for this type of integration is as follows.

$\frac{1}{b-a}\int_a^b x^n dx=\frac{1}{b-a}\left[\frac{x^{n+1}}{n+1}\right]\left |_a^b$

So we have:

$\frac{1}{2-0}\int_{0}^2 (3x^2+2x) dx=\frac{1}{2}(x^3+x^2)|_{0}^2=\frac{1}{2}(2^2+2^3)=\frac{12}{2}=6$

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Example Question #321 : Integrals

Find the length L of the graph of the function f(x) on the given interval.

$f(x)=2\sqrt{2}x+7\sqrt{3}$

$3 \leq x \leq 5$

Possible Answers:

L=1

L=312

L=6

$L=4\sqrt{39}$

L=18

Correct answer:

L=6

Explanation:

To find the length of a line along a given function f(x) from a to b, utilize the equation

$L=\int^{b}_{a}\sqrt{1+[f'(x)]^{2}} dx$

Applying the equation for this problem, we first find the first derivative of the function:

$f'(x)=2\sqrt{2}$

Then, we use this expression in the above formula:

$L=\int^{5}_{3}\sqrt{1+(2\sqrt{2})^{2}} dx=L=\int^{5}_{3}3dx=3x\left.\begin{matrix} \\ \end{matrix}\right|^{5}_{3}=6$

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Example Question #14 : Average Values And Lengths Of Functions

Let f(x) bet a function continuous on the defined below and having the derivative

$f'(x)=\sqrt{x^{4}-1}$

Find the length L of the graph of f(x) on the interval $1 \leq x \leq 4$

Possible Answers:

L=7

L=21

L=17

L=63

L=13

Correct answer:

L=21

Explanation:

To find the length of a line along a given function f(x) from a to b, utilize the equation

$L=\int^{b}_{a}\sqrt{1+[f'(x)]^{2}} dx$

Applying the equation for this problem and using the given expression for f'(x):

$L=\int^{4}_{1}\sqrt{1+(\sqrt{x^{4}-1})^{2}} dx=\int^{4}_{1}x^{2}dx=\frac{x^{3}}{3}\left.\begin{matrix} \\ \end{matrix}\right|^{4}_{1}=\frac{63}{3}=21$

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Example Question #15 : Average Values And Lengths Of Functions

Find the average value of the function $y = x\sin(x)$ over the interval $[\frac{\pi}{2}, \pi]$ .

Possible Answers:

None of the other answers

$2\pi-3$

$\frac{\pi}{2}$

Correct answer:

None of the other answers

Explanation:

The correct answer is $\frac{2(\pi-1)}{\pi}$ .

To find the average value of the function, we use

Average value $= \frac{1}{b-a}\int_{a}^{b} x\sin(x) dx$ ,

with $a = \frac{\pi}{2}, b=\pi$ in this case.

So we have

$= \frac{1}{\pi - \frac{\pi}{2}}\int_{\pi/2}^{\pi} x\sin(x) dx$

$= \frac{2}{\pi}\int_{\pi/2}^{\pi} x\sin(x) dx$

Using integration by parts with $u = x, dv = \sin(x)dx$ and hence $du=dx, v = -\cos(x)$ , we have

$=\frac{2}{\pi}\{[-x\cos(x)]_{\pi/2}^{\pi}-\int_{\pi/2}^{\pi}-\cos(x)dx\}$

$=\frac{2}{\pi}\{(\pi)- [-\sin(x)]_{\pi/2}^{\pi}\}$

$=\frac{2}{\pi}\{\pi- (0-(-1))\}$

$=\frac{2(\pi-1)}{\pi}$

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Example Question #16 : Average Values And Lengths Of Functions

Find the arc length of the graph $f(x)=\ln(\sec(x))$ on the interval $[-\frac{\pi}{4},\frac{\pi}{4}]$ .

Possible Answers:

$\frac{1}{2}$

$\frac{\sqrt2}{2}$

$\ln\biggr(\frac{\sqrt2-1}{\sqrt2+1}\biggr)$

$\ln\biggr(\frac{\sqrt2+1}{\sqrt2-1}\biggr)$

Correct answer:

$\ln\biggr(\frac{\sqrt2+1}{\sqrt2-1}\biggr)$

Explanation:

The arc length for a function f(x) on the interval [a,b] is given by the following integral:

$L[a,b]=\int_a^b\sqrt{1+(f^{'}(x))^2}dx$ .

Thus we must compute the value of $f^{'}(x)$ for the function f(x)=ln(sec(x)) .

To do so, we use the chain rule:

$f^{'}(x)=\frac{1}{sec(x)}(\sec(x))^{'}=\frac{1}{\sec(x)}\sec(x)\tan(x)=\tan(x)$ .

Now plugging $f^{'}(x)$ to the arc length formula, and using the interval $[-\frac{\pi}{4},\frac{\pi}{4}]$ , we obtain,

$L[-\frac{\pi}{4},\frac{\pi}{4}]=\int_\frac{-\pi}{4}^\frac{\pi}{4}\sqrt{1+\tan^2(x)}dx$ .

Using the trig identity $1+tan^{2}(x)=\sec^{2}(x)$ , our integral becomes:

$\int_\frac{-\pi}{4}^\frac{\pi}{4}\sqrt{\sec^2(x)}dx=\int_\frac{-\pi}{4}^\frac{\pi}{4}\sec(x)dx$

$=\ln\left | \sec(x)+tan(x)) \right |\biggr|_\frac{-\pi}{4}^\frac{\pi}{4}=\ln\left | \sec(\frac{\pi}{4})+tan(\frac{\pi}{4})) \right |-\ln\left | \sec(\frac{-\pi}{4})+tan(\frac{-\pi}{4})) \right |$

$=\ln\left | \sqrt2+1 \right |-\ln\left | \sqrt2-1\right |=\ln\left |\frac{\sqrt2+1}{\sqrt2-1} \right |$ .

In the last step above, we invoked the property

$\ln(a)-\ln(b)=\ln\frac{a}{b}$ .

Since the expression in the absolute value is positive, we can get rid of the absolute value brackets.

Thus we obtain

$L[-\frac{\pi}{4},\frac{\pi}{4}]=\ln\biggr(\frac{\sqrt2+1}{\sqrt2-1}\biggr)$ .

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Example Question #17 : Average Values And Lengths Of Functions

What is the average value of f(x)=x^3+1 on the interval [0,10] ?

Possible Answers:

502

2510

251

1255

1250

Correct answer:

251

Explanation:

To find the average value of a function, you must use the following equation:

$\frac{1}{b-a}\int_{a}^{b}f(x)dx$ . Now, we simply substitute our values, and integrate.

$\frac{1}{10}\int_{0}^{10}x^3+1dx=\frac{1}{10}[\frac{x^4}{4}+x]$ . Now, we plug in our bounds (upper - lower).

$\frac{1}{10}[\frac{(10^4}{4}+10)-(\frac{0^4}{4}+0)]=\frac{1}{10}[2510]=251.$

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Example Question #11 : Average Values And Lengths Of Functions

Find the average value of the function $f(x)=\frac{3}{x}$ from to .

Possible Answers:

$\frac{9}{1-e}$

$\frac{3}{1-e}$

$\frac{3}{e-1}$

$\frac{9}{e-1}$

$\frac{x^3}{e-1}$

Correct answer:

$\frac{3}{e-1}$

Explanation:

The average value equation is $\frac{1}{b-a}\int_{a}^{b}f(x)dx=\frac{1}{e-1}\int_{1}^{e}\frac{3}{x}dx.$

$=\frac{1}{e-1}[3ln(x)].$ Now, all we have to do is plug in our limits.

$\frac{1}{e-1}[3ln(e)-3ln(1)]=\frac{3}{e-1}.$

The second term drops out because the natural log of equals zero. In the first term, natural log and cancel each other out, just leaving the out front.

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Example Question #19 : Average Values And Lengths Of Functions

Determine the average value of the function
f(x)=x^2-3x+9
On the interval [0,5]

Possible Answers:

$\frac{11}{4}$

$\frac{295}{30}$

Correct answer:

$\frac{295}{30}$

Explanation:

The average value of a function on the interval [a,b] is defined as

$\frac{\int_{a}^{b}f(x)\,dx}{b-a}$

As such, we solve the expression

$\frac{\int_{0}^{5}x^2-3x+9\,dx}{5-0}$

$=\frac{(\frac{1}{3}x^3-\frac{3}{2}x^2+9x|_0^5}{5}$

$=\frac{[\frac{1}{3}(5)^3-\frac{3}{2}(5)^2+9(5)]-[\frac{1}{3}(0)^3-\frac{3}{2}(0)^2+9(0)]}{5}$

$=\frac{\frac{295}{6}}{5}$

As such, the average value is

$\frac{295}{30}$

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Example Question #20 : Average Values And Lengths Of Functions

Determine the average value of the function
f(x)=x^2
On the interval [2,6]

Possible Answers:

$\frac{217}{4}$

$\frac{52}{3}$

Correct answer:

$\frac{52}{3}$

Explanation:

The average value of a function on the interval [a,b] is defined as

$\frac{\int_{a}^{b}f(x)\,dx}{b-a}$

As such, we solve the expression

$\frac{\int_{2}^{6}x^2\,dx}{6-2}$

$=\frac{(\frac{1}{3}x^3|_2^6}{4}$

$=\frac{[\frac{1}{3}(6)^3]-[\frac{1}{3}(2)^3]}{4}$

$=\frac{\frac{208}{3}}{4}$

As such, the average value is

$\frac{52}{3}$

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Calculus 2 : Average Values and Lengths of Functions

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #11 : Average Values And Lengths Of Functions

Example Question #12 : Average Values And Lengths Of Functions

Example Question #321 : Integrals

Example Question #14 : Average Values And Lengths Of Functions

Example Question #15 : Average Values And Lengths Of Functions

Example Question #16 : Average Values And Lengths Of Functions

Example Question #17 : Average Values And Lengths Of Functions

Example Question #11 : Average Values And Lengths Of Functions

Example Question #19 : Average Values And Lengths Of Functions

Example Question #20 : Average Values And Lengths Of Functions

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