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Calculus 2 : Integrals

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Example Questions

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 #11 : 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

$\frac{\pi}{2}$

$2\pi-3$

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 #11 : Average Values And Lengths Of Functions

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

Possible Answers:

1255

1250

502

2510

251

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 #151 : Integral Applications

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

Possible Answers:

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

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

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

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

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

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 #161 : Integral Applications

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

Possible Answers:

$\frac{295}{30}$

$\frac{11}{4}$

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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Example Question #21 : Average Values And Lengths Of Functions

You will need a calculator for this question.

Find the surface area of revolution formed by the function, f(x)=x^2+1 , from x=0 to x=2, when revolved around the y axis. Round to the nearest hundredth.

Possible Answers:

$\approx 13.61$

$\approx 36.18$

$\approx 18.09$

$\approx 72.36$

Correct answer:

$\approx 36.18$

Explanation:

To find the surface area of a revolution, we must use the following formula.

$SA= 2\pi \int_{a}^{b} r \cdot ds$ , where "r" is the variable radius of the curve from the axis of revolution, and "ds" is the differential of the arc length of the curve at that radius. For this problem, "r" is simply the variable "x", because every point's radius from the axis of revolution (the y axis), is its x coordinate.

You can think of "ds" as the arc length formula after it is differentiated. Arc length is an integral, so the derivative of an integral is just what's inside that integral. Here that means $ds=\sqrt{1+(f'(x))^{2}} \cdot dx$ .

"a" and "b" will be the x=0 and x=2 part of the question.

From this information we can rewrite the surface area formula as

$SA = 2 \pi \int_{0}^{2} x \cdot \sqrt{1 +(f'(x))^2} \cdot dx$

Now we will find f'(x).

$f'(x)= \frac{\mathrm{d} }{\mathrm{d} x}(x^2 + 1)=2x$

Plug in to the formula.

$SA = 2 \pi \int_{0}^{2} x \cdot \sqrt{1 +(2x)^2} \cdot dx$

Now simplify (2x)^2 .

$SA = 2 \pi \int_{0}^{2} x \cdot \sqrt{1 +4x^2} \cdot dx$

Notice that the is part of the derivative of what is under the radical. This almost follows the basic integral form $\int{u^n \cdot du}$ ,with u= 1 + 4x^2 , and du = 8x . Let's force it to match the form exactly. First I will rewrite the square root as an exponent.

$SA = 2 \pi \int_{0}^{2} x \cdot (1 +4x^2)^{\frac{1}{2}} \cdot dx$

Now multiply by a form of 1 (highlighted in red) to force the "du" part of the integration form.

$SA = 2 \pi \cdot {\color{Red} \frac{1}{8}} \int_{0}^{2} {\color{Red} 8} \cdot x \cdot (1 +4x^2)^{\frac{1}{2}} \cdot dx$

It may help to rewrite the expression in terms of to make it easier to apply integration rules. We just need to remember to convert back to x after integrating, before we plug in the bounds.

$2\pi \cdot \frac{1}{8} \int u^\frac{1}{2} du$

Now integrate.

$\frac{\pi}{4} \cdot [\frac{2}{3}u^{\frac{3}{2}}]$

Now convert u back to x. Recall that u= 1 + 4x^2

$SA = \frac{\pi}{4} \cdot [\frac{2(1 + 4x^2)^{\frac{3}{2}}}{3}]|_{0}^{2}$

Now plug in the upper and lower bounds.

$SA = \frac{\pi}{4} \cdot [(\frac{2(1 +4(2)^2)^{\frac{3}{2}}}{3})-(\frac{2(1 + 4(0)^2)^{\frac{3}{2}}}{3})]$

Now use a calculator to find the approximate answer of

$SA \approx 36.18$

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Example Question #21 : Average Values And Lengths Of Functions

Evaluate the average value of the function on the interval [0,10] .

f(x)=2x

Possible Answers:

Correct answer:

Explanation:

To solve for the average value on the interval [a,b] we follow the following formula

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

For this problem we evaluate

$\frac{\int_{0}^{10}2x\,dx}{10-0}$

$=\frac{x^2|_0^{10}}{10}$

$=\frac{10^2-0^2}{10}$

$=\frac{100}{10}$

=10

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Example Question #22 : Average Values And Lengths Of Functions

Evaluate the average value of the function on the interval [0,3] .

f(x)=3x^2

Possible Answers:

Correct answer:

Explanation:

To solve for the average value on the interval [a,b] we follow the following formula

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

For this problem we evaluate

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

$=\frac{x^3|_0^{3}}{3}$

$=\frac{3^3-0^3}{3}$

$=\frac{27}{3}$

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Calculus 2 : Integrals

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #14 : Average Values And Lengths Of Functions

Example Question #11 : Average Values And Lengths Of Functions

Example Question #16 : Average Values And Lengths Of Functions

Example Question #11 : Average Values And Lengths Of Functions

Example Question #151 : Integral Applications

Example Question #161 : Integral Applications

Example Question #20 : Average Values And Lengths Of Functions

Example Question #21 : Average Values And Lengths Of Functions

Example Question #21 : Average Values And Lengths Of Functions

Example Question #22 : Average Values And Lengths Of Functions

All Calculus 2 Resources

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