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Calculus 2 : Integrals

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Example Questions

Example Question #132 : Integral Applications

What is the average value of the function $f(x) = \frac{1}{x + 4}$ on the interval $\left [ -2, 2\right ]$ ?

Possible Answers:

$\frac{1}{4} \ln 3$

$\ln 3$

$3 \ln 4$

$4 \ln 3$

$\frac{1}{3} \ln 4$

Correct answer:

$\frac{1}{4} \ln 3$

Explanation:

The average value of the function $f(x) = \frac{1}{x + 4}$ on the interval $\left [ -2, 2\right ]$ is equal to

$\frac{1}{2-(-2)} \int_{-2}^{2} \frac{\mathrm{d}x}{x + 4}$

$=\frac{1}{4} \int_{-2}^{2} \frac{\mathrm{d}x}{x + 4}$

Substitute u = x+ 4 . Then $\frac{\mathrm{d} u}{\mathrm{d} x} = 1$ , and $\mathrm{d} x = \mathrm{d} u$ ; the bounds of integration become 2 and 6, and the above expression becomes

$=\frac{1}{4} \int_{2}^{6} \frac{\mathrm{d}u}{u}$

$=\frac{1}{4} \cdot \left.\begin{matrix} \ln |u| \\ \\ \end{matrix}\right| \begin{matrix} 6\\ \\ 2 \end{matrix}$

$=\frac{1}{4} \cdot \left ( \ln 6 - \ln 2 \right )$

$=\frac{1}{4} \ln 3$

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Example Question #131 : Integral Applications

What is the average value of the function f(x) = xe^x on the interval $\left [2,3 \right ]$ ?

Possible Answers:

e^2

2e^2 -e

e^3

3e^3 -2e^2

2e^3 -e^2

Correct answer:

2e^3 -e^2

Explanation:

The average value of the function f(x) = xe^x on the interval $\left [0, 1 \right ]$ is equal to

$\frac{1}{ 3-2 }\int_{ 2}^{3} xe^x \; \mathrm{d}x$

$=\int_{ 2}^{3} xe^x \; \mathrm{d}x$

For now, we look at the indefinite integral $\int xe^x \; \mathrm{d}x$ . The integral can be changed by setting u = x and $\mathrm{d}v = e^x \mathrm{d}x$ . Then

$\frac{\mathrm{d} u}{\mathrm{d} x} = 1$ , or $\mathrm{d} u = \mathrm{d} x$

and

$v = \int e^x \mathrm{d}x = e^x$

$\int xe^x \; \mathrm{d}x$

$= \int u\; \mathrm{d} v$

$= uv -\int v\; \mathrm{d} u$

$=xe^x -\int e^x \; \mathrm{d} x$

=xe^x - e^x

which is the antiderivative.

Now, we can find the definite integral, and, subsequently, the average value:

$\int_{ 2}^{3} xe^x \; \mathrm{d}x$

$= \left.\begin{matrix} xe^x - e^x\\ \\ \end{matrix}\right| \begin{matrix} 3\\ \\ 2 \end{matrix}$

=(3e^3 - e^3) - (2e^2 - e^2) = 2e^3 -e^2

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Example Question #132 : Integral Applications

Approximate the length of the curve of $f (x) = \frac{1}{4^{x}}$ on the interval [0,2] . Use Simpson's Parabolic Rule with n = 4 to make your estimate to the nearest thousandth.

Possible Answers:

1.082

2.162

1.468

2.304

2.293

Correct answer:

2.293

Explanation:

The length of the curve of f(x) on the interval [a,b] can be determined by evaluating the integral

$\int_{a}^{b} \sqrt{1 + f'(x) ^{2}} \; \mathrm{d}x$ .

$f (x) = \frac{1}{4^{x}} = e ^{ \ln \frac{1}{4} \cdot x} = e ^{ -\ln 4 \cdot x}$ , so

$f'(x) =-\ln 4 e ^{ -\ln 4 \cdot x} = - \frac{\ln 4}{4^{x}}$ , and the integral to be approximated is

$\int_{0}^{2} \sqrt{1 + \left ( - \frac{\ln 4}{4^{x}} \right ) ^{2}} \; \mathrm{d}x$ , or simplified,

$\int_{0}^{2} \sqrt{1 + \frac{\left ( \ln 4 \right )^{2}}{16^{x}} } \; \mathrm{d}x$ .

We divide the interval [0,2] into four subintervals of width $\Delta x = 2 \div 4 = 0.5$ each, so

$x_{0} = 0, x_{1} = 0.5, x_{2} = 1, x_{3} = 1.5, x_{4} = 2$ .

By Simpson's rule, we can estimate the integral by evaluating

$\frac{\Delta x}{3} \left ( g (x_{0}) + 4g(x_{1}) + 2g(x_{2}) + 4 g (x_{3}) + g (x_{4}) \right )$

$= \frac{0.5 }{3} \left ( g (0) + 4g(0.5) + 2g(1) + 4 g (1.5) + g (2) \right )$

where $g(x) = \sqrt{1 + \frac{\left ( \ln 4 \right )^{2}}{16^{x}} }$ .

We evaluate at these points, then substitute:

$g(0) = \sqrt{1 + \frac{\left ( \ln 4 \right )^{2}}{16^{0}} } = \sqrt{1 + \frac{\left ( \ln 4 \right )^{2}}{1} } \approx \sqrt{1 + \frac{\left ( 1.3863)^{2}}{1} }$

$\approx \sqrt{1 + \frac{1.9218 }{1} } \approx 1.7093$

$g(0.5) = \sqrt{1 + \frac{\left ( \ln 4 \right )^{2}}{16^{0.5}} }= \sqrt{1 + \frac{\left ( \ln 4 \right )^{2}}{4} } \approx \sqrt{1 + \frac{\left ( 1.3863)^{2}}{4} }$

$\approx \sqrt{1 + \frac{1.9218 }{4} } \approx 1.2167$

$g(1) = \sqrt{1 + \frac{\left ( \ln 4 \right )^{2}}{16^{1}} }= \sqrt{1 + \frac{\left ( \ln 4 \right )^{2}}{16 } } \approx \sqrt{1 + \frac{\left ( 1.3863)^{2}}{16} }$

$\approx \sqrt{1 + \frac{1.9218 }{16} } \approx 1.0584$

$g(1.5) = \sqrt{1 + \frac{\left ( \ln 4 \right )^{2}}{16^{1.5}} }= \sqrt{1 + \frac{\left ( \ln 4 \right )^{2}}{64} } \approx \sqrt{1 + \frac{\left ( 1.3863)^{2}}{64} }$

$\approx \sqrt{1 + \frac{1.9218 }{64} } \approx 1.0149$ $g(2) = \sqrt{1 + \frac{\left ( \ln 4 \right )^{2}}{16^{2}} } = \sqrt{1 + \frac{\left ( \ln 4 \right )^{2}}{256} } \approx \sqrt{1 + \frac{\left ( 1.3863)^{2}}{256} }$

$\approx \sqrt{1 + \frac{1.9218 }{256} } \approx 1.0037$

The approximation is therefore

$\approx \frac{0.5 }{3} \left ( 1.7093 +4 \cdot 1.2167 + 2\cdot 1.0584 +4 \cdot 1.0149 +1.0037 \right )$

$\approx \frac{0.5 }{3} \left ( 1.7093 +4.8668 + 2.1168 +4.0596 +1.0037 \right )$

$\approx \frac{0.5 }{3} \left (13.7562 \right ) \approx 2.293$

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Example Question #133 : Integral Applications

Approximate the length of the curve of $f (x) = x^{2}+ x$ on the interval [0,2] . Use Simpson's Parabolic Rule with n = 4 to make your estimate to the nearest thousandth.

Possible Answers:

4.733

4.667

17.068

6.379

5.531

Correct answer:

6.379

Explanation:

The length of the curve of f(x) on the interval [a,b] can be determined by evaluating the integral

$\int_{a}^{b} \sqrt{1 + f'(x) ^{2}} \; \mathrm{d}x$ .

$f(x) = x^{2} + x$ , so

f'(x) =2 x + 1 , and the integral to be approximated is

$\int_{0}^{2} \sqrt{1 + \left ( 2x + 1\right ) ^{2}} \; \mathrm{d}x$

$= \int_{0}^{2} \sqrt{4x ^{2} + 4x + 2} \; \mathrm{d}x$

We divide the interval [0,2] into four subintervals of width $\Delta x = 2 \div 4 = 0.5$ each, so

$x_{0} = 0, x_{1} = 0.5, x_{2} = 1, x_{3} = 1.5, x_{4} = 2$ .

By Simpson's rule, we can estimate the integral by evaluating

$\frac{\Delta x}{3} \left ( g (x_{0}) + 4g(x_{1}) + 2g(x_{2}) + 4 g (x_{3}) + g (x_{4}) \right )$

$= \frac{0.5 }{3} \left ( g (0) + 4g(0.5) + 2g(1) + 4 g (1.5) + g (2) \right )$

where $g(x) = \sqrt{4x ^{2} + 4x + 2}$

We evaluate at these points, then substitute:

$g(0) = \sqrt{4 \cdot 0^{2} + 4\cdot 0 + 2} = \sqrt{ 2}$

$g(0.5) = \sqrt{4 \cdot 0.5^{2} + 4\cdot 0.5 + 2} = \sqrt{ 5}$

$g(1) = \sqrt{4 \cdot 1^{2} + 4\cdot 1 + 2} = \sqrt{ 10}$

$g(0.5) = \sqrt{4 \cdot 1.5^{2} + 4\cdot 1.5 + 2} = \sqrt{ 17}$

$g(2) = \sqrt{4 \cdot 2^{2} + 4\cdot 2 + 2} = \sqrt{ 26}$

The approximation is therefore

$\frac{0.5 }{3} \left ( \sqrt{2} +4 \sqrt{5} +2 \sqrt{10} +4\sqrt{17} +\sqrt{26} \right )$

$\approx \frac{0.5 }{3} \left ( 1.4142 +4 \cdot 2.2361 + 2\cdot 3.1623 +4 \cdot 4.1231 +5.0990 \right )$

$\approx \frac{0.5 }{3} \left ( 1.4142 + 8.9444 + 6.3246 +16.4924 +5.0990 \right )$

$\approx \frac{0.5 }{3} \left ( 38.2745\right ) \approx 6.379$

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Example Question #134 : Integral Applications

What is the average value of the function f(x) = | x^2 - 4 | on the interval [0, 4] ?

Possible Answers:

$3\sqrt{2}$

$2\sqrt{2}$

$5 - \sqrt{2}$

Correct answer:

Explanation:

The average value of the function f(x) = | x^2 - 4 | on the interval [0, 4] is equal to

$\frac{1}{4-0}\int_{0}^{4} | x^2 - 4 | \; \mathrm{d}x$ , or

$\frac{1}{4 }\int_{0}^{4} | x^2 - 4 | \; \mathrm{d}x$ .

To evaluate this integral, we note that x^2 - 4 has negative value on (0, 2) and positive value on (2, 4] , so on [0,4] , f(x) can be defined as

$f(x) = \left\{\begin{matrix} -\left ( x^2 - 4\right ) & x \in [0, 2) \\ x^2 - 4 & x \in \left [ 2, 4\right ] \end{matrix}\right.$

$f(x) = \left\{\begin{matrix} - x^2 + 4 & x \in [0, 2) \\ x^2 - 4 & x \in \left [ 2, 4\right ] \end{matrix}\right.$

Therefore, the average value of is equal to

$\frac{1}{4 }\left [ \int_{0}^{2}\left ( - x^2 + 4 \right ) \; \mathrm{d}x + \int_{2}^{4}\left ( x^2 - 4 \right ) \; \mathrm{d}x \right ]$

$=\frac{1}{4 }\left [ \left.\begin{matrix} \left ( - \frac{x^3}{3} + 4x \right ) \\ \\ \end{matrix}\right| \begin{matrix} 2\\ \\ 0 \end{matrix} + \left.\begin{matrix} \left ( \frac{x^3}{3} - 4x \right ) \\ \\ \end{matrix}\right| \begin{matrix} 4\\ \\ 2 \end{matrix} \right \]$

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

$=\frac{1}{4 }\left [ \left ( - \frac{8}{3} + 8 \right ) -0 + \left ( \frac{64}{3} - 16\right ) - \left ( \frac{8}{3} - 8 \right ) \right \]$

$=\frac{1}{4 }\left [ \frac{16}{3} + \frac{16}{3} - \left (- \frac{16}{3} \right ) \right \]$

$=\frac{1}{4 }\left ( \frac{16}{3} + \frac{16}{3} + \frac{16}{3} \right \) = \frac{1}{4 } \cdot16 = 4$

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Example Question #141 : Integral Applications

What is the average value of the function $f(x) = x \ln x$ on the interval $\left [1,2 \right ]$ ?

Possible Answers:

$\ln 4 - \frac{1}{4}$

$\ln 4 - \frac{3 }{4}$

$\ln 3$

$\ln 2 + \frac{1 }{4}$

$\ln 2 + \frac{3 }{4}$

Correct answer:

$\ln 4 - \frac{3 }{4}$

Explanation:

The average value of the function $f(x) = x \ln x$ on the interval $\left [1,2 \right ]$ is

$\frac{1}{2-1}\int_{1}^{2} x \ln x = \int_{1}^{2} x \ln x$ .

We can integrate the indefinite integral

$\int x \ln x$

as follows:

Set $u = \ln x$ , $\mathrm{d}v = x \; \mathrm{d}x$ .

Then

$\frac{\mathrm{d} u}{\mathrm{d} x} = \frac{\mathrm{d} }{\mathrm{d} x} \ln x = \frac{1}{x}$ or $\mathrm{d} u = \frac{1}{x} \; \mathrm{d} x$

and

$\mathrm{d}v = x \; \mathrm{d}x$ .

$\int \mathrm{d}v = \int x \; \mathrm{d}x$

$v = \frac{x^2}{2}$

Therefore,

$\int x \ln x$

$= \int u \; \mathrm{d}v$

$= uv -\int v\; \mathrm{d} u$

$= \frac{x^2}{2} \cdot \ln x -\int \frac{x^2}{2} \cdot \frac{1}{x} \; \mathrm{d} x$

$= \frac{x^2 \ln x}{2} - \frac{1}{2}\int x \; \mathrm{d} x$

$= \frac{x^2 \ln x}{2} - \frac{1}{2} \cdot \frac{x^2}{2} = \frac{x^2 \ln x}{2} - \frac{x^2}{4}= \frac{2x^2 \ln x - x^2}{4}$

Therefore, the average value is:

$\int_{1}^{2} x \ln x$

$=\left.\begin{matrix} \frac{2x^2 \ln x - x^2}{4}\\ \\ \end{matrix}\right|\begin{matrix} 2\\ \\ 1 \end{matrix}$

$= \frac{2\cdot 2^2 \ln 2 - 2^2}{4} - \frac{2 \cdot 1^2 \ln 1 - 1^2}{4}$

$= \frac{8 \ln 2 -4}{4} - \frac{0 - 1 }{4}$

$= 2 \ln 2 -1 + \frac{1 }{4}$

$= \ln 4 - \frac{3 }{4}$

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Example Question #142 : Integral Applications

What is the average value of the function $f(x) =\frac{\ln x }{x}$ on the interval $\left [1, e\right ]$ ?

Possible Answers:

$\frac{1}{ 2e-1 }$

$\frac{1}{ 2e }$

$\frac{1}{ e-2 }$

$\frac{1}{ e }$

$\frac{1}{ 2e-2 }$

Correct answer:

$\frac{1}{ 2e-2 }$

Explanation:

The average value of the function $f(x) =\frac{\ln x }{x}$ on the interval $\left [1, e\right ]$ is equal to

$\frac{1}{ e-1 }\int_{ 1}^{e} \frac{\ln x }{x} \; \mathrm{d}x$ .

Substitute $u = \ln x$ . Subsequently, $\frac{\mathrm{d} u}{\mathrm{d} x} = \frac{1}{x}$ and $\mathrm{d} u = \frac{1}{x} \mathrm{d} x$ . The bounds of integration become $u(1) = \ln 1 = 0$ and $u (e) = \ln e = 1$ , so the integral can be rewritten as

$\frac{1}{ e-1 }\int_{ 0}^{1} u \; \mathrm{d}u = \frac{1}{ e-1 } \cdot \left.\begin{matrix} \frac{u^2}{2}\\ \\ \end{matrix}\right|\begin{matrix} 1\\ \\ 0 \end{matrix} = \frac{1}{ e-1 } \left ( \frac{1^2}{2} - \frac{0^2}{2} \right ) = \frac{1}{ e-1 } \left ( \frac{1}{2} \right )= \frac{1}{ 2e-2 }$ .

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Example Question #1 : Average Values And Lengths Of Functions

What is the arc length if x=4cos(t) , y=4sin(t) from $[0,2\pi]$ ?

Possible Answers:

$16\pi$

$2\pi$

$8\pi$

$4\pi$

$10\pi$

Correct answer:

$8\pi$

Explanation:

Write the formula for arc length.

$s=\int_{a}^{b}\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}$

Calculate the derivatives.

$\frac{dx}{dt}=-4sin(t)$

$\frac{dy}{dt}=4cos(t)$

Substitute the derivatives and the bounds into the integral.

$s=\int_{0}^{2\pi}\sqrt{(-4sin(t))^2+(4cos(t))^2}=\int_{0}^{2\pi}\sqrt{16sin^2(t)+16cos^2(t)}$

$s=\int_{0}^{2\pi} 4 dt= 4[t]_{0}^{2\pi} = 8\pi$

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Example Question #1 : Average Values And Lengths Of Functions

What is the average value of the function

$p(t)=\frac{4t}{t^2+1}$

from t = 0 to t = 2 ?

Possible Answers:

$\frac{4}{5}$

ln(5)

ln(2)-1

e^2-1

Correct answer:

ln(5)

Explanation:

The average value of a function p(t) from t=a to t=b is found with the integral

$\frac{1}{b-a}\int_{a}^{b}p(t)dt$ .

In this case, we must compute the value of the integral

$\frac{1}{2-0}\int_{0}^{2}\frac{4t}{t^2+1}dt=\frac{1}{2}\int_{0}^{2}\frac{4t}{t^2+1}dt$ .

A substitution makes this integral clearer. Let u=t^2+1 . Then du=2tdt . We should also rewrite the limits of integration in terms of u. When t = 0, u=1, and when t = 2, u = 5. Making these substitutions results in the integral

$\frac{1}{2}\int_{1}^{5}\frac{2du}{u}=\frac{2}{2}\int_{1}^{5}\frac{du}{u}=\int_{1}^{5}\frac{du}{u}$

Evaluating this integral using the fact that

$\int \frac{du}{u}=ln(u)+c$ yields

$\int_{1}^{5}\frac{du}{u}=ln(5)-ln(1)=ln(5)$

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Example Question #1 : Average Values And Lengths Of Functions

What is the average value of the function $f(\theta)=2cos(\theta)$ over the interval $0\le\theta\le\4\pi$ ?

Possible Answers:

$\frac{\sqrt2}{2}$

$\sqrt3$

Correct answer:

Explanation:

In general, the average value of a function f(x) over the interval [a, b] is

$\frac{1}{b-a}\int_{a}^{b}f(x)dx$

This means that the average value of $f(\theta)$ over $0\le\theta\le4\pi$ is

$\frac{1}{4\pi}\int_{0}^{4\pi}2cos(\theta)d\theta$ .

Since the antiderivative of $cos(\theta)$ is $\int cos(\theta)d\theta=sin(\theta)+c$ , the integral evaluates to

$\frac{1}{4\pi}\int_{0}^{4\pi}2cos(\theta)d\theta=\frac{2}{4\pi}sin(\theta)|_{0}^{4\pi}$

$=\frac{1}{2\pi}(sin(4\pi)-sin(0))=0$ .

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Calculus 2 : Integrals

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #132 : Integral Applications

Example Question #131 : Integral Applications

Example Question #132 : Integral Applications

Example Question #133 : Integral Applications

Example Question #134 : Integral Applications

Example Question #141 : Integral Applications

Example Question #142 : Integral Applications

Example Question #1 : Average Values And Lengths Of Functions

Example Question #1 : Average Values And Lengths Of Functions

Example Question #1 : Average Values And Lengths Of Functions

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