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Calculus 2 : Integral Applications

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Example Questions

Example Question #301 : Integrals

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

Possible Answers:

2.486

2.303

2.198

2.442

2.667

Correct answer:

2.303

Explanation:

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

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

$f (x) = \ln (x + 1)$ , so

$f'(x) = \frac{1}{x+1}$ , and the integral to be approximated is

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

$= \int_{0}^{2} \sqrt{\frac{x^{2}+2x+1}{x^{2}+2x+1} + \frac{1}{x^{2}+2x+1} } \; \mathrm{d}x$

$= \int_{0}^{2} \sqrt{\frac{x^{2}+2x+2}{x^{2}+2x+1} } \; \mathrm{d}x$

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 $g(x) = \sqrt{\frac{x^{2}+2x+2}{x^{2}+2x+1}$

We evaluate at these points, then substitute:

$g(0) = \sqrt{\frac{0^{2}+2 \cdot 0+2}{0^{2}+2 \cdot 0+1} } \approx 1.4142$

$g(0.5) = \sqrt{\frac{0.5^{2}+2 \cdot 0.5+2}{0.5^{2}+2 \cdot 0.5+1} } \approx 1.2019$

$g(1) = \sqrt{\frac{1^{2}+2 \cdot 1+2}{1^{2}+2 \cdot 1+1} } \approx 1.1180$

$g(1.5) = \sqrt{\frac{1.5^{2}+2 \cdot 1.5+2}{1.5^{2}+2 \cdot 1.5+1} } \approx 1.0770$

$g(2) = \sqrt{\frac{2^{2}+2 \cdot 2+2}{2^{2}+2 \cdot 2+1} } \approx 1.0541$

The approximation is therefore

$\approx \frac{0.5 }{3} \left ( 1.4142 +4 \cdot 1.2019 + 2\cdot 1.1180 +4 \cdot 1.0770 +1.0541 \right )$

$\approx \frac{0.5 }{3} \left ( 1.4142 +4.8076 + 2.2360 +4.3080 +1.0541 \right )$

$\approx \frac{0.5 }{3} \left (13.8199 \right ) \approx 2.303$

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Example Question #1 : Length Of Curve, Distance Traveled, Accumulated Change, Motion Of Curve

Give the arclength of the graph of the function $f(x) = \frac{2}{3} x \sqrt{x}$ on the interval $\left [ 0,3 \right ]$ .

Possible Answers:

$\frac{11}{3}$

$\frac{14}{3}$

$\frac{16}{3}$

Correct answer:

$\frac{14}{3}$

Explanation:

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

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

$f(x) = \frac{2}{3} x \sqrt{x} = \frac{2}{3} x^{\frac{3}{2}}$

$f(x) = \frac{2}{3} \cdot \frac{3}{2} x^{\frac{1}{2}} = \sqrt{x}$ .

The above integral becomes

$\int_{0}^{3} \sqrt{1 + (\sqrt{x}) ^{2}} \; \mathrm{d}x$

$\int_{0}^{3} \sqrt{x+1} \; \mathrm{d}x$

Substitute u = x + 1 . Then $\frac{\mathrm{d} u}{\mathrm{d} x} = 1$ , $\mathrm{d} u = \mathrm{d} x$ , and the integral becomes

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

$= \left.\begin{matrix} \frac{u^{\frac{3}{2}}}{\frac{3}{2}} \\ \\ \end{matrix}\right| \begin{matrix} 4\\ \\ 1 \end{matrix}$

$= \left.\begin{matrix} \frac{2u \sqrt{u}}{3} \\ \\ \end{matrix}\right| \begin{matrix} 4\\ \\ 1 \end{matrix}$

$= \frac{2 \cdot 4 \sqrt{4}}{3} - \frac{2 \cdot 1 \sqrt{1}}{3}$

$= \frac{2 \cdot 4 \cdot 2 }{3} - \frac{2 \cdot 1 \cdot 1}{3} = \frac{16}{3} - \frac{2}{3} =\frac{14}{3}$

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Example Question #1 : Length Of Curve, Distance Traveled, Accumulated Change, Motion Of Curve

Give the arclength of the graph of the function $f(x) = \frac{1}{2} \ln \left ( \cos 2 x \right )$ on the interval $\left [ 0, \frac{\pi}{8} \right ]$ .

Possible Answers:

$1 + \sqrt{2}$

$\ln \left ( 1 + \sqrt{2} \right )$

$\frac{1}{2} \ln \left ( 1 + \sqrt{2} \right )$

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

$\frac{1 }{2}+\frac{1 }{2} \sqrt{2}$

Correct answer:

$\frac{1}{2} \ln \left ( 1 + \sqrt{2} \right )$

Explanation:

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

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

$f(x) = \frac{1}{2} \ln \left ( \cos 2 x \right )$ , so

$f'(x) = \frac{1}{2} \cdot \frac{\frac{\mathrm{d} }{\mathrm{d} x}\cos 2 x}{\cos 2 x} = \frac{1}{2} \cdot \frac{-2 \sin 2x}{\cos 2 x} = - \tan 2x$

The integral becomes

$\int_{0}^{ \frac{\pi}{8}} \sqrt{1 + \left ( - \tan 2x \right )^{2}} \; \mathrm{d}x$

$= \int_{0}^{ \frac{\pi}{8}} \sqrt{1 + \tan ^{2} 2x } \; \mathrm{d}x$

$= \int_{0}^{ \frac{\pi}{8}} \sqrt{\sec ^{2} 2x } \; \mathrm{d}x$

$= \int_{0}^{ \frac{\pi}{8}} \sec 2x \; \mathrm{d}x$

Use substitution - set u = 2x . Then $\frac{\mathrm{d} u}{\mathrm{d} x} = 2$ , and $\mathrm{d} x = \frac{1}{2}\mathrm{d} u$ . The bounds of integration become $2 \cdot 0 = 0$ and $2 \cdot \frac{\pi}{8} = \frac{\pi}{4}$ , and the integral becomes

$\frac{1}{2} \int_{0}^{ \frac{\pi}{4}} \sec u \; \mathrm{d}u$

$=\left.\begin{matrix}\frac{1}{2} \ln | \sec u + \tan u | \\ \\ \end{matrix}\right|\begin{matrix} \frac{\pi}{4}\\ \\ 0 \end{matrix}$

$=\frac{1}{2} \ln \left | \sec \frac{\pi}{4} + \tan \frac{\pi}{4} \right | - \frac{1}{2} \ln | \sec 0 + \tan 0 |$

$=\frac{1}{2} \ln \left | \sqrt{2} + 1 \right | -\frac{1}{2} \ln | 1 + 0 |$

$=\frac{1}{2} \ln \left ( 1 + \sqrt{2} \right ) -\frac{1}{2} \ln 1$

$=\frac{1}{2} \ln \left ( 1 + \sqrt{2} \right ) - 0$

$=\frac{1}{2} \ln \left (1 + \sqrt{2} \right )$

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Example Question #131 : Integral Applications

Give the arclength of the graph of the function $f(x) = \frac{1}{3} \left ( \cosh 3 x \right )$ on the interval $\left [ 0, 2 \right ]$ .

Possible Answers:

$\frac{ e^{6} +1 }{6e^{3}}$

$\frac{ e^{12} + 1 }{6e^{6}}$

$\frac{ e^{6} - 1 }{6e^{3}}$

$2e^{6}$

$\frac{ e^{12} - 1 }{6e^{6}}$

Correct answer:

$\frac{ e^{12} - 1 }{6e^{6}}$

Explanation:

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

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

$f(x) = \frac{1}{3} \left ( \cosh 3 x \right )$ , so

$f'(x) = \frac{1}{3} \cdot 3 \sinh 3 x = \sinh 3 x$

and the integral becomes

$\int_{0}^{2} \sqrt{1 + \sinh^{2} 3 x } \; \mathrm{d}x$

$= \int_{0}^{2} \sqrt{ \cosh^{2} 3 x } \; \mathrm{d}x$

$= \int_{0}^{2} \cosh 3 x \; \mathrm{d}x$

Use substitution - set u =3x . Then $\frac{\mathrm{d} u}{\mathrm{d} x} = 3$ , and $\mathrm{d} x = \frac{1}{3}\mathrm{d} u$ . The bounds of integration become 0 and 6; the integral becomes

$= \int_{0}^{6} \frac{1}{3} \cosh u \; \mathrm{d}u$

$=\frac{1}{3} \int_{0}^{6} \cosh u \; \mathrm{d}u$

$= \frac{1}{3} \cdot \left.\begin{matrix} \sinh u\\ \\ \end{matrix}\right|\left.\begin{matrix} 6\\ \\ 0 \end{matrix}\right$

$=\frac{1}{3}\left ( \sinh 6 - \sinh 0 \right ) = \frac{1}{3} \sinh 6 =\frac{1}{3 } \cdot \frac{ e^{6} - e^{-6}}{2} = \frac{ \left (e^{6} - e^{-6} \right )e^{6}}{6e^{6}} = \frac{ e^{12} - 1 }{6e^{6}}$

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Example Question #133 : Integral Applications

What is the average value of the function $f(x) = \tan x$ on the interval $\left [\frac{ \pi }{6} , \frac{ \pi }{3}\right ]$ ?

Possible Answers:

$\frac{ \pi \ln 3}{6}$

$\frac{3 \ln 3}{\pi}$

$\frac{6 \ln 3}{\pi}$

$\frac{ \ln 3}{\pi}$

$\frac{ \pi \ln 3}{3}$

Correct answer:

$\frac{3 \ln 3}{\pi}$

Explanation:

The average value of the function $f(x) = \tan x$ on the interval $\left [\frac{ \pi }{6} , \frac{ \pi }{3}\right ]$ is equal to

$\frac{1}{ \frac{ \pi}{3}-( \frac{ \pi}{6})} \int_{ \frac{ \pi}{6}}^{ \frac{ \pi}{3}}\tan x \cdot \mathrm{d}x$

$= \frac{1}{ \frac{ \pi}{6}} \int_{ \frac{ \pi}{6}}^{ \frac{ \pi}{3}}\tan x \cdot \mathrm{d}x$

$= \frac{6}{ \pi} \int_{ \frac{ \pi}{6}}^{ \frac{ \pi}{3}}\tan x \cdot \mathrm{d}x$

$= \frac{6}{ \pi} \cdot \left.\begin{matrix} \ln\left | \sec x \right |\\ \\ \end{matrix} \right | \begin{matrix} \frac{\pi}{3}\\ \\ \frac{ \pi}{6} \end{matrix}$

$= \frac{6}{ \pi} \left ( \ln\left | \sec \frac{ \pi}{3} \right | - \ln\left | \sec \frac{ \pi}{6} \right | \right )$

$= \frac{6}{ \pi} \left ( \ln 2 - \ln \frac{2\sqrt{3}}{3} \right )$

$= \frac{6}{ \pi} \ln \left (2 \div \frac{2\sqrt{3}}{3} \right )$

$= \frac{6}{ \pi} \ln \sqrt{3}= \frac{6}{ \pi} \cdot \frac{1}{2} \cdot \ln 3 = \frac{3 \ln 3}{\pi}$

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Example Question #131 : 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:

$4 \ln 3$

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

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

$3 \ln 4$

$\ln 3$

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:

2e^2 -e

3e^3 -2e^2

e^2

e^3

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 #313 : Integrals

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:

2.304

1.468

2.162

1.082

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 #132 : 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 #138 : 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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Calculus 2 : Integral Applications

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Example Question #301 : Integrals

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