Answer to Question #313340 in Calculus for Jhonny

Question

Find favg for the functions given on the interval and determine the value of c in the given

interval for which f(c) = favg.

a) f(x) = 9 − 2e^(4x+1) on [2,6]

b) 8 − cos (x/4) on [0 4π]

Answer 1

Answer

We know that

${f_{avg}} = \frac{1}{{b - a}}\int\limits_a^b {f\left( x \right)\,dx} \$

Solution (a) 

$f\left( x \right) = 9 - 2{e^{4x + 1}}$ for the interval $[2, 6]$

$\begin{gathered} {f_{avg}} = \frac{1}{{6 - 2}}\int\limits_2^6 {\left( {9 - 2{e^{4x + 1}}} \right)dx} \\ {f_{avg}} = \frac{9}{4}\int\limits_2^6 {\left( 1 \right)dx} - \frac{2}{4}\int\limits_2^6 {\left( {{e^{4x + 1}}} \right)dx} \\ {f_{avg}} = \frac{9}{4}\left[ x \right]_2^6 - \frac{1}{2}\left[ {\frac{{{e^{4x + 1}}}}{4}} \right]_2^6 \\ {f_{avg}} = \frac{9}{4}\left[ {6 - 2} \right] - \frac{1}{8}\left[ {{e^{4x + 1}}} \right]_2^6 \\ {f_{avg}} = \frac{9}{4}\left[ 4 \right] - \frac{1}{8}\left[ {{e^{4\left( 6 \right) + 1}} - {e^{4\left( 2 \right) + 1}}} \right] \\ {f_{avg}} = 9 - \frac{1}{8}\left[ {{e^{25}} - {e^9}} \right] \\ {f_{avg}} = 9 - \frac{1}{8}\left[ {{e^{25}} - {e^9}} \right] \\ {f_{avg}} = - 9000611395 \\ \end{gathered}$

Then for $f(c)=f_{avg}$ means

$f\left( c \right) = {f_{avg}} \\ 9 - 2{e^{4c + 1}} = - 9000611395 \\ 2{e^{4c + 1}} = 9 + 9000611395 \\ 2{e^{4c + 1}} = 9000611404 \\ {e^{4c + 1}} = 4500305702 \\ 4c + 1 = \ln \left( {4500305702} \right) \\ 4c + 1 = 22.22741117 \\ 4c = 21.22741117 \\ c = 5.306852791 \\$

Solution (b) 

$f\left( x \right) = 8 - \cos \left( {\tfrac{x}{4}} \right)$ for the interval $[0, 4\pi]$

${f_{avg}} = \frac{1}{{4\pi - 0}}\int\limits_0^{4\pi } {\left( {8 - \cos \left( {\tfrac{x}{4}} \right)} \right)dx} \\ {f_{avg}} = \frac{1}{{4\pi }}\int\limits_0^{4\pi } {\left( 8 \right)dx} + \frac{1}{{4\pi }}\int\limits_0^{4\pi } {\left( {\cos \left( {\tfrac{x}{4}} \right)} \right)dx} \\ {f_{avg}} = \frac{8}{{4\pi }}\left[ x \right]_0^{4\pi } + \frac{1}{{4\pi }}\left[ {\frac{{\sin \left( {\tfrac{x}{4}} \right)}}{{\tfrac{1}{4}}}} \right]_0^{4\pi } \\ {f_{avg}} = \frac{2}{\pi }\left[ {4\pi - 0} \right] + \frac{4}{{4\pi }}\left[ {\sin \left( {\tfrac{x}{4}} \right)} \right]_0^{4\pi } \\ {f_{avg}} = 8 + \frac{1}{\pi }\left[ {\sin \left( {\tfrac{{4\pi }}{4}} \right) - \sin \left( {\tfrac{0}{4}} \right)} \right] \\ {f_{avg}} = 8 + \frac{1}{\pi }\left[ {0 - 0} \right] \\ {f_{avg}} = 8 \\$

Now for

$f\left( c \right) = {f_{avg}}\\ 8 - \cos \left( {\tfrac{c}{4}} \right) = 8 \\ \cos \left( {\tfrac{c}{4}} \right) = 0 \\ \frac{c}{4} = \frac{\pi }{2} \\ c = 2\pi \\$