Question

The mean value of the function $f\left(x\right)=\frac{2}{e^x+1}$ on the interval $[0,2]$ is :

• A. $-2+{\log }_e\left(\frac{2}{e^2-1}\right)$
• B. $2-{\log }_e\left(\frac{2}{e^2+1}\right)$
• C. $2+{\log }_e\left(\frac{2}{e^2+1}\right)$
• D. $2+{\log }_e\left(\frac{2}{e^2-1}\right)$

Answer 1

The correct option is C $2+{\log }_e\left(\frac{2}{e^2+1}\right)$

Mean value of the function $f_{\text{avg}}=\frac{1}{2-0}\underset{0}{\overset{2}{\int }}\frac{2}{1+e^x}dx$
$=\underset{0}{\overset{2}{\int }}\frac{e^{-x}}{1+e^{-x}}dx$
Putting $1+e^{-x}=t$
$\Rightarrow dt=-e^{-x}dx$

$f_{\text{avg}}=-\underset{2}{\overset{1+e^{-2}}{\int }}\frac{1}{t}dt=\ln t{|}_{1+e^{-2}}^2=\ln 2-\ln (1+e^{-2})=\ln \left(\frac{2}{1+e^{-2}}\right)=\ln \left(\frac{2e^2}{e^2+1}\right)$
Mean value $=2+{\log }_e\frac{2}{e^2+1}$