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

Answer: (B)

$2+{\log }_e\left(\frac{2}{e^2+1}\right)$

Given, $f\left(x\right)=\frac{2}{e^x+1}$
we know, if $f\left(x\right)$ is continuous function on $[a,b]$ then
Mean value of the function is
$M=\frac{1}{b-a}{\int }_a^{b}f\left(x\right)dx$
$\therefore$ Mean value of the given function is
$M=\frac{1}{2-0}{\int }_0^2\frac{2}{e^x+1}dx={\int }_0^2\frac{e^{-x}}{e^{-x}+1}dx$
Let $1+e^{-x}=t$
On differentiating
$e^{-x}(-1)dx=dt$
$\therefore M=-{\int }_2^{1+\frac{1}{e^2}}\frac{dt}{t}={\int }_{1+\frac{1}{e^2}}^2\frac{dt}{t}$ (By changing limit of integration)
$={\left[\log \left|t\right|\right]}_{1+\frac{1}{e^2}}^2=\log 2-\log |1+\frac{1}{e^2}|=\log 2-\log (1+\frac{1}{e^2})$
(Here $1+\frac{1}{e^2}$ is always positive, so we can omit modulus size)
$=\log 2-\log \left(\frac{1+e^2}{e^2}\right)=2+\log \left(\frac{2}{1+e^2}\right)$