Question

If $f(x)=f(\pi +e-x)$ and $\underset{e}{\overset{\pi }{\int }}f(x)dx=\frac{2}{e+\pi }$, then $\underset{e}{\overset{\pi }{\int }}xf(x)dx$ is equal to

• A. $\frac{\pi +e}{2}$
• B. $\frac{\pi -e}{2}$
• C. $\pi -e$
• D. $1$

Answer 1

Answer: (D)

Given, $f(x)=f(\pi +e-x)$
and $\underset{e}{\overset{\pi }{\int }}f(x)dx=\frac{2}{e+\pi }$...(i)
Let $I=\underset{e}{\overset{\pi }{\int }}xf(x)dx$
$=\underset{e}{\overset{\pi }{\int }}(e+\pi -x)f(e+\pi -x)dx$
$=\underset{e}{\overset{\pi }{\int }}(e+\pi )f(e+\pi -x)dx$
$-\underset{e}{\overset{\pi }{\int }}xf(e+\pi -x)dx$
$=\underset{e}{\overset{\pi }{\int }}(e+\pi )f(x)dx-{\int }_e^{\pi }xf(x)dx$
[from Eq.(I)]
$\Rightarrow I=(e+\pi )\underset{e}{\overset{\pi }{\int }}f(x)dx-I$
$\Rightarrow 2I=(e+\pi )\times \frac{2}{e+\pi }$ [from Eq. (i)]
$\therefore I=1$