Question

If $f(x)=f(\pi +e-x)$ and $\int\limits_{e}^{\pi} f(x) d x=\frac{2}{e+\pi}$, then $\int\limits_{e}^{\pi} x f(x) d x$ 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 $\int\limits_{e}^{\pi} f(x) d x=\frac{2}{e+\pi}$...(i)
Let $I=\int\limits_{e}^{\pi} x f(x) d x$
$=\int\limits_{e}^{\pi}(e+\pi-x) f(e+\pi-x) d x$
$=\int\limits_{e}^{\pi}(e+\pi) f(e+\pi-x) d x$
$-\int\limits_{e}^{\pi} x f(e+\pi-x) d x$
$=\int\limits_{e}^{\pi}(e+\pi) f(x) d x-\int_{e}^{\pi} x f(x) d x$
[from Eq.(I)]
$\Rightarrow I=(e+\pi) \int\limits_{e}^{\pi} f(x) d x-I$
$\Rightarrow 2 I=(e+\pi) \times \frac{2}{e+\pi}$ [from Eq. (i)]
$\therefore I=1$