Question

If $f\left(x\right)=f(\pi +e-x)$ and ${\int }_e^{\pi }f\left(x\right)dx=\frac{2}{e+\pi }$,then ${\int }_e^{\pi }xf\left(x\right)dx$ is equal to.

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

Answer 1

The correct option is D 1

Let $I={\int }_e^{\pi }xf\left(x\right)dx$......(1)

Now using ${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$

$I={\int }_e^{\pi }(\pi +e-x)f(\pi +e-x)dx$

$\Rightarrow I={\int }_e^{\pi }(\pi +e-x)f\left(x\right)dx$, using given condition.....(2)

Now add (1) and (2)

$2I=(\pi +e){\int }_e^{\pi }f\left(x\right)dx=(\pi +e)\cdot \frac{2}{e+\pi }=2$, use given values

$\therefore I=1$