Question:
$\int_{-\pi}^{\pi}|\pi-| x|| d x$ is equal to :
Correct Option: , 3
Solution:
$I=\int_{-\pi}^{\pi}|\pi-| x|| d x \quad[\because|\pi-| x||$ is even $]$
$=2 \int_{0}^{\pi}|\pi-| x|| d x$
$=2 \int_{0}^{\pi}(\pi-x) d x$
$=2\left[\pi x-\frac{x^{2}}{2}\right]_{0}^{\pi}=2\left(\pi^{2}-\frac{\pi^{2}}{2}\right)=\pi^{2}$