Let f(x) be a function satisfying f(x)+f(-x)=^2, xR.

Question

Let $f(x)$ be a function satisfying $f(x)+f(\pi-x)=\pi^2$, $\forall x\in\mathbb{R}$. Then $\displaystyle\int_0^\pi f(x)\sin x\,dx$ is equal to:

Accepted Answer

By King's rule, $I=\displaystyle\int_0^\pi f(\pi-x)\sin x\,dx$ (since $\sin(\pi-x)=\sin x$). Adding:

$$2I = \int_0^\pi\big[f(x)+f(\pi-x)\big]\sin x\,dx = \pi^2\int_0^\pi\sin x\,dx = \pi^2\cdot2 = 2\pi^2$$

$$I = \pi^2$$

Solutions · 1