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

Question

Let $f(x)$ be a function satisfying $$f(x)+f(\pi-x)=\pi^{2}, \forall x \in \mathbb{R}$$. Then $$\int_\limits{0}^{\pi} f(x) \sin x d x$$ is equal to :

Accepted Answer

Let $I=\int\limits_0^\pi f(x) \sin x d x$ ..........(i)

$$
\begin{aligned}
& =\int\limits_0^\pi f(\pi-x) \sin (\pi-x) d x \\
& =\int\limits_0^\pi f(\pi-x) \sin x d x ........(ii)
\end{aligned}
$$

On adding Equations (i) and (ii), we get

$$
\begin{aligned}
& 2 I=\int\limits_0^\pi[f(x)+f(\pi-x)] \sin x d x \\
& \Rightarrow 2 I=\int\limits_0^\pi \pi^2 \sin x d x=\pi^2 \int\limits_0^\pi \sin x d x \\
& =\pi^2(-\cos x)_0^\pi=-\pi^2(-1-1)=2 \pi^2 \\
& \Rightarrow I=\pi^2
\end{aligned}
$$

Solutions · 1