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 :
Solution
Let $I=\int\limits_0^\pi f(x) \sin x d x$ ..........(i)
<br/><br/>$$
\begin{aligned}
& =\int\limits_0^\pi f(\pi-x) \sin (\pi-x) d x \\\\
& =\int\limits_0^\pi f(\pi-x) \sin x d x ........(ii)
\end{aligned}
$$
<br/><br/>On adding Equations (i) and (ii), we get
<br/><br/>$$
\begin{aligned}
& 2 I=\int\limits_0^\pi[f(x)+f(\pi-x)] \sin x d x \\\\
& \Rightarrow 2 I=\int\limits_0^\pi \pi^2 \sin x d x=\pi^2 \int\limits_0^\pi \sin x d x \\\\
& =\pi^2(-\cos x)_0^\pi=-\pi^2(-1-1)=2 \pi^2 \\\\
& \Rightarrow I=\pi^2
\end{aligned}
$$
About this question
Subject: Mathematics · Chapter: Definite Integration · Topic: Properties of Definite Integrals
This question is part of PrepWiser's free JEE Main question bank. 216 more solved questions on Definite Integration are available — start with the harder ones if your accuracy is >70%.