JEE MAIN - Mathematics (2023 - 6th April Evening Shift - No. 8)
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 :
$$\pi^{2}$$
$$\frac{\pi^{2}}{2}$$
$$2 \pi^{2}$$
$$\frac{\pi^{2}}{4}$$
Explanation
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
$$ \begin{aligned} & 2 I=\int\limits_0^\pi[f(x)+f(\pi-x)] \sin x d x \\\\ & \Rightarrow 2 I=\int\limits_0^\pi \pi^2 \sin x d x=\pi^2 \int\limits_0^\pi \sin x d x \\\\ & =\pi^2(-\cos x)_0^\pi=-\pi^2(-1-1)=2 \pi^2 \\\\ & \Rightarrow I=\pi^2 \end{aligned} $$
$$ \begin{aligned} & =\int\limits_0^\pi f(\pi-x) \sin (\pi-x) d x \\\\ & =\int\limits_0^\pi f(\pi-x) \sin x d x ........(ii) \end{aligned} $$
On adding Equations (i) and (ii), we get
$$ \begin{aligned} & 2 I=\int\limits_0^\pi[f(x)+f(\pi-x)] \sin x d x \\\\ & \Rightarrow 2 I=\int\limits_0^\pi \pi^2 \sin x d x=\pi^2 \int\limits_0^\pi \sin x d x \\\\ & =\pi^2(-\cos x)_0^\pi=-\pi^2(-1-1)=2 \pi^2 \\\\ & \Rightarrow I=\pi^2 \end{aligned} $$
Comments (0)
