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

Question

Let $f(x)$ be a function satisfying $f(x)+f(\pi -x)={\pi }^{2},\forallx\in R$. Then $\int _{0}^{\pi }f(x)\sin xdx$ is equal to :

Accepted Answer

Let $I=\int _{0}^{\pi }f(x)\sin xdx$ ..........(i)

$\begin{gathered} =\int _{0}^{\pi }f(\pi -x)\sin (\pi -x)dx \ =\int _{0}^{\pi }f(\pi -x)\sin xdx........(ii) \end{gathered}$

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

$\begin{gathered} 2I=\int _{0}^{\pi }[f(x)+f(\pi -x)]\sin xdx \ \Rightarrow 2I=\int _{0}^{\pi }{\pi }^{2}\sin xdx={\pi }^{2}\int _{0}^{\pi }\sin xdx \ ={\pi }^{2}(-\cos x{)}_{\pi }^{0}=-{\pi }^{2}(-1-1)=2{\pi }^{2} \ \Rightarrow I={\pi }^{2} \end{gathered}$

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