If $f$ and $g$ are continuous functions on $[0, \pi]$ satisfying $f(x)+f(\pi-x)=1=g(x)+g(\pi-x)$, then $\int\limits_0^\pi\{f(x)+g(x)\} d x$ is equal to

$\pi$

We have,

$I=\int\limits_0^\pi\{f(x)+g(x)\} d x$

$\Rightarrow I=\int\limits_0^\pi\{f(\pi-x)+g(\pi-x)\} d x$

$\Rightarrow I=\int\limits_0^\pi\{1-f(x)+1-g(x)\} d x$            $\left[\begin{array}{c}∵ f(\pi-x)+f(x)=1 \text { and } \\ g(\pi-x)+g(x)=1\end{array}\right]$

$\Rightarrow I=\int\limits_0^\pi[2-\{f(x)+g(x)\}] d x$

$\Rightarrow I=2 \pi-I \Rightarrow 2 I=2 \pi \Rightarrow I=\pi$