Show that


Show that $\int_{0}^{u} f(x) g(x) d x=2 \int_{0}^{n} f(x) d x$, if $f$ and $g$ are defined as $f(x)=f(a-x)$ and $g(x)+g(a-x)=4$


Let $I=\int_{0}^{a} f(x) g(x) d x$           ...(1)

$\Rightarrow I=\int_{0}^{a} f(a-x) g(a-x) d x$         $\left(\int_{0}^{\infty} f(x) d x=\int_{0}^{a} f(a-x) d x\right)$

$\Rightarrow I=\int_{0}^{a} f(x) g(a-x) d x$         ...(2)

Adding (1) and (2), we obtain

$2 I=\int_{0}^{a}\{f(x) g(x)+f(x) g(a-x)\} d x$

$\Rightarrow 2 I=\int_{0}^{a} f(x)\{g(x)+g(a-x)\} d x$

$\Rightarrow 2 I=\int_{0}^{a} f(x) \times 4 d x$     $[g(x)+g(a-x)=4]$

$\Rightarrow I=2 \int_{0}^{a} f(x) d x$

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