Question

By using the properties of definite integrals, Show that ${\int }_0^{a}f\left(x\right)g\left(x\right)dx=2{\int }_0^{a}f\left(x\right)dx$, if $f$ and $g$ are defined as $f\left(x\right)=f(a-x)$ and $g\left(x\right)+g(a-x)=4$

Answer 1

Let $I={\int }_0^{a}f\left(x\right)g\left(x\right)dx$ ............ (1)
$\Rightarrow I={\int }_0^{a}f(a-x)g(a-x)dx,(\because {\int }_0^{a}f(x)dx={\int }_0^{a}f(a-x\left)dx\right)$
$\Rightarrow I={\int }_0^{a}f\left(x\right)g(a-x)dx$ ......... (2)
Adding (1) and (2), we obtain
$2I={\int }_0^a\left\{f\right(x\left)g\right(x)+f(x\left)g\right(a-x\left)\right\{dx$
$\Rightarrow 2I={\int }_0^{a}f\left(x\right)\left\{g\right(x)+g(a-x\left)\right\}dx$
$\Rightarrow 2I={\int }_0^{a}f\left(x\right)\times 4dx$ using $\left[g\right(x)+g(a-x)=4]$
$\Rightarrow I=2{\int }_0^{a}f\left(x\right)dx$