Question

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

$I$=${\int }_0^{a}f\left(x\right)g\left(x\right)dx={\int }_0^{a}f(a-x)g(a-x)dx$
$I$=${\int }_0^{a}f\left(x\right)(4-g(x\left)\right)dx$

$I={\int }_0^{a}4f\left(x\right)-{\int }_0^{a}f\left(x\right)g\left(x\right)$
$2I={\int }_0^{a}4f\left(x\right)dx$
$I=2{\int }_0^{a}f\left(x\right)dx$

$hence$ $proved.$