Mathematics - Definite Integration Question with Solution | TestHub

Given:$g\left(x\right)$is an even function, then
$f\left(x\right)=\underset{0}{\overset{x}{\int }}g\left(t\right)dt$will be an odd function,
$\Rightarrow f\left(-x\right)=-f\left(x\right)$
Let$I=\underset{0}{\overset{x}{\int }}f\left(t\right)dt$put$t=z+5$
$\Rightarrow I=\underset{-5}{\overset{x-5}{\int }}f\left(z+5\right)dz$
$\Rightarrow I=\underset{-5}{\overset{x-5}{\int }}g\left(z\right)dz$$\left(\because f\left(x+5\right)=g\left(x\right)\right)$
$\Rightarrow I=\underset{-5}{\overset{x-5}{\int }}{f}^{\text{'}}\left(z\right)dz (\because f^{\text{'}}(x)=g(x\left)\right)$
$\Rightarrow I=f\left(z\right){|}_{-5}^{x-5}=f(x-5)-f\left(-5\right)$
$\Rightarrow I=f\left(5\right)-f\left(5+x\right)$($f$is an odd function and$g$is given as an even function)
$\Rightarrow I=\underset{5+x}{\overset{5}{\int }}{f}^{\text{'}}\left(t\right)dt=\underset{5+x}{\overset{5}{\int }}g\left(t\right)dt$