Mathematics - Definite Integration Question with Solution | TestHub

$I=\frac{1}{\left(a+b\right)}\underset{a}{\overset{b}{\int }}x\left[f\left(x\right)+f\left(x+1\right)\right]dx \dots \left(1\right)$
$x\to a+b-x$
$I=\frac{1}{\left(a+b\right)}\underset{a}{\overset{b}{\int }}\left(a+b-x\right)\left[f\left(a+b-x\right)+\left(a+b+1-x\right)\right]dx$
$I=\frac{1}{\left(a+b\right)}\underset{a}{\overset{b}{\int }}a\left(a+b-x\right)\left[f\left(x+1\right)+f\left(x\right)dx\right] \dots \left(2\right)$
$\because$Keeping$x\to x+1$in given relation
$\left(1\right)+\left(2\right)$
$2I=\underset{a}{\overset{b}{\int }}\left[f\left(x+1\right)+f\left(x\right)\right]dx$
$2I=\underset{a}{\overset{b}{\int }}f\left(x+1\right)dx+\underset{a}{\overset{b}{\int }}f\left(x\right)dx$
$\underset{a}{\overset{b}{\int }}f\left(a+b+1-x\right)dx+\underset{a}{\overset{b}{\int }}f\left(x\right)dx$
$2I=2\underset{a}{\overset{b}{\int }}f\left(x\right)dx$
$I=\underset{a}{\overset{b}{\int }}f\left(x\right)dx$
Put$x=t+1\Rightarrow I=\underset{a-1}{\overset{b-1}{\int }}f\left(t+1\right)dt=\underset{a-1}{\overset{b-1}{\int }}f\left(x+1\right)dx$