Question

Prove that ${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$

Answer 1

It is just substitution, if we let $u=a+b-x,$ we have $du=-dx$ and hence $u=b$ when $x=a$ and vice versa.

$\Rightarrow {\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f\left(x\right)dx$

$={\int }_b^{a}f(a+b-x)(-dx)$

$=-{\int }_b^{a}f(a+b-x)dx$

$={\int }_a^{b}f(a+b-x)dx.$

Hence, proved.