Question

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

Answer 1

consider $a+b-x=t⟹-dx=dt$
$a+b-x=t\Rightarrow a+b-t=x$
$\therefore f(a+b-x)=f\left(t\right)$

${\int }_a^{b}f(a+b-x)dx=-{\int }_b^{a}f\left(t\right)dt$

Using the property of ${\int }_a^{b}f\left(t\right)dt=-{\int }_b^{a}f\left(t\right)dt$

${\int }_a^{b}f(a+b-x)dx={\int }_a^{b}f\left(t\right)dt$
Now using property,${\int }_a^{b}f(a+b-x)dx={\int }_b^{a}f(a+b-t)dt$

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

Replacing $t=a+b-x$

$⟹{\int }_a^{b}f(a+b-a-b+x)dx$

$⟹{\int }_a^{b}f\left(x\right)dx$

Hence Proved..