Question

If $f\left(x\right)$ is a monotonic and differentiable function on $[a,b]$,

then $2{\int }_{f\left(a\right)}^{f\left(b\right)}x(b-f^{-1}\left(x\right))dx$ is equal to?

• A. $f^2\left(x\right)-f^2\left(a\right)$
• B. ${\int }_a^b{f}^2\left(x\right)-f^2\left(a\right)dx$
• C. ${\int }_a^{b}f\left(x\right)-f\left(a\right)dx$
• D. None of these

Answer 1

The correct option is B ${\int }_a^b{f}^2\left(x\right)-f^2\left(a\right)dx$

Let $f^{-1}\left(x\right)=t\Rightarrow x=f\left(t\right)$
$\therefore 2{\int }_{f\left(a\right)}^{f\left(b\right)}x(b-f^{-1}(x\left)\right)dx=2{\int }_a^b(b-t)f\left(t\right)f^{\prime }\left(t\right)dt$
$={\left[\right(b-t\left)\right(f\left(t{)}^2{)}^2\right]}_a^b+{\int }_a^b\left(f\right(t){)}^{2}dt$
$=-(b-a)\left(f\right(a){)}^2+{\int }_a^b\left(f\right(t){)}^{2}dt$
$=-{\int }_a^b\left(f\right(a){)}^{2}dt+{\int }_a^b(f\left(t\right){)}^{2}dt$
$={\int }_a^b\left(f^2\right(t)-(f^2\left(a\right)\left)\right)dt$
Replace $t\to x$,we get choice (b)