Question

If $f\left(x\right)$ and $g\left(x\right)$ are inverse function of each other such that $f\left(1\right)=3$ & $f\left(3\right)=1$, then ${\int }_1^3(g\left(x\right)+\frac{x}{f^{\prime }\left(g\right(x\left)\right)})dx$ is equal to

Answer 1

$f^{-1}\left(x\right)=g\left(x\right)\Rightarrow f\left(g\right(x\left)\right)=x$
$\Rightarrow f^{\prime }\left(g\right(x\left)\right).g^{\prime }\left(x\right)=1$
We have,
${\int }_1^3(g\left(x\right)+\frac{x}{f^{\prime }\left(g\right(x\left)\right)})dx$
$={\int }_1^3\left(g\right(x)+x{g}^{\prime }(x\left)\right)dx=[xg\left(x\right){]}_1^3$
$=3\left(g\right(3\left)\right)-1\left(g\right(1\left)\right)$
We know that,
$f\left(3\right)=1\Rightarrow g\left(1\right)=3$ &
$f\left(1\right)=3\Rightarrow g\left(3\right)=1$
After substituting we get,
$=3\left(1\right)-1\left(3\right)=0$