Q. If $\int_{}^{}{f(x)dx = g(x) + c}$, then $\int_{}^{}{f^{- 1}(x)dx}$ is equal to
A) $xf^{- 1}(x) + c$
B) $f\left\{ g^{- 1}(x) \right\} + c$
C) $xf^{- 1}(x) - g\left\{ f^{- 1}(x) \right\} + c$
D) $g^{- 1}(x) + c$
Explanation
(c)
Let $f^{- 1}(x) = u\ \ \ \Rightarrow \ \ \ \ x = f(u)$
$\Rightarrow \ \ dx = f^{'}(u)du$
$\therefore\ I = \int_{}^{}{f^{- 1}(x)}dx = \int_{}^{}uf^{'}(u)du$
$\Rightarrow I = uf(u) - \int_{}^{}{f(u)}du$
$= uf(u) - g(u) + c$ [$\because\int_{}^{}{f(x)}dx = g(x) + c$, given]
$\therefore$ On putting $u = f^{- 1}(x)$ and $f(u) = x$
We get, $I = xf^{- 1}(x) - g\left\{ f^{- 1}(x) \right\} + c$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.