Functions and Integrals 1

It is a well-known fact that

$\int f^{-1}(x) dx = x f^{-1}(x) - F \left( f^{-1}(x) \right) + C,$

where $F$ is the antiderivative of $f$ . Let $f$ be our function and observe that the antiderivative is really simple to find:

$F(x) = \frac{x^2}{2} - \cos(x) + K.$

Then whatever $f^{-1}$ is we have

$\int_0^\pi f^{-1}(x) dx = \left[ xf^{-1}(x) - \frac{\left( f^{-1}(x) \right)^2}{2} + \cos \left( f^{-1}(x) \right) \right]_0^\pi.$

But observe that the only $x \in \mathbb{R}$ such that $f(x) = \pi$ is $x = \pi$ itself. Same happens for $f(x) = 0$ , such that $x = 0$ . Therefore

$\int_0^\pi f^{-1}(x) dx = \frac{\pi^2}{2} - 2.$