If f(x) is a continuous, 1-to-1 function and $f(3)=5$, $f(10)=1$, and $\int_{3}^{10} f(x) d x=20$, how do I find $$
\int_{1}^{5} f^{-1}(x) d x
$$
? I have poorly attempted to draw the situation in paint:
where the y-axis is incrementing in 5's and the x is incrementing in 1's. I can't really think of a way to approach this.
Using the change of variables $x = f(u)$ and integration by parts:
\begin{align*} \int_{1}^5 f^{-1}(x) dx = & \int_{10}^3 f'(u) \cdot u du = [u f(u)]_{10}^3-\int_{10}^3 f(u) du = 3 f(3)-10 f(10)+20 \\ = & 15-10+20=25. \end{align*}