What is the value of $\int_{0}^{+x} f(x) d x$ in terms of $f^{-1}$?

Question

I was trying to solve the following problem, which came up in a Calculus exam.

Let $f$ be a continuous, strictly decreasing, real-valued function such that $\int_{0}^{+\infty} f(x) d x$ is finite and $f(0)=1$. In terms of $f^{-1}($ the inverse function of $f), \int_{0}^{+x} f(x) d x$ is:

(A) less than $\int_{1}^{+\infty} f^{-1}(y) d y$

(B) greater than $\int_{0}^{1} f^{-1}(y) d y$

(C) equal to $\int_{1}^{+\infty} f^{-1}(y) d y$

(D) equal to $\int_{0}^{1} f^{-1}(y) d y$

(E) equal to $\int_{0}^{+\infty} f^{-1}(y) d y$

We know that the graph of $f^{-1}(x)$ on the $xy$-plane is just reflecting the graph of $f(x)$, with respect to the line $x = y$. In this way I chose (E), as the graph of $f^{-1}(x)$ was defined for $y \in (0,\infty)$ or in the $x \in (0,1)$ region. However, the solution is (D). So firstly, what is wrong with my approach? And secondly, how would you go about solving this, in an intuitive manner? (One way, using substitutions, is written here The integral on the inverse of a function. I'm trying to find a quicker, graphical way of solving this problem under time constraints.)

Answer 1

For a visual approach: We know $f(0)=1$ and $f(\infty)=0$ (if you'll excuse the notation). The latter is to make sure $\int_0^\infty f(x)dx$ is finite. So the graph of $f$ makes a sort of big triangle consisting of the positive $x$-axis, $[0,1]$ on the $y$-axis and the graph itself.

The area of this infinite "triangle" is either the "sum" of heights over the $x$-axis, which is $\int_0^\infty f(x)dx$, or the "sum" of widths over the $y$-axis, which is $\int_0^1 f^{-1}(y)dy$.

See https://en.wikipedia.org/wiki/Integral_of_inverse_functions for a nice picture. EDIT: the picture shows the increasing case, but the ideas are the same in the decreasing case.

Answer 2

Using the aforementioned Inverse function integral theorem, the integral would use these formulas from the link. The second formula uses a slightly generalized form. Notice the determinant property of the first expression: $$\mathrm{ad-bc=\int_c^ d f^{-1}(x)+\int_{a=0}^{b=x}f(x)dx\implies\int_0^x f(x)dx=0d-xc-\int_c^d f^{-1}(x)dx=-xc-\int_c^d f^{-1}(x)dx}$$

For the generalized formula using the link. The “$\circ$“ symbol means a combination of the integral of f(x) function and the inverse f(x) function. Notice the integral of f(x) from 0 to x indeed is in terms of the inverse here. This assumes $x f^{-1}(x)|_{x=0}$ is defined which just means the function evaluated at x=0:

$$\mathrm{\int f^{-1}(x)dx=x\,f^{-1}(x)-\int f(x)dx \circ f^{-1}(x)\implies \int_0^x f^{-1}(x)= x\,f^{-1}(x)-\int f(x)dx \circ f^{-1}(x)+\left[\int f(x)dx \circ f^{-1}(x)\right]\bigg|_{x=0} }$$

Please correct me and give me feedback!