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The Problem
Suppose $f$ is continuous, $f(0)=0$, $f(1)=1$, $f'(x)>0$, and $\int_0^1{f(x)dx}=\frac{1}{3}$. Find the value of the integral $\int_0^1{f^{-1}(y)dy}$. [Calculus: Early Transcendentals, Chapter 5, Review, Exercise 71]
What I Have Done
I first tried evaluating the integral myself but got stuck on the step to prove that \begin{equation} 1-\int_0^1{f(x)dx}=\int_0^1{f^{-1}(x)dx}. \end{equation} I then searched the web but the answers I got seemed to be wrong.
My Questions
How can I prove the equation above? Otherwise, is there an alternative solution to this problem?

Thanks in advance.

Math Is Fun
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3 Answers3

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Maybe try a substitution $u = f^{-1}(x)$: Then (note $f^{-1}(0)=0$, $f^{-1}(1)=1$) $$ \int^1_0 f^{-1}(x) ~\mathrm{d}x = \int^1_0 uf'(u)~\mathrm{d}u $$ Integrate by parts: $$ \int^1_0 uf'(u)~\mathrm{d}u = uf(u) \bigg\vert^{u=1}_{u=0} - \int^1_0 f(u)~\mathrm{d}u = f(1) - \int^1_0 f(u)~\mathrm{d}u = 1- \frac{1}{3} = \frac{2}{3} $$

Hyperbolic PDE friend
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1

Let's prove the following:

Proposition. Let $f:[a,b]\to\mathbb{R}$ be a strictly increasing differentiable function. Then

$$\int_a^bf(x)~\mathrm{d}x=bf(b)-af(a)-\int_{f(a)}^{f(b)}f^{-1}(x)~\mathrm{d}x.$$

Recall first the identity

$$\frac{\mathrm{d}}{\mathrm{d}x}f^{-1}(x)=\frac{1}{f'(f^{-1}(x))}.$$

If we make the substitution $\xi=f^{-1}(x)$, then $f'(\xi)\mathrm{d}\xi=\mathrm{d}x$, and

$$\int_{f(a)}^{f(b)}f^{-1}(x)~\mathrm{d}x=\int_a^b\xi f'(\xi)~\mathrm{d}\xi\overset{\substack{\text{Integration}\\\text{by parts}}}=\biggl[\xi f(\xi)\biggr]_a^b-\int_a^bf(\xi)~\mathrm{d}\xi=bf(b)-af(a)-\int_a^bf(\xi)~\mathrm{d}\xi.$$

Rearranging we get the result in the proposition.

For your problem, using the proposition we get that

$$\int_{0}^{1}f^{-1}(x)~\mathrm{d}x=\int_{f(0)}^{f(1)}f^{-1}(x)~\mathrm{d}x=1f(1)-0f(0)-\int_0^1 f(x)~\mathrm{d}x=1-\frac{1}{3}=\frac{2}{3}.$$

Lorago
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From Young's inequality on the integrals we have $\int_{0}^{1} h(x) dx + \int_{0}^{1} h^{-1}(x) dx \geq 1$,whatever $h$ is a continuous and bijective function on $[0,1]$ and $h(0)=0$, with equality for $h(1)=1$.
From the fact that $f'(x)>0$ it follows that $f$ is strictly increasing, therefore injective, and from the fact that f is continuous and $f(0)=0$ and $f(1)=1$ it follows that $f$ is bijective, so replacing $h$ with $f$ we even get equality in Young's inequality, so $\int_{0}^{1} f(x) dx + \int_{0}^{1} f^{-1}(x) dx =1$ and from here you can finish it.

Gary
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bEtAVs
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