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Let $f:[0,1]\rightarrow [0,1]$ be a strictly decreasing continuous function with $f(0)=1$ and $f(1)=0$. Then show that $\int_{0}^{1}f(x)dx=\int_{0}^{1}f^{-1}(y)dy$

I was thinking to do it by constructing a function then using Lagrange MVT on it. But I am undone.

Sanajit Patra
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    That not true... – Surb May 09 '22 at 14:28
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    I think yours is a special case of this: https://math.stackexchange.com/questions/1199283/let-f-be-a-strictly-decreasing-function-then-int-abf-1-bf-1b-af?rq=1 – Randall May 09 '22 at 14:28
  • The proof isn't so simple especially if we don't assume that $f$ is differentiable. I'd suggest you check the picture here and try to understand why the formula in the article makes sense. – bjorn93 May 09 '22 at 14:50
  • Essentially a duplicate of https://math.stackexchange.com/q/1115222. – Martin R May 09 '22 at 15:17

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$\int_0^1f(x)dx$ is the area under the curve $y=f(x)$ for $0\le x\le 1$. But $f$ is strictly decreasing, so this is the same as the area to the left of the curve. If we reflect this picture about the line $y=x$, the curve traces the function $y=f^{-1}(x)$; and the area is unchanged, but it is now seen to be the region under the curve $y=f^{-1}(x)$.

TonyK
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  • I think most instructors would not accept this as an answer. "Area" is defined by the integral itself. Before one learns the integral the concept of area is undefined except for rectangles and triangles, etc. You are assuming properties of "area" that would need to be proved by properties of the "integral." It is, however, a good intuitive explanation of why this should be true. – B. S. Thomson May 09 '22 at 18:12
  • @B.S.Thomson: I was not treating it as a homework assignment. – TonyK May 09 '22 at 18:48
  • I would tend to interpret your answer as a good hint of how to "show that" this statement holds. The other two answers here took your hint and went with it. – B. S. Thomson May 09 '22 at 18:54
  • @B.S.Thomson It is an answer from high above. I bet Newton and Leibnitz would have accepted it as an obvious proof. Doubts came later. – Gribouillis May 10 '22 at 07:48
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Almost a rephrasing of @TonyK 's answer. As $f$ and $f^{-1}$ decrease \begin{equation} \int_{[0, 1]^2}[y \le f(x)] dx dy = \int_{[0, 1]^2}[f^{-1}(y) \ge x] dx dy \end{equation} where $[]$ is Iverson's bracket.

Gribouillis
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Here is another proof base on Fubini's theorem. Assuming $f$ is strictly monotone decreasing and continuous with $f(0)=1$ and $f(1)=1$, $$A=\{(x,y)\in[0,1]^2:0\leq x\leq 1,\,0\leq y\leq f(x)\}=\{(x,y)\in[0,1]^2: 0\leq y\leq 1,\,0\leq x\leq f^{-1}(y)\}$$ Hence

$$\begin{align}\int^1_0f(x)\,dx&=\int^1_0\Big(\int^{f(x)}_{0}\,dy\Big)\,dx=\int_{[0,1]\times[0,1]}\mathbb{1}_{A}(x,y)\,dx\,dy\\ &=\int^1_0\Big(\int^{f^{-1}(y)}_0dx\Big)\,dy=\int^1_0f^{-1}(y)\,dy \end{align}$$

Edit: I was typing this when TonyK's solution pop-up. My posting is just to formalized his geometric idea.

Mittens
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