Proof of integral identity using integration by parts

Question

If $f:[a,b] \to \mathbb{R}$ is continuous and increasing, then the inverse $f^{-1}$ exists and

$$\int_a^b f(x)dx + \int_{f(a)}^{f(b)} f^{-1}(x)dx = f(b)b - f(a)a.$$

This is obvious by looking at a graph where the integrals are areas of regions that cover a rectangle of area $f(b)b$ with a rectangle of area $f(a)a$ removed.

I recall this is proved rigorously using integration by parts but I can't reconstruct the proof. Can anyone please help?

Your limits for the second integral should be $$\int\limits_{f(a)}^{f(b)}$$ — Rushabh Mehta, Nov 06 '18 at 17:11
Integration by parts seems would not work, since $f$ is not necessarily differentiable. — xbh, Nov 06 '18 at 17:11
The correct question: https://math.stackexchange.com/questions/1115222/show-rigorously-that-the-sum-of-integrals-of-f-and-of-its-inverse-is-bfb-af — Robert Z, Nov 06 '18 at 17:11

score 2 · Accepted Answer · answered Nov 06 '18 at 17:31

2

Hint:

$$I=\int f^{-1}(x)dx$$ $x=f(u)\Rightarrow dx=f'(u)du$: $$I=\int f'(u)\cdot (f^{-1}\circ f)(u)du$$ $$I=\int uf'(u)du$$ Integration by parts: $$dV=f'(u)du\qquad U=u\\V=f(u)\qquad dU=du$$ $$\therefore I=UV-\int VdU$$ $$\therefore I=uf(u)-\int f(u)du$$ $$\therefore I=xf^{-1}(x)-(F\circ f^{-1})(x)+C$$ You can go from here

answered Nov 06 '18 at 17:31

clathratus

17,161

Thank you very much.. This is exactly what I was trying to remember. Naturally we need another condition that $f$ is differentiable. – WoodWorker Nov 06 '18 at 19:06
None of the "duplicate" answers take this approach. – WoodWorker Nov 06 '18 at 19:08
You're very welcome – clathratus Nov 06 '18 at 19:16

Proof of integral identity using integration by parts

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