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Let $f(x)$ be an increasing invertible function. Prove that $$\int_{a}^{b}f(x)dx+\int_{f(a)}^{f(b)}f^{-1}(x)dx=bf(b)-af(a)$$ We have proved by a variable shift in the second integral. It comes like $\int_{a}^{b}(f(x)+xf'(x))dx.$

But it unclear as to how the increasing property helps. Also we tried using geometry and compute the areas but failed.

Grobber
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  • Actually I would like if there is some geometric proof of the original integral. – Grobber Mar 15 '14 at 18:36
  • Geometric? Interesting. Mind if I steal that idea? and post a question with some analysis of my own? – Guy Mar 15 '14 at 18:40

2 Answers2

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$$\int_{a}^{b}(f(x)+xf'(x))dx$$

Let $y=xf(x)$

$$\frac{dy}{dx}=xf'(x)+f(x)$$

Thus our integral becomes,

$$\int_{x=a}^{x=b}dy.$$

$$bf(b)-af(a)$$

The increasing property seems misleading, there is no need for it.

As per OP's request here is a graphical explanation from here by David Mitra.

enter image description here

This is so brilliant, no explanation required.

Community
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Guy
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Hint: Since $y=f(x)$ is invertible, the function $x=f^{-1}(y)$ is defined, $x=a\iff y=f(a)$ and $x=b\iff y=f(b)$. Thus $$\int_{f(a)}^{f(b)} f^{-1}(y)dy=\int_{f(a)}^{f(b)} xdy=\int_a^b (x \frac{dy}{dx}) dx=\int_a^b x f'(x) dx$$ Now, it is clear that $(xf(x))'=f(x)+xf'(x)$ and hence it is easy to complete.

Semsem
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