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$f(x)$ is a strictly increasing, and continuous, function on $[0,+\infty)$. Consider the integral,

$$\int_0^a f^{-1}(x) dx.$$

Reasoning

The interval $[0,a]$ is "reflected" across the line $y=x$ onto the interval $0 \leq y \leq a.$

$f(x)$ and $f^{-1}(x)$ are symmetric across the line $y=x$. So, the integral is equivalent to the area bounded by $y = 0$, $y=a$, and $f(x)$.

So, "integrating $f$ along an interval on the $y$-axis" is the same as integrating $f^{-1}$ along the same interval on the $x$-axis?

Question

Is my reasoning correct?

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1 Answer

up vote 2 down vote accepted

In your setup, if $f(0)=0$ and $f(b)=a$ then $$\int_0^a f^{-1}(x)\, d x + \int_0^b f(x)\, d x = ab$$ that is the two integrals combine to the area of a rectangle.

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Thanks for the insight. –  Legendre Oct 29 '12 at 8:09
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