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How to simplify expression

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

The answer is $bf(b)-af(a)$ but I am wondering how to get the answer.

1 Answers 1

1

Here is a graphical proof (that doesn't assume that $f$ is differentiable).

Take a look at the following figure :

enter image description here

The colored areas constitute a rectangle whose area can be computed in two ways :

A) globally, as :

$$\text{length} \times \text{width} \ = \ b.f(b)\tag{1}$$

B) by considering it as partitionned into three non-overlapping regions :

  • the blue region (deep blue + light blue) : its area is the area under the curve of $f$ starting at $a$ and ending at $b$ i.e., $\int_a^b f(x)dx$,

  • the red region (deep red + light red) : its area is the area under the curve of $f^{-1}$, starting at $f(a)$ and ending at $f(b)$, i.e., $\int_{f(a)}^{f(b)} f^{-1}(x)dx$ (one has to turn his head 90° !),

  • the green region, a rectangle whose area is : $a.f(a)$.

Equating the two ways of computing this area, we get :

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

which is equivalent to the relationship you want to establish.

Addendum : It has taken me some time to spot an analogous question. Here it is, on this site : Show rigorously that the sum of integrals of $f$ and of its inverse is $bf(b)-af(a)$ with some very interesting answers (the accepted answer of Julian Aguirre, and the answer of Kobe ; the answer of Vim is the same as mine). Another reference : Prove that $\int_a^bf(x)dx+ \int_{f(a)}^{f(b)} f^{-1}(y)dy=bf(b)-af(a)$.

Last but not least, a generalisation of the equation at hand is given by Young's inequality : (http://mathworld.wolfram.com/YoungsInequality.html) as remarked in this answer (https://math.stackexchange.com/q/1154557).