# Integration of some root functions.

while trying to help the daughter of my cousin to understand Integrals, I see that I forget about some of the integral calculations. Can someone help me with that one, please?

$$\int_0^1 \Bigl(\sqrt[3]{(1-x^7)}-\sqrt[7]{(1-x^3)} \Bigr) dx$$

(this is not a homework, I'm trying to understand and explain to her how things work.)

I have a clue like..

$$\sqrt[3]{(1-x^7)} = y$$ $$(1-x^7) = y^3$$ $$(1-y^3) = x^7$$ $$x = \sqrt[7]{(1-y^3)}$$

but that is all...

Hint:

What's is the area of curve $$x^3+y^7=1$$ in $$x\in[0,1]$$

$$\int_0^1 y\ dx=\int_{x=0}^{x=1} x \ dy$$

• Ah yes, we missed that it is equals to 1. Great, thanks! – tanaydin Dec 14 '18 at 13:44
• @tanaydin, What is $=1?$ – lab bhattacharjee Dec 14 '18 at 13:50
• area of the curve is 0, (if I'm not wrong) – tanaydin Dec 14 '18 at 14:03
• @tanaydin, You are correct – lab bhattacharjee Dec 14 '18 at 14:05
• ok, we will study a bit more :) – tanaydin Dec 14 '18 at 14:39

There is a nice insight to be gleaned here.

First of all, note that the two integrands are inverse functions of each other, that is, if $$f(x) = (1-x^7)^{\frac 13}$$, then $$f^{-1}(x) = (1-x^3)^{\frac 17}$$.

Also note that $$f(0) = 1$$ and $$f(1) = 0$$. If you represent a function and its inverse on a graph (the exact shapes don't matter), they will be superimposed on each other by reflection about the line of identity $$y=x$$. A simple curve sketch (and your job is easy because the shapes don't matter that much, just where they start and end and the fact that one is the mirror image of the other about $$y=x$$) should convince you and your student that the difference in area under the curve is zero for the interval $$[0,1]$$.

Need something more rigorous and "mathematical"? Well, there's this nice identity which can be proven with integration by parts:

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

In this problem, $$a = 0, b = 1, f(a) = 1, f(b) = 0$$. You can now prove directly that the required integral vanishes.