# Definite integral involving a square and a cube root

I have the following integral to solve $$\int_0^1 \left(\sqrt{1-x^3}-\sqrt[3]{1-x^2}\right) \, dx.$$

I tried the substitutions $u=x^2$, $u=x^3$, $u=\sqrt{1-x^3}$, and $u=\sqrt[3]{1-x^2}$ but the integral would net get any simpler. Any hint, how to tackle the integral is welcome (maybe I should try integration by parts?).

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Here is a similar problem (the answer there includes your integral as a particular case). – David Mitra Dec 19 '12 at 17:58

## 2 Answers

You can use two substitutions and one integration by parts to prove that $$\begin{equation*} I=\int_{0}^{1}\sqrt[3]{1-x^{2}}dx=\int_{0}^{1}\sqrt{1-x^{3}}dx=J. \end{equation*}$$ Starting with $I$ make the substitution $t=1-x^{2}$ to obtain $$\begin{equation*} I=\int_{0}^{1}\frac{\sqrt[3]{t}}{2\sqrt{1-t}}dt. \end{equation*}$$ Now integrate by parts choosing the factors $u(t)=\sqrt[3]{t}$ and $v'(t)=\frac{1}{2\sqrt{1-t}}$ $$\begin{eqnarray*} I &=&\left. \sqrt[3]{t}\left( -\sqrt{1-t}\right) \right\vert _{0}^{1}-\int_{0}^{1}\frac{1}{3\sqrt[3]{t^{2}}}\left( -\sqrt{1-t}\right) \,dt =\int_{0}^{1}\frac{1}{3\sqrt[3]{t^{2}}}\sqrt{1-t}\,dt. \end{eqnarray*}$$ Finally use the substitution $t=v^{3}$ $$\begin{equation*} I=\int_{0}^{1}\sqrt{1-v^{3}}\,dv=J. \end{equation*}$$

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How did you get 2*sqrt(1-t)? – Vinicius L. Beserra Oct 6 '13 at 15:56

Here is an informal argument:

First note that your integral is of the form $\int_0^1 f(x)-f^{-1}(x)\,dx$, where $f$ is a decreasing function with $f(0)=1$, $f(1)=0$, and $f^{-1}$ is the inverse of $f$.

Then

• $\int_0^1 f(x)\,dx$ is the area of the region bounded above by the graph of $f$ over the interval $0\le x\le 1$;

and, noting that the graph of the equation $x=f^{-1}(y)$ is precisely the graph of the equation $y=f(x)$,

• $\int_0^1 f^{-1}(y)\,dy$ is the area of the region bounded to the right by the graph of $f$, below by the interval $0\le x\le 1$, and to the left by the interval $0\le y\le1$.

The two aforementioned regions coincide; thus, we have $\int_0^1 f(x)\,dx =\int_0^1 f^{-1}(x)\,dx$.

And so, $\int_0^1 f(x)-f^{-1}(x)\,dx=0$.

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