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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](http://math.stackexchange.com/questions/139393/how-to-show-that-int-01-left-sqrt31-x7-sqrt71-x3-right-dx) is a similar pro$b$lem (the answer there includes your integral as a particular case).2012-12-19

2 Answers 2

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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)?2013-10-06
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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, $\int_0^1 f(x)-f^{-1}(x)\,dx=0$.