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Evaluate the integral: $ \int_0^1 \left(\sqrt[3]{1-x^7} - \sqrt[7]{1-x^3}\right)\;dx$

The answer is $0,$ but I am unable to get it. There is some symmetry I can not see.

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    Generally speaking, $\int_0^1 \sqrt[m]{1-x^n}dx = \int_0^1 \sqrt[n]{1-x^m}dx = \frac{\frac{1}{m} ! \cdot \frac{1}{n} !}{\left(\frac{1}{m} + \frac{1}{n}\right) !}$ where $\frac{1}{n} ! = \int_0^\infty e^{-x^n}dx$ See [here](http://en.wikipedia.org/wiki/Wikipedia:Reference_desk/Archives/Mathematics/2013_May_7#Linking_Factorials_to_Geometric_Shapes) and [here](http://en.wikipedia.org/wiki/Wikipedia:Reference_desk/Archives/Mathematics/2013_June_10#Basic_Demonstration_Required) for more details.2013-10-14

3 Answers 3

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Let $m, n > 0$. Then observe that $ \int_{0}^{1} \sqrt[n]{1-x^m} \; dx$ is the area of the region given by inequalities $ 0 \leq x \leq 1 \quad \text{and} \quad 0 \leq y \leq \sqrt[n]{1-x^m}.$ But the last inequality is equivalent to $0 \leq x^m + y^n \leq 1$. Thus $ \int_{0}^{1} \sqrt[n]{1-x^m} \; dx = [\text{Area given by} \ 0 \leq x^m + y^n \leq 1, \ 0 \leq x, y \leq 1 ]$ Thus by interchanging the role of $x$ and $y$, we have $ \int_{0}^{1} \sqrt[n]{1-x^m} \; dx = \int_{0}^{1} \sqrt[m]{1-x^n} \; dx.$


Of course, we can give a purely analytic approach. Let $y = \sqrt[3]{1 - x^7}$. Then $x = \sqrt[7]{1 - y^3}$ and hence by integration by substitution, $\begin{align*} \int_{0}^{1} \sqrt[3]{1 - x^7} \; dx &= \int_{0}^{1} y(x) \; dx \\ &= \int_{1}^{0} y \; dx(y) \\ &= [y x(y)]_{1}^{0} - \int_{1}^{0} x(y) \; dy \\ &= \int_{0}^{1} \sqrt[7]{1 - y^3} \; dy. \end{align*}$

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    @Siminore: The way you demonstrated is very clear and systematic. All the scattered understandings on this kind of integral are now integrated. Thanks.2012-05-01
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Another way to go is to use $\beta$-function

$ \mathrm{\beta}(x,y) = \int_0^1t^{x-1}(1-t)^{y-1}\,dt,\quad {Re}(x),{Re(y)}>0. $

$ \int_0^1 \sqrt[3]{1-x^7} \mathrm dx = \frac{1}{7}\int_0^1 u^{-6/7}(1-u)^{1/3} \mathrm dx=\dots\,. $

Now, you can finish the problem.

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Observing the inverse of the function $f(x) = \sqrt[3]{1 - x^7}$ on the interval $[0,1]$ is $f^{-1} (x) = \sqrt[7]{1 - x^3}$, using the result $\int^b_a f(x) \, dx + \int^{f(b)}_{f(a)} f^{-1} (x) \, dx = b f(b) - a f(a),$ since $f(a) = f(0) = 1$ and $f(b) = f(1) = 0$ it immediately follows that $\int^1_0 \sqrt[3]{1 - x^7} \, dx + \int^0_1 \sqrt[7]{1 - x^3} \, dx = 0,$ or $\int^1_0 \left \{\sqrt[3]{1 - x^7} - \sqrt[7]{1 - x^3} \right \}\, dx = 0.$