show that $\int_0^1 (\sqrt[3]{1-x^7}-\sqrt[7]{1-x^3}) \, dx=0$ [duplicate]

Solution 1:

Both parts of the integral express the area under the curve given by $x^7+y^3=1$.

Solution 2:

You can write the two parts of the integrals as follows: $$ \int_0^1 \sqrt[3]{1-x^7} \, dx = \int_0^1 \int_0^\sqrt[3]{1-x^7} \, dy\,dx $$ and $$ \int_0^1 \sqrt[7]{1-y^3}) \, dy = \int_0^1 \int_0^\sqrt[7]{1-y^3} \, dx\,dy. $$ Now you can show that the regions of both these integral are the same. Namely the sets $\{(x, y): 0 \leq x \leq 1, 0 \leq \sqrt[3]{1-x^7}\}$ and $\{(x, y): 0 \leq y \leq 1, 0 \leq x \leq \sqrt[7]{1-y^3}\}$ are the same.

Solution 3:

The antiderivative of $(1-x^a)^{1/b}$, at least for integer values of $a$ and $b$, is given by $x{\rm Hypergeometric2F1}[1/a, -(1/b), 1 + 1/a, x^a]$.
The integral of $(1-x^a)^{1/b}$ between 0 and 1 is then given by
$(\Gamma[1 + 1/a] \Gamma[1 + 1/b]) / \Gamma[1 + 1/a + 1/b]$
Then, the two integrals are equal and your identity is proved.