Question

\[\int\limits_0^1 \sqrt[3]{1-x^3}dx\]

Answer 1

eeewwwww ick. are you sure? maybe you are supposed to use geometry to get this

Answer 2

the x^3 makes it messy fer sure

Answer 3

It's not a homework problem... I'm just curious haha. WolframAlpha solves it with the Gamma function, and I'm not at all familiar with how that works with integrals.

Answer 4

Easy peasy
$\frac{{\sqrt \pi Gamma[\frac{4}{3}]}}{{{2^{2/3}}Gamma[\frac{5}{6}]}}$

Answer 5

\[\frac{{\sqrt \pi \Gamma[\frac{4}{3}]}}{{{2^{2/3}}\Gamma[\frac{5}{6}]}}\]

Answer 6

\[\frac{{\sqrt \pi \Gamma[\frac{4}{3}]}}{{{2^{2/3}}\Gamma[\frac{5}{6}]}}
\]

Answer 7

where does that come from??

Answer 8

I'm honestly more interested in how to get that by hand haha...of course one could look that up.

Answer 9

if anything it would come from something similar to:
\[\frac{3(1-x^3)^{4/3}}{-12x^2}\]

Answer 10

the gamma function is used in place of cleaner stuff

Answer 11

lol, don't worry about it

Answer 12

\[x^3+y^3=1\] makes a nice picture. you can visualize the area, but i have no idea how to compute it.

Answer 13

What do you mean that the gamma function is used in place of cleaner stuff? And yes, actually that graph was the motivation for me wanting to find this integral. I noticed that for all positive integer values of \(n\), the equation \(|x|^n+|y|^n=1\) forms round-ish shapes. \(n=1\) forms a square tilted \(45^\circ\), \(n=2\) makes a circle, \(n=3\) makes more of a squareish circle, and as \(n\rightarrow \infty\), it becomes even more square like. The first tilted square has area \(2\), the circle has area \(\pi\) and in the limiting case, the area approaches \(4\). The integral above is \(\frac 14\) the area of the \(n=3\) case. Have there been defined any functions like sine and cosine that describe the x and y coordinates of \(|x|^3+|y|^3=1\)? If so, do their inverse functions have power series, and can we use that to calculate the area via an infinite sum?