1. ## tricky integral

Hi all,

2 3 3
∫ [x √(8-x ) ] dx
0

( Sorry I am not good at writing this integral clear enough. In words it's Integral of x and cube root of (8 - x cube) dx with integral range from 0 to 2. Hope this helps.)

2. Originally Posted by niz
Hi all,

2 3 3
∫ [x √(8-x ) ] dx
0

( Sorry I am not good at writing this integral clear enough. In words it's Integral of x and cube root of (8 - x cube) dx with integral range from 0 to 2. Hope this helps.)
So it's $\displaystyle \int_0^2{x\sqrt[3]{8 - x^3}\,dx}$?

3. Originally Posted by Prove It
So it's $\displaystyle \int_0^2{x\sqrt[3]{8 - x^3}\,dx}$?
Yes that's the integral. Thanks

4. $\displaystyle I \int_0^2 x \sqrt[3]{8-x^3}~dx$

By substituting $\displaystyle x = 2 \sqrt[3]{t}$

$\displaystyle dx = \frac{2}{3} t^{-\frac{2}{3}} dt$

we have $\displaystyle I = \int_0^1 2 \sqrt[3]{t} \cdot 2 \sqrt[3]{1 -t } ~ \cdot \frac{2}{3} t^{-\frac{-2}{3}} dt$

$\displaystyle = \frac{8}{3} \int_0^1 t^{ \frac{1}{3} - \frac{2}{3} } (1 - t)^{\frac{1}{3}}$

$\displaystyle = \frac{8}{3} \int_0^1 t^{-\frac{1}{3}} (1 - t)^{\frac{1}{3}}$

$\displaystyle = \frac{8}{3} B( \frac{2}{3} , \frac{4}{3} )$

$\displaystyle = \frac{16\pi}{9\sqrt{3}}$

5. Hi Simplependulum,
Thanks a lot for your help but I am bit unclear about the 2nd last step which consists of a function B(2/3, 4/3). What's this function? Can you please throw some light on it? Thanks once again.

6. Originally Posted by niz
Hi Simplependulum,
Thanks a lot for your help but I am bit unclear about the 2nd last step which consists of a function B(2/3, 4/3). What's this function? Can you please throw some light on it? Thanks once again.
Beta Function -- from Wolfram MathWorld

7. Thanks a lot.