# tricky integral

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• Nov 23rd 2009, 08:52 PM
niz
tricky integral
Hi all,
Here's a integral problem which has been baffling me. Please help me find its solution.

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.)
Thanks in advance.
• Nov 23rd 2009, 11:13 PM
Prove It
Quote:

Originally Posted by niz
Hi all,
Here's a integral problem which has been baffling me. Please help me find its solution.

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.)
Thanks in advance.

So it's $\int_0^2{x\sqrt[3]{8 - x^3}\,dx}$?
• Nov 24th 2009, 12:27 AM
niz
Quote:

Originally Posted by Prove It
So it's $\int_0^2{x\sqrt[3]{8 - x^3}\,dx}$?

Yes that's the integral. Thanks
• Nov 24th 2009, 02:26 AM
simplependulum
$I \int_0^2 x \sqrt[3]{8-x^3}~dx$

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

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

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

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

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

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

$= \frac{16\pi}{9\sqrt{3}}$
• Nov 24th 2009, 06:41 AM
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.
• Nov 24th 2009, 07:44 AM
Drexel28
Quote:

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
• Nov 24th 2009, 08:57 PM
niz
Thanks a lot.