double integration

**ewander** · December 1st 2009, 11:31 AM

can someone help me solve it?
(integral from 0 to 4) dx (integral form x/2 to x^1/2) dy (x^2 - y^1/2)=

I ended up with:

(76/56) - (64*2^1/2)/21 + 64/(15*2^1/2)

help!!!

**Dogod11** · December 1st 2009, 12:12 PM

Hey, maybe it's your integral?
$<br /> \displaystyle\int_{0}^{4} dx \displaystyle\int_{x/2}^{x^{1/2}} x^2 - y^{1/2}.dy$

**ewander** · December 1st 2009, 12:30 PM

Originally Posted by Dogod11

Hey, maybe it's your integral?
$<br /> \displaystyle\int_{0}^{4} dx \displaystyle\int_{x/2}^{x^{1/2}} x^2 - y^{1/2}.dy$

yes, this is exactly my integral, can you help me with this?

**Dogod11** · December 1st 2009, 12:54 PM

Hello, well, not so difficult, is integrated first with respect to $y$ , we can separate into several integrals as follows:

$\displaystyle\int_{x/2}^{x^{1/2}} x^2.dy - \displaystyle\int_{x/2}^{x^{1/2}}y^{1/2}.dy$ =

$x^2y\bigr]_{x/2}^{x^{1/2}} - \displaystyle\frac{2}{3} .y^{3/2}\bigr]_{x/2}^{x^{1/2}}$ .

Evaluating......

$x^2( \sqrt[ ]{x} - \displaystyle\frac{x}{2} ) - (- \displaystyle\frac{1}{6} \sqrt[ ]{2} x^{3/2} + \displaystyle\frac{2}{3} x^{3/4}).<br />$
Simplify the latter expression, and the integrity of zero to four.

Greetings

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