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!!!

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- Dec 1st 2009, 11:31 AMewanderdouble integration
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!!! - Dec 1st 2009, 12:12 PMDogod11
Hey, maybe it's your integral?

$\displaystyle

\displaystyle\int_{0}^{4} dx \displaystyle\int_{x/2}^{x^{1/2}} x^2 - y^{1/2}.dy$ - Dec 1st 2009, 12:30 PMewander
- Dec 1st 2009, 12:54 PMDogod11
Hello, well, not so difficult, is integrated first with respect to $\displaystyle y$, we can separate into several integrals as follows:

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

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

Evaluating......

$\displaystyle x^2( \sqrt[ ]{x} - \displaystyle\frac{x}{2} ) - (- \displaystyle\frac{1}{6} \sqrt[ ]{2} x^{3/2} + \displaystyle\frac{2}{3} x^{3/4}).

$

Simplify the latter expression, and the integrity of zero to four.

Greetings