# double integration

• Dec 1st 2009, 11:31 AM
ewander
double 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 PM
Dogod11
Hey, maybe it's your integral?
$
\displaystyle\int_{0}^{4} dx \displaystyle\int_{x/2}^{x^{1/2}} x^2 - y^{1/2}.dy$
• Dec 1st 2009, 12:30 PM
ewander
Quote:

Originally Posted by Dogod11
Hey, maybe it's your integral?
$
\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?
• Dec 1st 2009, 12:54 PM
Dogod11
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}).
$

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

Greetings