double integration

**tulip** · May 18th 2008, 02:23 PM

∫01 ∫y√y (2x2 + y) dy dx

**galactus** · May 18th 2008, 02:34 PM

I am going to take aleap here and try to decipher that.

$\int_{0}^{1}\int_{y}^{\sqrt{y}}(2x^{2}+y)dydx$ ?.

**tulip** · May 18th 2008, 02:50 PM

That's right! I have confused myself on how to work it out.

**Chris L T521** · May 18th 2008, 06:29 PM

Originally Posted by tulip

∫01 ∫y√y (2x2 + y) dy dx

$\int_{0}^{1}\int_{y}^{\sqrt{y}}(2x^2+y)\,dy\,dx$

Its not going to work out...could it be this?

$\int_{0}^{1}\int_{y}^{\sqrt{y}}(2x^2+y)\,dx\,dy$
$\implies \int_{0}^{1}\left.\left[\frac{2}{3}x^3+yx\right]\right|_{y}^{\sqrt{y}}\,dy$
$\implies \int_{0}^{1}\left[\frac{5}{3}y^{\frac{3}{2}}-\frac{2}{3}y^3-y^2\right]\,dy$
$\implies \left.\left[\frac{2}{3}y^{\frac{5}{2}}-\frac{1}{6}y^4-\frac{1}{3}y^3\right]\right|_{0}^{1}$
$\implies \frac{2}{3}-\frac{1}{6}-\frac{1}{3}=\color{red}\boxed{\frac{1}{6}}$ .

Hope this helped you out!

**mr fantastic** · May 18th 2008, 06:51 PM

Originally Posted by tulip

∫01 ∫y√y (2x2 + y) dy dx

Are we meant to guess that you mean $\int_{0}^{1} \int_{y}^{\sqrt{y}} (2x^2 + y) \, dy \, dx$ .

If this is the question, where exactly are you stuck? Please show your working.

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