# double integration

• May 18th 2008, 02:23 PM
tulip
double integration
∫01 ∫y√y (2x2 + y) dy dx
• May 18th 2008, 02:34 PM
galactus
I am going to take aleap here and try to decipher that.

$\int_{0}^{1}\int_{y}^{\sqrt{y}}(2x^{2}+y)dydx$?.
• May 18th 2008, 02:50 PM
tulip
That's right! I have confused myself on how to work it out.
• May 18th 2008, 06:29 PM
Chris L T521
Quote:

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! :D
• May 18th 2008, 06:51 PM
mr fantastic
Quote:

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.