1. double integration

∫01 ∫y√y (2x2 + y) dy dx

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

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

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

4. Originally Posted by tulip
∫01 ∫y√y (2x2 + y) dy dx
$\displaystyle \int_{0}^{1}\int_{y}^{\sqrt{y}}(2x^2+y)\,dy\,dx$

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

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

Hope this helped you out!

5. Originally Posted by tulip
∫01 ∫y√y (2x2 + y) dy dx
Are we meant to guess that you mean $\displaystyle \int_{0}^{1} \int_{y}^{\sqrt{y}} (2x^2 + y) \, dy \, dx$.

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