I'm a Computer Engineer Students, and Try to get familiar with Integral, but I couldn't get the point with three form of $x$ and $y$ variables. How we can easily get the point for following equation to solve it easily?
$1) E[x]=\int_{-1}^1 \int_0^1 xdxdy $
$2) E[x]=\int_{-1}^1 \int_0^1 ydxdy $
$3) E[x]=\int_{-1}^1 \int_0^1 xydxdy $
I am going to show you the second one, and you can try the others yourself.
A double integral is an integral of an integral. The inner one is in terms of the first of the $\mathrm{d}...$ variables, the outer - in terms of the second. For you this means $$ \begin{split} \int_{-1}^1 \int_0^1 y \mathrm{d}x \mathrm{d}y &= \int_{y=-1}^{y=1} \int_{x=0}^{x=1} y \mathrm{d}x \mathrm{d}y \\ &= \int_{y=-1}^{y=1} \left(\int_{x=0}^{x=1} y \mathrm{d}x\right) \mathrm{d}y \\ &= \int_{y=-1}^{y=1} \left(\int_{x=0}^{x=1} \mathrm{d}x\right) y \mathrm{d}y \\ &= \int_{y=-1}^{y=1} \left(1-0\right) y \mathrm{d}y \\ &= \int_{y=-1}^{y=1} y \mathrm{d}y \\ &= \left. \frac{y^2}{2} \right|_{y=-1}^{y=1} \\ &= 0. \end{split} $$
HINT FOR #3
$$ \begin{split} \int_{-1}^1 \int_0^1 y \mathrm{d}x \mathrm{d}y &= \int_{y=-1}^{y=1} \int_{x=0}^{x=1} xy \mathrm{d}x \mathrm{d}y \\ &= \int_{y=-1}^{y=1} \left(\int_{x=0}^{x=1} xy \mathrm{d}x\right) \mathrm{d}y \\ &= \int_{y=-1}^{y=1} \left(\int_{x=0}^{x=1} x \mathrm{d}x\right) y \mathrm{d}y \end{split} $$