Non rectangular regions of integration
Compute the integral of the function $f(x)$, $f(x,y)=y \cdot e^y$ on a triangle with vertices $(0,0)$, $(1,0)$ and $(1,1)$.
First, we must write the integration limits in the integral.
In this case, this is an integral in a region with horizontal cross-sections, therefore the integral is: $$\int_R f(x,y) \ dxdy = \int_a^b \Big(\int_{g(x)}^{h(x)} f(x,y) \ dy \Big) \ dx=\int_0^1\int_0^x e^x\cdot y \ dydx$$ where for a given $x$, the variable $y$ takes values between $0$ and the straight line $y=x$ (the straight line that passes through the points $(0,0)$ and $(1,1)$, the $45$ degree line.
Therefore, $h (x) =x$, $g (x) =0$.
We calculate the integral: $$int_0^1\int_0^x e^x\cdot y \ dydx=int_0^1 e^x\Big(\int_0^x y\cdot dy \Big)\cdot dx=int_0^1 e^x\cdot\dfrac{x^2}{2} \ dx=$$ $$=\dfrac{1}{2}\int_0^1 e^x\cdot x^2 \ dx=$$ $$=\mbox{using integration by parts two times}=$$ $$=\dfrac{1}{2}[e^x(x^2-2x+2)]_0^1=\dfrac{e}{2}$$
$$\displaystyle \int_R f(x,y) \ dxdy=\dfrac{e}{2}$$