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 $$(1,1)$$.
Development:
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: $$$int_0^1\int_0^x e^x\cdot y \ dydx=int_0^1 e^x\Big(\int_0^x y\cdot dy \Big)\cdot dx=int_0^1 e^x\cdot\dfrac{x^2}{2} \ dx=$$$ $$$=\dfrac{1}{2}\int_0^1 e^x\cdot x^2 \ dx=$$$ $$$=\mbox{using integration by parts two times}=$$$ $$$=\dfrac{1}{2}[e^x(x^2-2x+2)]_0^1=\dfrac{e}{2}$$$
Solution:
$$\displaystyle \int_R f(x,y) \ dxdy=\dfrac{e}{2}$$