Compute the integral of the function $$f(x,y)=xy+3y$$ in the region $$R=[1,2] \times [0,4]$$: $$$\int_0^4\int_ 1^2 xy+3y \ dxdy$$$
Development:
We will follow the usual procedure:
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We will compute the integral with respect to $$x$$, assuming $$y$$ is constant $$$\int_0^4\Big(\int_ 1^2 (xy+3y) \ dx\Big)dy = \int_0^4\Big[ y \cdot \dfrac{x^2}{2}+3yx\Big]_1^2 \ dy =$$$ $$$=\int_0^4 y\Big(2-\dfrac{1}{2}\Big) +3y\cdot(2-1) \ dy= \int_0^4 \dfrac{9}{2}y \ dy$$$
- We compute the integral with respect to $$y$$ $$$\int_0^4 \dfrac{9}{2}y \ dy= \dfrac{9}{2}\int_0^4 y \ dy=\dfrac{9}{2} \cdot 8=36$$$
There is another way to solve the problem: we will use Fubini's theorem to rearrange the integral and will arrive at the same result:
$$$\int_0^4\Big(\int_ 1^2 (xy+3y) \ dx\Big)dy = \int_1^2\Big(\int_ 0^4 (xy+3y) \ dy\Big)dx=$$$ $$$\int_1^2\Big[x\dfrac{y^2}{2}+3\dfrac{y^2}{2}]_0^4 \ dx=\int_1^2 8x+24 \ dx=36$$$
Solution:
$$\displaystyle \int_0^4\int_ 1^2 (xy+3y) \ dxdy = 36$$