Fubini's theorem

We will follow the usual procedure:

• 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:

$$\int_0^4\Big(\int_ 1^2 (xy+3y) \ dx\Big)dy = \int_1^2\Big(\int_ 0^4 (xy+3y) \ dy\Big)dx=$$ $$\int_1^2\Big[x\dfrac{y^2}{2}+3\dfrac{y^2}{2}]_0^4 \ dx=\int_1^2 8x+24 \ dx=36$$