Question

O resultado da integral dupla

Answer 1

            $\displaystyle I=\int_{1}^{2}\int_{0}^{1}3x-y^2dx\,dy\\ \\ \text{Fubini Theorem:}\\ \\ I=\int_{1}^{2} \left.\left(\dfrac{3}{2}x^2-xy^2\right)\right|_{x=0}^{x=1}dy\\ \\ I=\int_{1}^{2} \dfrac{3}{2}-y^2\;dy\\ \\ I=\left.\left(\dfrac{3}{2}y-\dfrac{y^3}{3}\right)\right|_{y=1}^{y=2}\\ \\ I=\dfrac{3}{2}-\dfrac{7}{3}\\ \\ \\ \boxed{I=-\dfrac{5}{6}}$

Answer 2

$\int\limits^2_1 \int\limits^1_0 {(3x-y^2)} \, dx dy \\ \\ \int\limits^2_1 [ \frac{3x^2}{2} - y^2x ] |^1_0} \, dy \\ \\\int\limits^2_1 [ \frac{3.1^2}{2} - y^2.1 ]-[ \frac{3.0^2}{2} - y^2.0 ] } \, dy \\ \\ \int\limits^2_1 ( \frac{3}{2} - y^2)} \, dy \\ \\ \ [ \frac{3}{2}y- \frac{y^3}{3}] |^2_1 \\ \\ \ [ \frac{3}{2}.2- \frac{2^3}{3}]-[ \frac{3}{2}1- \frac{1^3}{3}] \\ \\ \[3- \frac{8}{3}- \frac{3}{2}+ \frac{1}{3}= \boxed{-\frac{5}{6} }$