Question

O valor da Integral dupla

Answer 1

$\displaystyle I=\int_{0}^{1}\int_{0}^{2}(x^3y+x)\,dx\,dy\\\\ I=\int_{0}^{1}\left[\int_{0}^{2}(x^3y+x)\,dx\right]\,dy\\\\ I=\int_{0}^{1}\left[\frac{x^4}{4}\cdot y+\frac{x^2}{2}\right] _{0}^{2}\,dy\\\\ I=\int_{0}^{1}\left[\left(\frac{2^4}{4}\cdot y+\frac{2^2}{2}\right)- \underbrace{\left(\frac{0^4}{4}\cdot y+\frac{0^2}{2}\right)}_{=\,0}\right] \,dy\\\\ I=\int_{0}^{1}\left[\frac{16}{4}\cdot y+\frac{4}{2}\right] \,dy\\\\ I=\int_{0}^{1}(4y+2) \,dy$

$\displaystyle I=\left[4\cdot\frac{y^2}{2}+2y\right]_{0}^{1}\\\\ I=\left[\left(4\cdot\frac{1^2}{2}+2\cdot1\right)-\underbrace{\left(4\cdot\frac{0^2}{2}+2\cdot0\right)}_{=\,0}\right]\\\\ I=\left[4\cdot\frac{1}{2}+2\right]\\\\ I=[2+2]\\\\ I=4\\\\ \boxed{\int_{0}^{1}\int_{0}^{2}(x^3y+x)\,dx\,dy =4}$