Answer to Question #150522 in Calculus for Randal Rodriguez

Question

1. \int _0^2\int _0^1\:\left(2x+y\right)^8dx\:dy

2. \:\int _0^4\int _0^{\sqrt{y}}\:xy^2\:dx\:dy

3. \int \int R\:\left(6x^2y^3-5y^4\right)\:dA,\:R=\left\{\left(x,y\right)\left|0\le x\le 3,0\le y\le 1\right|\:\right\}

4. \int \int R\:xye^{x^2y}\:dA,\:R=\left\{\left(x,y\right)\left|0\le x\le 1,0\le y\le 2\right|\:\right\}

Answer 1

$\int _0^2\int _0^1\:\left(2x+y\right)^8dx\:dy=$

$=\int _0^2(\frac{1}{18}(2x+y)^9|_{x=0}^{x=1})\:dy=$

$=\int _0^2(\frac{1}{18}((2+y)^9-y^9))\:dy=$

$=\frac{1}{180}((2+y)^{10}-y^{10})|_{y=0}^{y=2}=$

$=\frac{1}{180}(2^{20}-2^{10}-2^{10})=\frac{1}{45}(2^{18}-2^9)=$

$=\frac{2^9}{45}*(2^{9}-1)$

$\:\int _0^4\int _0^{\sqrt{y}}\:xy^2\:dx\:dy$ =

$=\:\int _0^4(\frac{1}{2}x^2y^2)|_{x=0}^{x=\sqrt{y}}\:dy=$

$=\:\int _0^4\frac{1}{2}y^3\:dy=\frac{1}{8}y^4|_{y=0}^{y=4}=$

$=32$

$\int \int R\:\left(6x^2y^3-5y^4\right)\:dA,\:R=\left\{\left(x,y\right)\left|0\le x\le 3,0\le y\le 1\right|\:\right\}$

$=\int _0^1\int _0^3\:(6x^2y^3-5y^4)\:dx\:dy=$

$=\int _0^1(2x^3y^3-5xy^4)|_{x=0}^{x=3}\:dy=$

$=\int _0^1(54y^3-15y^4)\:dy=$

$=(\frac{27}{2}y^4-3y^5)|_{y=0}^{y=1}=$

$=\frac{27}{2}-3=10\frac{1}{2}$

$\int \int R\:xye^{x^2y}\:dA,\:R=\left\{\left(x,y\right)\left|0\le x\le 1,0\le y\le 2\right|\:\right\}$

$=\int _0^2\int _0^1\:xye^{x^2y}\:dx\:dy=$

$=\int_0^2(\frac{1}{2}e^{x^2y}|_{x=0}^{x=1})\:dy=$

$=\int_0^1(\frac{1}{2}(e^y-1))\:dy=$

$=\frac{1}{2}(e^y-y)|_{y=0}^{y=1}=$

$=\frac{1}{2}(e-1-1)=\frac{e}{2}-1$