Question

find the integral of f(x,y)=x^4+y^2 over the region bounded by y=x,y=2x and x=2

Answer 1

Consider the given region :

$\begin{aligned} & \text{The given region is} \\ & \\ & x\le y\le 2x \\ & \\ & 0\le x\le 2 \\ & \\ & \text{Find the integral as} \\ & \\ & \int\limits_{0}^{2}{\int\limits_{x}^{2x}{\left( {{x}^{4}}+{{y}^{2}} \right)dydx}}=\int\limits_{0}^{2}{\left[ {{x}^{4}}y+\frac{{{y}^{3}}}{3} \right]_{x}^{2x}dx} \\ & \\ & =\int\limits_{0}^{2}{\left( \left( {{x}^{4}}\left( 2x \right)+\frac{{{\left( 2x \right)}^{3}}}{3} \right)-\left( {{x}^{4}}\left( x \right)+\frac{{{\left( x \right)}^{3}}}{3} \right) \right)}dx \\ & \\ & =\int\limits_{0}^{2}{\left( \left( 2{{x}^{5}}+\frac{8{{x}^{3}}}{3} \right)-\left( {{x}^{5}}+\frac{{{x}^{3}}}{3} \right) \right)}dx \\ & \\ & =\int\limits_{0}^{2}{\left( {{x}^{5}}+\frac{7{{x}^{3}}}{3} \right)}dx \\ & \\ & \text{Apply}\,\text{the}\,\text{Power}\,\text{Rule}:\quad \int{{{x}^{a}}}dx=\frac{{{x}^{a+1}}}{a+1} \\ & \\ & =\left[ \frac{{{x}^{6}}}{6}+\frac{7{{x}^{4}}}{12} \right]_{0}^{2} \\ & \\ & =\frac{{{2}^{6}}}{6}+\frac{7\cdot {{2}^{4}}}{12}-0 \\ & \\ & =\text{20} \\ \end{aligned}$