Answer to Question #219919 in Calculus for Kimutai

Question

Let R be the region bounded by the lines y=2x-1,y=2x-2,y=0 and y=4. By making the substitutions u=(2x-y)/2 ,v=y/2 and integrating over a suitable region evaluate the integral

$\intop$₀⁴ $\intop$^x=y/2+1 _x=y/2. (2x-y)/2 dxdy

Answer 1

$y=2x-1,y=2x-2,y=0, y=4\\ u=\frac{2x-y}{2}, v=\frac{y}{2} \implies x=u+v, y=2v\\ J= \frac{\partial (x,y)}{\partial (u,v)}=\begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial y}{\partial u} \\ \frac{\partial x}{\partial v} & \frac{\partial y}{\partial v} \end{vmatrix} =\begin{vmatrix} 1 & 0 \\ 1 & 2 \end{vmatrix} =2$

The integral is $I= \int_0^4 \int_{x=\frac{y}{2}}^{x={\frac{y}{2}}+1}(2x-y) \frac{1}{2}dxdy\\$

$I= \int_{u=0}^{u=1} \int_{v=0}^{v={2}}u|J|dudv\\ I= \int_{0}^{1} \int_{0}^{{2}}u(2)dudv\\ I=2[\frac{u^2}{2}]_0^1[v]_0^2 \\ I=1(2)\\ I=2$