Past exam of the mathematical methods II course of the University of Oxford / 2013 / 6

The change of variables has inverse $x=(u+3v)/7$, $y=(v-2u)/7$, with absolute inverse Jacobian determinant $1/7$. The two lines $y=-2x+1,-2x+4$ become $v=1,4$, and $y=x/3,(x-7)/3$ become $u=0,7$. Hence the parallelogram maps to a rectangle andThe denominator $v$ is bounded away from zero throughout the rectangle, so this transformation introduces no improper-integral issue.