Past exam of the mathematical methods II course of the University of Oxford / 2016 resit / 2

The integration region lies between the line $y=x$ and the parabola $y=x^2$, with $-1\le x\le 0$. It has a portion below the $x$-axis bounded above by $y=0$, and a portion above it bounded below by $y=0$. Horizontal slices change form at $y=0$; keeping that split is essential when reversing the order of integration.

For $-1\le y\le 0$, a horizontal slice runs from $x=-1$ to the line $x=y$. For $0\le y\le 1$, it runs from $x=-1$ to the negative branch $x=-\sqrt{y}$ of the parabola. Thus the reversed iterated integral isThe positive branch $x=\sqrt{y}$ is outside the original integration region.

In the original order of integration,In the reversed order, the below-axis contribution is ${\int }_{-1}^{0}y(1+y)dy=-1/6$, and the above-axis contribution is ${\int }_0^{1}y(1-\sqrt{y})dy=1/10$. Their sum is again $-1/15$. A negative answer is possible because the integrand $y$ changes sign; the integral is not the area of the region.