Solution
= Solution
In the original <order of integration>,
$$
I=\frac12\int_{-1}^0(x^4-x^2)\,dx
=\frac12\left(\frac15-\frac13\right)=\boxed{-\frac1{15}.}
$$
In the reversed order, the below-axis contribution is $\int_{-1}^0y(1+y)\,dy=-1/6$, and the above-axis contribution is $\int_0^1y(1-\sqrt y)\,dy=1/10$. 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.