<s>[INST] Iterated Integral and Sign Change in Answer Given the iterated integral $\int_0^1\int_x^{2-x}(x^2-y) \, dy \, dx$, the value for the type I integral is,
\begin{align*}
& \int_0^1\int_x^{2-x}(x^2-y)\,dy\,dx \\
= {} & \int_0^1 x^2y\Big|_x^{2-x} \, dx - \int_0^1 \left.\frac{y^2}{2}\right|_x^{2-x} \, dx \\
= {} & \int_0^1 x^2(2-2x) \, dx - \int_0^1(2-2x) \, dx \\
= {} & \int_0^1 2x^2 \, dx - \int_0^1 2x^3 \, dx - \int_0^1 2 \, dx + \int_0^1 2x \, dx \\
= {} & 2 \left.\frac{1}{3}x^3\right|_0^1 - \left.2 \frac{x^4}{4}\right|_0^1 - 2x\Big|_0^1 + 2\frac{1}{2} x^2\Big|_0^1 \\
= {} & \frac{2}{3} - \frac{1}{2} -2 + 1 \\
= {} & \frac{2}{3} - \frac{3}{2} = -\frac{5}{6}
\end{align*}
We then calculate the type II integral, thus,
$$
\int_0^1\int_y^{2-y}(x^2-y) \, dx \, dy = \frac{5}{6}
$$
The two integrals differ by the presence of a negative sign. Does this mean that these integrals are not iterated?
EDIT: It would seem that the issue has to deal with the order of the integration limits. Why are the limits reversed when you take the iterated integral with respect to $dx \, dy$?
Thank you for your time. 
 [/INST] The integral for $dxdy$ has to be split. You are integrating a different region in your actual set up. You should be integrating
$$\int_0^1\int_0^y(x^2-y) \, dx\,dy+\int_1^2\int_0^{2-y}(x^2-y)\,dx\,dy$$
instead.
The area you were integrating for $dx \, dy$ was the triangle below $y=x$ and $y=2-x$ and above $y=0$. The one you were integrating for $dy \, dx$ was to the left of $y=x$ and $y=2-x$ and to the right of $x=0$. (the first is the lower triangle, and the second is the upper triangle)

</s>