<s>[INST] Double integral over square shaped region $$\iint_R(y-2x^2)dxdy$$ where $R$ is the region inside the square $|x|+|y|=1$.
So the area is:
\begin{align}
4×\left[\int_{x=0}^{x=1}\int_{y=0}^{y=1-x}(y-2x^2)dydx\right]
&=4×\left[\int_0^1\int_{y=0}^{y=1-x}ydydx-2\int_0^1\int_{y=0}^{y=1-x}x^2dydx\right]\\
&=4×\left[\frac{1}{2}\int_0^1(1-x)^2dx-2\int_0^1x^2(1-x)dx\right]\\
&=4×\left[\int_0^1(1-x)^2dx-2\int_0^1x^2dx+2\int_0^1x^3dx\right]\\
&=4×\left[-\frac{1}{6}[(1-x)^3]_0^1-\frac{2}{3}[x^3]_0^1+\frac{2}{4}[x^4]_0^1\right]\\
&=4×\left[\frac{1}{6}-\frac{2}{3}+\frac{1}{2}\right]\\
&=0
\end{align}
I can't find my fault please check this..
 [/INST] HINT
The proposed integral can be expressed as follows:
\begin{align*}
I = \int_{-1}^{0}\int_{-x-1}^{x+1}(y - 2x^{2})\mathrm{d}y\mathrm{d}x + \int_{0}^{1}\int_{x - 1}^{-x + 1}(y - 2x^{2})\mathrm{d}y\mathrm{d}x
\end{align*}
Can you take it from here?
</s>