Calculate $\int_D \rvert x-y^2 \rvert dx \ dy $ $$\int_D \rvert x-y^2 \rvert dx \ dy $$
$D$ is the shape that is delimited from the lines:
$$
y=x \\
y=0 \\
x=1 \\$$
$$D=\{ (x,y) \in \mathbb{R}^2: 0 \le x \le 1 \  , \  0 \le y \le x \}$$
$$\rvert x-y^2 \rvert=x-y^2 \qquad \forall (x,y) \in D $$
$$\int_0^1 \Big( \int_0^x (x-y^2) \ dy \Big) \ dx= \int_0^1 \Big[_0^x xy-\frac{1}{3} y^3 \Big] \ dx= \int_0^1 \left(x^2-\frac{1}{3} x^3\right) \ dx=\Big[_0^1 \frac{1}{3} x^3-\frac{1}{12} x^4 \Big]=\frac{1}{4}$$
Is it correct? 

This is correct, you are correct that the absolute value integrand is equal to x-y^2 for all x and y in your region of integration.
