Official solution
\[\int_0^2|2x^2-3x|\,dx = \int_0^{3/2}(3x-2x^2)\,dx+\int_{3/2}^2(2x^2-3x)\,dx\]
\[\int_0^2\left[x-\frac12\right]dx = \int_0^{1/2}(-1)\,dx+\int_{1/2}^{3/2}0\,dx+\int_{3/2}^2 1\,dx = -\frac12+\frac12=0\]
Wait --- combining properly per the source's piecewise breakdown gives a net contribution; carrying out all pieces:
\[\left[\frac{3x^2}2-\frac{2x^3}3\right]_0^{3/2}+\left[\frac{2x^3}3-\frac{3x^2}2\right]_{3/2}^2 - \frac12+0+\frac12 = \frac{27}8-\frac{27}{12}+\frac{16}3-6-\frac{27}{12}+\frac{27}8 = \frac{19}{12}\]