Question
The value of $12 \int \limits_0^3\left|x^2-3 x+2\right| d x$ is $.............$
$={ }_{12} \int_0^3\left|\left( x -\frac{3}{2}\right)^2-\frac{1}{4}\right| dx$
If $x-\frac{3}{2}=t$
$dx = dt$
$=24 \int \limits_0^{3 / 2}\left| t ^2-\frac{1}{4}\right| dt$
$=24\left[-\int^{1 / 2}\left(t^2-\frac{1}{4}\right) d t+\int \limits_0^{3 / 2}\left(t^2-\frac{1}{4}\right) d t\right]=22$
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