Question

If the area bounded by $y=\left|\left|x\right|^{2} - 4 \left|x + 3\right|\right|$ and the $x$ -axis from $x=1$ to $x=3$ is $\frac{p}{q}$ square units (where, $p$ & $q$ are coprime), then the value of $p+q$ is

Answer 1

Area $=\left|\displaystyle \int _{1}^{3} \left(4 x - x^{2} + 12\right) d x\right|$ $=\frac{94}{3}$ square units.