Find the area of the region bounded by curve $\displaystyle 4 x^{2}=\mathrm{y}$ and the line $\displaystyle \mathrm{y}=8 x+12$, using integration.

Correct Figure Point of intersection \[\begin{aligned} & 4 x^{2}=8 x+12 \\ & x^{2}-2 x-3=0 \\ & (x-3)(x+1)=0 \\ & x=3,-1 \end{aligned} \] \[\begin{aligned} \text { Area } & =\int_{-1}^{3}\left[(8 x+12)-4 x^{2}\right] d x \\ & \left.=4 x^{2}+12 x-\frac{4}{3} x^{3}\right]_{-1}^{3} \\ & =36+36-36-\left(4-12+\frac{4}{3}\right) \\ & =44-\frac{4}{3}=\frac{128}{3} \end{aligned} \]