The area of the region S = ( x , y ) : 3 x 2 ≤ 4 y ≤ 6 x + 24 is______.

We have,

$S=\left\{(x,y):3x^2\le 4y\le 6x+24\right\}$

Then,

$y=\frac{3x^2}{4}$

$4y=6x+24$

Solving both equations, we get $\left(-2,3\right)$ and $\left(4,12\right)$.

Required area

$A={\int }_{-2}^4 \left(\frac{6x+24}{4}-\frac{3x^2}{4}\right)dx$

$\Rightarrow A=\frac{1}{4}{\int }_{-2}^4 \left(6x+24-3x^2\right)dx$

$\Rightarrow A={\left(\frac{3}{4}{x}^2+6x-\frac{x^3}{4}\right)}_{-2}^4$

$\Rightarrow A=\left(12+24-16-3+12-2\right)$

$\Rightarrow A=27 \mathrm{sq}.\mathrm{units}$