The area of the region S = {(x, y): 3x^{2} le | vedclass

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(C) The region is bounded by the parabola $y = \frac{3}{4}x^{2}$ and the line $y = \frac{3}{2}x + 6$.To find the intersection points $A$ and $B$,we se

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$27$

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(C) The region is bounded by the parabola $y = \frac{3}{4}x^{2}$ and the line $y = \frac{3}{2}x + 6$.To find the intersection points $A$ and $B$,we set the equations equal:$\frac{3}{4}x^{2} = \frac{3}{2}x + 6$Multiply by $4$:$3x^{2} = 6x + 24$$3x^{2} - 6x - 24 = 0$$x^{2} - 2x - 8 = 0$$(x - 4)(x + 2) = 0$So,$x = -2$ and $x = 4$.The area is given by the integral of the upper curve minus the lower curve:$	ext{Area} = \int_{-2}^{4} \left( (\frac{3}{2}x + 6) - \frac{3}{4}x^{2} ight) dx$$= \left[ \frac{3x^{2}}{4} + 6x - \frac{x^{3}}{4} ight]_{-2}^{4}$$= \left( \frac{3(16)}{4} + 6(4) - \frac{64}{4} ight) - \left( \frac{3(4)}{4} + 6(-2) - \frac{-8}{4} ight)$$= (12 + 24 - 16) - (3 - 12 + 2)$$= 20 - (-7) = 27$.

The area of the region $S = \{(x, y): 3x^{2} \leq 4y \leq 6x + 24\}$ is $......$

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