Find the area enclosed by the parabola 4y = 3 | vedclass

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Question

(A) The area enclosed between the parabola $4y = 3x^{2}$ and the line $2y = 3x + 12$ is given by the integral of the upper curve minus the lower curve

Vedclass · Accepted Answer

$27 	ext{ sq. units}$

Vedclass · Answer

(A) The area enclosed between the parabola $4y = 3x^{2}$ and the line $2y = 3x + 12$ is given by the integral of the upper curve minus the lower curve between the points of intersection.First,find the points of intersection:$2y = 3x + 12 \implies y = \frac{3x + 12}{2}$Substitute this into the parabola equation $4y = 3x^{2}$:$4\left(\frac{3x + 12}{2}ight) = 3x^{2}$$2(3x + 12) = 3x^{2}$$6x + 24 = 3x^{2}$$x^{2} - 2x - 8 = 0$$(x - 4)(x + 2) = 0$So,$x = -2$ and $x = 4$.The area $A$ is given by:$A = \int_{-2}^{4} \left( \frac{3x + 12}{2} - \frac{3x^{2}}{4} ight) dx$$A = \left[ \frac{3x^{2}}{4} + 6x - \frac{3x^{3}}{12} ight]_{-2}^{4}$$A = \left[ \frac{3x^{2}}{4} + 6x - \frac{x^{3}}{4} ight]_{-2}^{4}$$A = \left( \frac{3(16)}{4} + 6(4) - \frac{64}{4} ight) - \left( \frac{3(4)}{4} + 6(-2) - \frac{-8}{4} ight)$$A = (12 + 24 - 16) - (3 - 12 + 2)$$A = (20) - (-7)$$A = 27 	ext{ sq. units}$

Find the area enclosed by the parabola $4y = 3x^{2}$ and the line $2y = 3x + 12$.

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