The area (in sq. units) of the region {(x, y) | vedclass

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(D) The region is bounded by the parabola $y = 4x^2$ and the line $y = 8x + 12$.To find the points of intersection,set $4x^2 = 8x + 12$.$4x^2 - 8x - 1

Vedclass · Accepted Answer

$\frac{128}{3}$

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(D) The region is bounded by the parabola $y = 4x^2$ and the line $y = 8x + 12$.To find the points of intersection,set $4x^2 = 8x + 12$.$4x^2 - 8x - 12 = 0 \implies x^2 - 2x - 3 = 0$.$(x - 3)(x + 1) = 0$,so $x = -1$ and $x = 3$.The points of intersection are $A(-1, 4)$ and $B(3, 36)$.The required area is given by the integral of the upper curve minus the lower curve from $x = -1$ to $x = 3$:Area $= \int_{-1}^{3} (8x + 12 - 4x^2) dx$$= [4x^2 + 12x - \frac{4x^3}{3}]_{-1}^{3}$$= (4(9) + 12(3) - \frac{4(27)}{3}) - (4(1) + 12(-1) - \frac{4(-1)}{3})$$= (36 + 36 - 36) - (4 - 12 + \frac{4}{3})$$= 36 - (-8 + \frac{4}{3}) = 36 - (-\frac{20}{3}) = 36 + \frac{20}{3} = \frac{108 + 20}{3} = \frac{128}{3}$ sq. units.

The area (in sq. units) of the region $\{(x, y) \in R^{2} | 4 x^{2} \leq y \leq 8 x+12\}$ is:

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