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

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(D) To find the area of the region bounded by the parabola $y = x^{2}$ and the line $y = 3 - 2x$,we first find their points of intersection by setting

Vedclass · Accepted Answer

$\frac{32}{3}$

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(D) To find the area of the region bounded by the parabola $y = x^{2}$ and the line $y = 3 - 2x$,we first find their points of intersection by setting $x^{2} = 3 - 2x$.$x^{2} + 2x - 3 = 0$$(x + 3)(x - 1) = 0$Thus,the points of intersection are $x = -3$ and $x = 1$.The area $A$ is given by the integral of the upper curve minus the lower curve from $x = -3$ to $x = 1$:$A = \int_{-3}^{1} ((3 - 2x) - x^{2}) dx$$A = [3x - x^{2} - \frac{x^{3}}{3}]_{-3}^{1}$$A = (3(1) - (1)^{2} - \frac{(1)^{3}}{3}) - (3(-3) - (-3)^{2} - \frac{(-3)^{3}}{3})$$A = (3 - 1 - \frac{1}{3}) - (-9 - 9 + 9)$$A = (2 - \frac{1}{3}) - (-9)$$A = \frac{5}{3} + 9 = \frac{5 + 27}{3} = \frac{32}{3}$ sq. units.

The area (in sq. units) of the region $\{(x, y) \in R^{2}: x^{2} \leq y \leq 3-2x\}$ is

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