The area (in sq. units) of the region bounded | vedclass

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(C) To find the area bounded by the curves $y=x^2$ and $y=8-x^2$,we first find their points of intersection by setting $x^2 = 8-x^2$. $2x^2 = 8 \impli

Vedclass · Accepted Answer

$\frac{64}{3}$

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(C) To find the area bounded by the curves $y=x^2$ and $y=8-x^2$,we first find their points of intersection by setting $x^2 = 8-x^2$. $2x^2 = 8 \implies x^2 = 4 \implies x = \pm 2$. The curves intersect at $x = -2$ and $x = 2$. In the interval $[-2, 2]$,the curve $y=8-x^2$ lies above $y=x^2$. The area $A$ is given by the integral: $A = \int_{-2}^{2} ((8-x^2) - x^2) \, dx = \int_{-2}^{2} (8-2x^2) \, dx$. Since the function is even,$A = 2 \int_{0}^{2} (8-2x^2) \, dx$. $A = 2 [8x - \frac{2x^3}{3}]_{0}^{2} = 2 [8(2) - \frac{2(8)}{3}] = 2 [16 - \frac{16}{3}] = 2 [\frac{48-16}{3}] = 2 [\frac{32}{3}] = \frac{64}{3}$ sq. units.

The area (in sq. units) of the region bounded by the curves $y=x^2$ and $y=8-x^2$ is

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