The area of the region bounded by the parabol | vedclass

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(D) The region is bounded by $y=x^2$ and $y=4$. First,find the points of intersection by setting $x^2 = 4$,which gives $x = \pm 2$. The area $A$ is gi

Vedclass · Accepted Answer

$\frac{32}{3}$

Vedclass · Answer

(D) The region is bounded by $y=x^2$ and $y=4$. First,find the points of intersection by setting $x^2 = 4$,which gives $x = \pm 2$. The area $A$ is given by the integral of the upper curve minus the lower curve from $x = -2$ to $x = 2$: $A = \int_{-2}^{2} (4 - x^2) dx$ Since the function is symmetric about the $y$-axis,$A = 2 \int_{0}^{2} (4 - x^2) dx$. $A = 2 [4x - \frac{x^3}{3}]_{0}^{2}$ $A = 2 [(4(2) - \frac{2^3}{3}) - (0)]$ $A = 2 [8 - \frac{8}{3}] = 2 [\frac{24-8}{3}] = 2 [\frac{16}{3}] = \frac{32}{3}$ sq. units.

The area of the region bounded by the parabola $y=x^2$ and the line $y=4$ is . . . . . . sq. units.

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