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(A) The given curve is $y=x^{2}$,which is a parabola symmetric about the $y$-axis with its vertex at the origin $(0,0)$.The region is bounded by the c

Vedclass · Accepted Answer

$32$

Vedclass · Answer

(A) The given curve is $y=x^{2}$,which is a parabola symmetric about the $y$-axis with its vertex at the origin $(0,0)$.The region is bounded by the curve $y=x^{2}$ and the line $y=4$. The intersection points of the curve and the line are found by substituting $y=4$ into $y=x^{2}$,which gives $x^{2}=4$,so $x=\pm 2$.Since the parabola is symmetric about the $y$-axis,the total area $A$ is twice the area of the region in the first quadrant bounded by the curve,the $y$-axis,and the line $y=4$.Using horizontal strips of width $dy$ and length $x = \sqrt{y}$,the area is given by:$A = 2 \int_{0}^{4} x \, dy = 2 \int_{0}^{4} \sqrt{y} \, dy$Evaluating the integral:$A = 2 \left[ \frac{y^{3/2}}{3/2} ight]_{0}^{4} = 2 	imes \frac{2}{3} \left[ y^{3/2} ight]_{0}^{4}$$A = \frac{4}{3} \left( 4^{3/2} - 0^{3/2} ight) = \frac{4}{3} 	imes 8 = \frac{32}{3}$Thus,the area of the region is $\frac{32}{3}$ square units.

Find the area of the region bounded by the curve $y=x^{2}$ and the line $y=4$. (in $/3$)

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