The area of the region bounded by the curve y | vedclass

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(B) The curve is $y=x^2$ and the line is $y=16$. The intersection points are found by setting $x^2=16$,which gives $x = \pm 4$.Since the region is sym

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$\frac{256}{3}$ sq. units

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(B) The curve is $y=x^2$ and the line is $y=16$. The intersection points are found by setting $x^2=16$,which gives $x = \pm 4$.Since the region is symmetric about the $y$-axis,the area $A$ is given by:$A = 2 \int_0^{16} x \, dy = 2 \int_0^{16} \sqrt{y} \, dy$$A = 2 \left[ \frac{y^{3/2}}{3/2} ight]_0^{16}$$A = 2 	imes \frac{2}{3} \left[ y^{3/2} ight]_0^{16}$$A = \frac{4}{3} \left( 16^{3/2} - 0^{3/2} ight)$$A = \frac{4}{3} 	imes (4^2)^{3/2} = \frac{4}{3} 	imes 4^3$$A = \frac{4}{3} 	imes 64 = \frac{256}{3} 	ext{ sq. units}$

The area of the region bounded by the curve $y=x^2$ and the line $y=16$ is

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