The area bounded between the curve x^{2}=y an | vedclass

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(A) The given curves are $x^{2}=y$ and $y=4x$. To find the intersection points,substitute $y=x^{2}$ into $y=4x$: $x^{2}=4x \implies x^{2}-4x=0 \implie

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$\frac{32}{3} 	ext{ sq unit}$

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(A) The given curves are $x^{2}=y$ and $y=4x$. To find the intersection points,substitute $y=x^{2}$ into $y=4x$: $x^{2}=4x \implies x^{2}-4x=0 \implies x(x-4)=0$. So,$x=0$ and $x=4$. When $x=0, y=0$ and when $x=4, y=16$. The intersection points are $(0,0)$ and $(4,16)$. The required area $A$ is given by the integral: $A = \int_{0}^{4} (4x - x^{2}) dx$. Integrating the terms: $A = \left[ \frac{4x^{2}}{2} - \frac{x^{3}}{3} ight]_{0}^{4} = \left[ 2x^{2} - \frac{x^{3}}{3} ight]_{0}^{4}$. Evaluating at the limits: $A = \left( 2(4)^{2} - \frac{(4)^{3}}{3} ight) - (0) = \left( 32 - \frac{64}{3} ight) = \frac{96-64}{3} = \frac{32}{3} 	ext{ sq unit}$.

The area bounded between the curve $x^{2}=y$ and the line $y=4x$ is

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