Find the area bounded by the curve x^{2}=4y a | vedclass

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(A) The area bounded by the curve $x^{2}=4y$ and the line $x=4y-2$ is found by determining the points of intersection.From the line equation,$4y = x+2

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$9$

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(A) The area bounded by the curve $x^{2}=4y$ and the line $x=4y-2$ is found by determining the points of intersection.From the line equation,$4y = x+2$. Substituting this into the parabola equation $x^{2}=4y$,we get $x^{2} = x+2$,which simplifies to $x^{2}-x-2=0$.Solving $(x-2)(x+1)=0$,we find the intersection points at $x=2$ and $x=-1$.For $x=2$,$y = (2^{2})/4 = 1$. Point $B$ is $(2, 1)$.For $x=-1$,$y = ((-1)^{2})/4 = 1/4$. Point $A$ is $(-1, 1/4)$.The required area is the integral of the upper curve minus the lower curve from $x=-1$ to $x=2$.Area $= \int_{-1}^{2} \left( \frac{x+2}{4} - \frac{x^{2}}{4} ight) dx$$= \frac{1}{4} \left[ \frac{x^{2}}{2} + 2x - \frac{x^{3}}{3} ight]_{-1}^{2}$$= \frac{1}{4} \left( \left( \frac{4}{2} + 4 - \frac{8}{3} ight) - \left( \frac{1}{2} - 2 - \frac{-1}{3} ight) ight)$$= \frac{1}{4} \left( \left( 6 - \frac{8}{3} ight) - \left( \frac{1}{2} - 2 + \frac{1}{3} ight) ight)$$= \frac{1}{4} \left( \frac{10}{3} - \left( \frac{3-12+2}{6} ight) ight)$$= \frac{1}{4} \left( \frac{10}{3} - \left( -\frac{7}{6} ight) ight) = \frac{1}{4} \left( \frac{20+7}{6} ight) = \frac{27}{24} = \frac{9}{8} 	ext{ sq. units.}$

Find the area bounded by the curve $x^{2}=4y$ and the line $x=4y-2$. (in $\pi$)

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