The area (in sq. units) of the region bounded | vedclass

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(A) Given curves are $x^2=4y$ and $x=4y-2$. From the second equation,$4y = x+2$. Substituting this into the first equation,we get $x^2 = x+2$,which im

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$\frac{9}{8}$

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(A) Given curves are $x^2=4y$ and $x=4y-2$. From the second equation,$4y = x+2$. Substituting this into the first equation,we get $x^2 = x+2$,which implies $x^2 - x - 2 = 0$. Solving for $x$,we get $(x-2)(x+1) = 0$,so $x=2$ and $x=-1$. When $x=2$,$y=1$. When $x=-1$,$y=1/4$. The points of intersection are $(2,1)$ and $(-1, 1/4)$. The required area is given by the integral of the upper curve minus the lower curve from $x=-1$ to $x=2$: Area $= \int_{-1}^{2} [\frac{x+2}{4} - \frac{x^2}{4}] dx$ $= \frac{1}{4} [\frac{x^2}{2} + 2x - \frac{x^3}{3}]_{-1}^{2}$ $= \frac{1}{4} [(\frac{4}{2} + 4 - \frac{8}{3}) - (\frac{1}{2} - 2 + \frac{1}{3})]$ $= \frac{1}{4} [(6 - \frac{8}{3}) - (\frac{1}{2} - \frac{5}{3})]$ $= \frac{1}{4} [\frac{10}{3} - (-\frac{7}{6})] = \frac{1}{4} [\frac{20+7}{6}] = \frac{27}{24} = \frac{9}{8} 	ext{ sq. units.}$

The area (in sq. units) of the region bounded by the curve $x^2=4y$ and the straight line $x=4y-2$ is

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