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

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(B) Given equations are $x^2 = 4y$ $(1)$ and $x = 4y - 2$,which implies $4y = x + 2$ $(2)$.To find the points of intersection,equate the expressions f

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$9/8$

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(B) Given equations are $x^2 = 4y$ $(1)$ and $x = 4y - 2$,which implies $4y = x + 2$ $(2)$.To find the points of intersection,equate the expressions for $4y$:$x^2 = x + 2$$x^2 - x - 2 = 0$$(x - 2)(x + 1) = 0$Thus,the intersection points are $x = -1$ and $x = 2$.The area $A$ is given by the integral of the upper curve minus the lower curve:$A = \int_{-1}^{2} \left( \frac{x + 2}{4} - \frac{x^2}{4} ight) dx$$A = \frac{1}{4} \left[ \frac{x^2}{2} + 2x - \frac{x^3}{3} ight]_{-1}^{2}$$A = \frac{1}{4} \left[ \left( \frac{4}{2} + 4 - \frac{8}{3} ight) - \left( \frac{1}{2} - 2 + \frac{1}{3} ight) ight]$$A = \frac{1}{4} \left[ \left( 6 - \frac{8}{3} ight) - \left( \frac{3 - 12 + 2}{6} ight) ight]$$A = \frac{1}{4} \left[ \frac{10}{3} - \left( -\frac{7}{6} ight) ight] = \frac{1}{4} \left[ \frac{20 + 7}{6} ight]$$A = \frac{1}{4} 	imes \frac{27}{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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