The area bounded by the curve {x^2} = 4y and | vedclass

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(B) To find the area bounded by the curve ${x^2} = 4y$ and the line $x = 4y - 2$,we first find their points of intersection.From the line equation,$4y

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$\frac{9}{8} \, 	ext{sq. unit}$

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(B) To find the area bounded by the curve ${x^2} = 4y$ and the line $x = 4y - 2$,we first find their 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$${x^2} - x - 2 = 0$$(x - 2)(x + 1) = 0$So,$x = 2$ and $x = -1$.When $x = 2$,$4y = 4 \implies y = 1$. Point $A(2, 1)$.When $x = -1$,$4y = 1 \implies y = \frac{1}{4}$. Point $B(-1, \frac{1}{4})$.The required area is the integral of the upper curve minus the lower curve from $x = -1$ to $x = 2$:$	ext{Area} = \int_{-1}^{2} \left( \frac{x+2}{4} - \frac{x^2}{4} ight) dx$$= \frac{1}{4} \int_{-1}^{2} (x + 2 - x^2) dx$$= \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{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{1}{4} 	imes \frac{27}{6} = \frac{27}{24} = \frac{9}{8} \, 	ext{sq. unit}$.

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

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