The area bounded by the curve x^2 = 8y and th | vedclass

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(C) Given equations are $x^2 = 8y$ $(1)$ and $x - 8y + 2 = 0$ $(2)$.From $(2)$,$8y = x + 2$. Substituting this into $(1)$,we get $x^2 = x + 2$,which i

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$\frac{9}{16}$ sq. units

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(C) Given equations are $x^2 = 8y$ $(1)$ and $x - 8y + 2 = 0$ $(2)$.From $(2)$,$8y = x + 2$. Substituting this into $(1)$,we get $x^2 = x + 2$,which implies $x^2 - x - 2 = 0$.Factoring the quadratic equation: $(x - 2)(x + 1) = 0$,so 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 from $x = -1$ to $x = 2$.$A = \int_{-1}^{2} (\frac{x+2}{8} - \frac{x^2}{8}) dx = \frac{1}{8} \int_{-1}^{2} (x + 2 - x^2) dx$.$A = \frac{1}{8} [\frac{x^2}{2} + 2x - \frac{x^3}{3}]_{-1}^{2}$.Evaluating at the limits: $A = \frac{1}{8} [(\frac{4}{2} + 4 - \frac{8}{3}) - (\frac{1}{2} - 2 + \frac{1}{3})]$.$A = \frac{1}{8} [(2 + 4 - \frac{8}{3}) - (\frac{1}{2} - 2 + \frac{1}{3})] = \frac{1}{8} [\frac{10}{3} - (-\frac{7}{6})] = \frac{1}{8} [\frac{20+7}{6}] = \frac{1}{8} 	imes \frac{27}{6} = \frac{9}{16}$ sq. units.

The area bounded by the curve $x^2 = 8y$ and the straight line $x - 8y + 2 = 0$ is

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