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The area of the region bounded by the curve x^2 = 4y and the straight line x = 4y – 2 is.

A. \frac{3}{8}  sq units

B.\frac{5}{8}  sq units

C. \frac{7}{8} sq units

D. \frac{9}{8} sq units

Answers (1)

The questions give equation for curve x^2 = 4y and straight line x = 4y – 2

Below is the figure,

\\\text{By substituting for} $4 y ; $\Rightarrow x=x^{2}-2$\\ $\Rightarrow x^{2}-x-2=0$\\ $\Rightarrow(x+1)(x-2)=0$\\ $\therefore x=-1,2$\\ \text{Required area}\\ $=\int_{-1}^{2}\left(\frac{\mathrm{x}+2}{4}\right) \mathrm{d} \mathrm{x}-\int_{-1}^{2} \frac{\mathrm{x}^{2}}{4}$ $=\frac{1}{4}\left[\frac{\mathrm{x}^{2}}{2}+2 \mathrm{x}\right]_{-1}^{2}-\frac{1}{4}\left[\frac{\mathrm{x}^{3}}{3}\right]_{-1}^{2}

\\ =\frac{1}{4}\left(\frac{4}{2}+4-\frac{1}{2}+2-\frac{8}{3}-\frac{1}{3}\right) \\ =\frac{9}{8} \text { sq. units }

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