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

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Let points of intersection of the curve and the line be $P$ and $Q$ $x^{2} = 4 (\frac{x + 2}{4})$ $x^{2} - x - 2 = 0$ $x = 2, - 1$ Point are (2,1) and $(- 1, \frac{1}{4})$ $Area = \int_{- 1}^{2} [(\frac{x + 2}{4}) - (\frac{x ^{2}}{4})] d x = [\frac{x ^{2}}{8} + \frac{1}{2} x - \frac{x ^{3}}{12}]_{- 1}^{2}$ $= (\frac{1}{2} + 1 - \frac{2}{3}) - (\frac{1}{8} - \frac{1}{2} + \frac{1}{12}) = \frac{9}{8}$

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