The area of the region bounded by the curve x2 = 4y and the straight line x = 4y – 2 is ______.

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The area of the region bounded by the curve x2 = 4y and the straight line x = 4y – 2 is $\frac{9}{8}$ sq.units. Explanation: Given that: The equation of parabola is x2 = 4y .....(i) And equation of straight line is x = 4y – 2 .....(ii) Solving equation (i) and (ii) We get y = x^$\frac{2}{4}$ x = 4(x^$\frac{2}{4}$) - 2 $\Rightarrow$ x = x2 – 2 $\Rightarrow$ x2 – x – 2 = 0 $\Rightarrow$ x2 – 2x + x – 2 = 0 $\Rightarrow$ x(x – 2) + 1(x – 2) = 0 $\Rightarrow$ (x – 2)(x + 1) = 0 $\therefore$ x = –1, x = 2 Required area = `int_(-1)^2 (x + 2)/4 "d"x - int_(-1)^2 x^$\frac{2}{4}$ "d"x = $\frac{1}{4}$ [x^$\frac{2}{2}$ + 2x]_-1^2 - $\frac{1}{4}$ * $\frac{1}{3}$ [x^3]_1^2 = $\frac{1}{4}$ [($\frac{4}{2}$ + 4) - ($\frac{1}{2}$ - 2)] - $\frac{1}{12}$[8 + 1] = $\frac{1}{4}$ [6 + $\frac{3}{2}$] - $\frac{1}{12}$ [9] = $\frac{1}{4}$ xx $\frac{15}{2}$ - $\frac{3}{4}$ = $\frac{15}{8}$ - $\frac{3}{4}$ = $\frac{9}{8}$` sq.units [Solution diagram: 251715_solution_1.png]

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