Question

The area of the region bounded bu the curve x² = 4y and the straight line x = 4y - 2 is
$$\begin{gathered} \left(\mathrm{a}\right)\frac{3}{8}\mathrm{sq}.\mathrm{units} \\ \left(\mathrm{b}\right)\frac{5}{8}\mathrm{sq}.\mathrm{units} \\ \left(\mathrm{c}\right)\frac{7}{8}\mathrm{sq}.\mathrm{units} \\ \left(\mathrm{d}\right)\frac{9}{8}\mathrm{sq}.\mathrm{units} \end{gathered}$$

Answer 1

To find: area of the region bounded by the curve x² = 4y and the straight line x = 4y − 2

x² = 4y ..(1)
x = 4y − 2 ..(2)

Solving (1) and (2), we find the coordinates of the point of intersection A and B.
i.e., A$\left(-1,\frac{1}{4}\right)$ and B(2, 1)



The area of the shaded region AOBA = Area of the region ACDBA − Area of the region under the parabola above x-axis
Thus,
$$\begin{gathered} \mathrm{Required}\mathrm{area}={\int }_{-1}^2\left(\frac{x+2}{4}\right)dx-{\int }_{-1}^2\frac{x^2}{4}dx \\ =\frac{1}{4}{\left(\frac{x^2}{2}+2x\right)}_{-1}^2-\frac{1}{4}{\left(\frac{x^3}{3}\right)}_{-1}^2 \\ =\frac{1}{4}\left(\frac{2^2}{2}+2\left(2\right)-\frac{{\left(-1\right)}^2}{2}-2\left(-1\right)\right)-\frac{1}{4}\left(\frac{2^3}{3}-\frac{{\left(-1\right)}^3}{3}\right) \\ =\frac{1}{4}\left(\frac{4}{2}+4-\frac{1}{2}+2\right)-\frac{1}{4}\left(\frac{8}{3}-\frac{-1}{3}\right) \\ =\frac{1}{4}\left(2+6-\frac{1}{2}\right)-\frac{1}{4}\left(\frac{8}{3}+\frac{1}{3}\right) \\ =\frac{1}{4}\left(8-\frac{1}{2}\right)-\frac{1}{4}\left(\frac{9}{3}\right) \\ =\frac{1}{4}\left(\frac{15}{2}\right)-\frac{1}{4}\left(3\right) \\ =\frac{15}{8}-\frac{3}{4} \\ =\frac{15-6}{8} \\ =\frac{9}{8} \\ \mathrm{Thus},\mathrm{Area}=\frac{9}{8}\mathrm{sq}.\mathrm{units} \end{gathered}$$


Hence, the correct option is (d).