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

$\frac{9}{8}$ sq unit

The correct answer is Option (4) → $\frac{9}{8}$ sq unit

Given equation of curve is $x^2 = 4y$ and the straight line $x = 4y - 2$.

On solving both equations, we get

$(4y - 2)^2 = 4y$

$\Rightarrow 16y^2 + 4 - 16y = 4y$

$\Rightarrow 16y^2 - 20y + 4 = 0 \Rightarrow 4y^2 - 5y + 1 = 0$

$\Rightarrow 4y^2 - 4y - y + 1 = 0 \Rightarrow 4y(y - 1) - 1(y - 1) = 0$

$\Rightarrow (4y - 1)(y - 1) = 0$

$∴y = 1, \frac{1}{4}$

For $y = 1, x = \sqrt{4 \cdot 1} = 2$ [since, negative value does not satisfy the equation of line]

For $y = \frac{1}{4}, x = -\sqrt{4 \cdot \frac{1}{4}} = -1$ [positive value does not satisfy the equation of line]

So, the intersection points are $(2, 1)$ and $\left(-1, \frac{1}{4}\right)$.

$∴\text{Area of shaded region} = \int\limits_{-1}^{2} \left( \frac{x + 2}{4} \right) dx - \int\limits_{-1}^{2} \frac{x^2}{4} dx$

$= \frac{1}{4} \left[ \frac{x^2}{2} + 2x \right]_{-1}^{2} - \frac{1}{4} \left[ \frac{x^3}{3} \right]_{-1}^{2}$

$= \frac{1}{4} \left[ \frac{4}{2} + 4 - \frac{1}{2} + 2 \right] - \frac{1}{4} \left[ \frac{8}{3} + \frac{1}{3} \right]$

$= \frac{1}{4} \cdot \frac{15}{2} - \frac{1}{4} \cdot \frac{9}{3} = \frac{45 - 18}{24}$

$= \frac{27}{24} = \frac{9}{8} \text{ sq. units}$