The area of the region bounded by the curve $x = 2y + 3$ and the lines $y = 1, y = -1$ is

$6$ sq units

The correct answer is Option (3) → $6$ sq units

We have the equations, $x = 2y + 3, y = 1$ and $y = -1$.

Required area, $A = \int_{-1}^{1} (2y + 3) \, dy$

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

$= [y^2 + 3y]_{-1}^{1}$

$= [1 + 3 - 1 + 3] = 6 \text{ sq. units}$