The area (in sq. units) of the region bounded by the lines y = 2x+3,…

The correct answer is Option (3) → $\frac{25}{2}$

Given line $y=2x+3$

The area between the curve, $x$-axis and $x=-2$ to $x=2$ is

$\int_{-2}^{2}|2x+3|\,dx$

The line cuts the $x$-axis at $2x+3=0$ giving $x=-\frac{3}{2}$

Hence split the integral

$=\int_{-2}^{-3/2}-(2x+3)\,dx+\int_{-3/2}^{2}(2x+3)\,dx$

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

$=\left[\left(-\frac{9}{4}+\frac{9}{2}\right)-\left(-4+6\right)\right]+\left[(4+6)-\left(\frac{9}{4}-\frac{9}{2}\right)\right]$

$=\left[\frac{9}{4}-2\right]+\left[10+\frac{9}{4}\right]$

$=\frac{1}{4}+\frac{49}{4}$

$=\frac{50}{4}$

$=\frac{25}{2}$

The required area is $\frac{25}{2}$ square units.