The area (in sq. units) of the region bounded by the lines $y = 2x + 3$, the x – axis and the ordinates $x = -2$ and $x = 2$ is equal to

The line $y = 2x + 3$ crosses the x-axis when $y = 0$:

$2x + 3 = 0$

$x = -\dfrac{3}{2}$

This lies inside the interval $[-2, 2]$, so:

From $x = -2$ to $x = -\dfrac{3}{2}$: the line is below the x-axis $(y < 0)$

From $x = -\dfrac{3}{2}$ to $x = 2$: the line is above the x-axis $(y > 0)$

---

$\text{Area} = \left|\displaystyle\int_{-2}^{-3/2}(2x+3)\,dx\right| + \int_{-3/2}^{2}(2x+3)\,dx$

The antiderivative is:

$\displaystyle\int(2x+3)\,dx = x^2 + 3x$

---

For $x = -2$ to $x = -\dfrac{3}{2}$:

$\displaystyle\int_{-2}^{-3/2}(2x+3)\,dx = \Big[x^2+3x\Big]_{-2}^{-3/2}$

$= \left(\dfrac{9}{4} - \dfrac{9}{2}\right) - \left(4 - 6\right)$

$= \left(-\dfrac{9}{4}\right) - (-2)$

$= -\dfrac{9}{4} + 2$

$= -\dfrac{1}{4}$

Taking absolute value → $\dfrac{1}{4}$

---

For $x = -\dfrac{3}{2}$ to $x = 2$:

$\displaystyle\int_{-3/2}^{2}(2x+3)\,dx = \Big[x^2+3x\Big]_{-3/2}^{2}$

$= (4+6) - \left(-\dfrac{9}{4}\right)$

$= 10 + \dfrac{9}{4}$

$= \dfrac{49}{4}$

---

$\text{Area} = \dfrac{1}{4} + \dfrac{49}{4}$

$= \dfrac{50}{4}$

$= \boxed{\dfrac{25}{2} \text{ sq. units}}$