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

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

The line cuts the x-axis at \(2x + 3 = 0\), that is \(x = -\frac{3}{2}\), which lies between −2 and 2. So the region is two triangles, one below and one above the x-axis, and their areas are added.

Below the axis, from x = −2 to \(x = -\frac{3}{2}\): base \(\frac{1}{2}\), height \(|2(-2) + 3| = 1\):

    \( \displaystyle A_1 = \frac{1}{2} \times \frac{1}{2} \times 1 = \frac{1}{4} \)

Above the axis, from \(x = -\frac{3}{2}\) to x = 2: base \(\frac{7}{2}\), height \(2(2) + 3 = 7\):

    \( \displaystyle A_2 = \frac{1}{2} \times \frac{7}{2} \times 7 = \frac{49}{4} \)

    \( \displaystyle \text{Area} = \frac{1}{4} + \frac{49}{4} = \frac{50}{4} = \frac{25}{2} \)

Why the other options are incorrect

• Option (1) 12 — Incorrect. This is \(\int_{-2}^{2} (2x + 3)\,dx = \frac{49}{4} - \frac{1}{4}\), where the part below the axis is subtracted instead of added.

• Option (2) \(\displaystyle\frac{49}{4}\) — Incorrect. This is only the part above the x-axis.

• Option (4) 25 — Incorrect. This is twice the actual area.