Consider the region bounded by the lines y - 1 = x, x = -2, x = 3 and…

The correct answer is Option (4) → (D) only

The line is y = x + 1. It crosses the x-axis at x = −1, which lies between −2 and 3. So y < 0 for −2 < x < −1 and y > 0 for −1 < x < 3. An area is always positive, so the part below the axis is taken with a minus sign:

    \(\displaystyle \text{Area} = \int_{-2}^{-1} -(x + 1)\,dx + \int_{-1}^{3} (x + 1)\,dx\)

The first part:

    \(\displaystyle \int_{-2}^{-1} -(x + 1)\,dx = -\left[\frac{x^2}{2} + x\right]_{-2}^{-1}\)

    \(\displaystyle = -\left[\left(\frac{1}{2} - 1\right) - (2 - 2)\right] = \frac{1}{2}\)

The second part:

    \(\displaystyle \int_{-1}^{3} (x + 1)\,dx = \left[\frac{x^2}{2} + x\right]_{-1}^{3}\)

    \(\displaystyle = \left(\frac{9}{2} + 3\right) - \left(\frac{1}{2} - 1\right) = \frac{15}{2} + \frac{1}{2} = 8\)

    \(\displaystyle \text{Area} = \frac{1}{2} + 8 = \frac{17}{2}\) sq. units

So the area is \(\frac{17}{2}\) sq. units: \(\frac{1}{2}\) below the x-axis and 8 above it.

Checking the statements

(A) The area of the bounded region is given by \(\int_{-2}^{3} (x + 1)\,dx\) — Incorrect. This integral subtracts the part below the x-axis:

    \(\displaystyle \int_{-2}^{3} (x + 1)\,dx = 8 - \frac{1}{2} = \frac{15}{2} \ne \frac{17}{2}\)

(B) The numerical value of the area is \(\frac{15}{2}\) sq. units — Incorrect. \(\frac{15}{2}\) is the value of the integral in (A), not the area.

(C) The numerical value of the area is 8 sq. units — Incorrect. 8 is only the part above the x-axis.

(D) The numerical value of the area is \(\frac{17}{2}\) sq. units — Correct. \(\frac{1}{2} + 8 = \frac{17}{2}\).

Shortcut: The region is two right triangles with their vertex at (−1, 0): the left one has base 1 and height 1, area \(\frac{1}{2}\); the right one has base 4 and height y(3) = 4, area 8. Total \(\frac{17}{2}\).