The area of the region bounded by the curve 2y = 3x - 6 , y-axis and…

The correct answer is Option (3) → \(\displaystyle\frac{25}{3}\) sq. units

The region lies between the y-axis and the line, from y = −3 to y = 2, so it is an area with respect to the y-axis: integrate x with respect to y.

    \( \displaystyle 2y = 3x - 6 \;\Rightarrow\; x = \frac{2y + 6}{3} \)

For \(-3 \le y \le 2\), \(x \ge 0\) (x = 0 at y = −3), so the line stays on the right of the y-axis.

    \( \displaystyle \text{Area} = \int_{-3}^{2} \frac{2y + 6}{3}\,dy \)

    \( \displaystyle = \frac{1}{3}\left[y^2 + 6y\right]_{-3}^{2} \)

    \( \displaystyle = \frac{1}{3}\left[(4 + 12) - (9 - 18)\right] \)

    \( \displaystyle = \frac{1}{3}(16 + 9) = \frac{25}{3} \text{ sq. units} \)

Shortcut: The region is a triangle with vertices (0, −3), (0, 2) and \(\left(\frac{10}{3}, 2\right)\). Its base on the y-axis is 5 and its height is \(\frac{10}{3}\), so the area is \(\frac{1}{2} \times 5 \times \frac{10}{3} = \frac{25}{3}\).