The area (in sq. units) of the region bounded by the curve y = x^2-2x…

The correct answer is Option (4) → \(\displaystyle\frac{8}{3}\)

Since \(x^{2}-2x = x(x-2)\), the curve crosses the x-axis at x = 2. It lies below the axis for 0 < x < 2 and above it for 2 < x < 3, so the two parts are found separately and added as positive numbers.

Part below the axis:

    \( \displaystyle \int_{0}^{2}(x^{2}-2x)\,dx = \left[\frac{x^{3}}{3}-x^{2}\right]_{0}^{2} \)

    \( \displaystyle = \frac{8}{3}-4 = -\frac{4}{3} \), so this part has area \(\displaystyle\frac{4}{3}\)

Part above the axis:

    \( \displaystyle \int_{2}^{3}(x^{2}-2x)\,dx = (9-9)-\left(\frac{8}{3}-4\right) = \frac{4}{3} \)

Adding the two parts:

    \( \displaystyle A = \frac{4}{3}+\frac{4}{3} = \frac{8}{3} \)

Why the other options are incorrect

• Option (1) 0 — Incorrect. This is \(\int_{0}^{3}(x^{2}-2x)\,dx\), in which the part below the axis cancels the part above it.

• Option (2) \(\displaystyle\frac{4}{3}\) — Incorrect. This is the area of only one of the two parts.

• Option (3) 2 — Incorrect. The two parts add up to \(\frac{8}{3}\), not 2.