The area (in square units) of the region enclosed between the lines \(x + y = 2, x = 0, x = 3\) and x-axis is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\frac{5}{2}\)
The line is \(y = 2 - x\). It crosses the x-axis at x = 2: it lies above the axis for \(0 \le x \lt 2\) and below it for \(2 \lt x \le 3\). So the region has two parts, and the part below the axis is counted with its sign changed.
\( \displaystyle \int_{0}^{2} (2 - x)\,dx = \left[2x - \frac{x^2}{2}\right]_{0}^{2} = 4 - 2 = 2 \)
\( \displaystyle \int_{2}^{3} (2 - x)\,dx = \left(6 - \frac{9}{2}\right) - (4 - 2) = -\frac{1}{2} \)
The second part has area \(\frac{1}{2}\). Total area:
\( \displaystyle 2 + \frac{1}{2} = \frac{5}{2} \)
Shortcut: Both parts are right triangles. From x = 0 to 2: base 2, height 2, area 2. From x = 2 to 3: base 1, height 1, area \(\frac{1}{2}\). Total \(\frac{5}{2}\).
Tip: Integrating straight from 0 to 3 gives \(2 - \frac{1}{2} = \frac{3}{2}\), because the part below the x-axis cancels instead of adding. Always split the area where the curve crosses the axis.