The area (in sq. units) of the region bounded by the line \(y = 2x - 4\), the x-axis and the ordinates \(x = 0\) and \(x = 3\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 5
The line \(y = 2x - 4\) meets the x-axis at x = 2. It is below the x-axis for x < 2 and above it for x > 2, so the area is found in two parts and the negative part is taken with its absolute value.
Since \(\int (2x - 4)\,dx = x^2 - 4x\):
\( \displaystyle \int_{0}^{2} (2x - 4)\,dx = \left[x^2 - 4x\right]_{0}^{2} = (4 - 8) - 0 = -4 \)
\( \displaystyle \int_{2}^{3} (2x - 4)\,dx = \left[x^2 - 4x\right]_{2}^{3} = (9 - 12) - (4 - 8) = 1 \)
So the area is
\(A = |-4| + 1 = 5\)
Why the other options are incorrect
Option (1) 1 — Incorrect. This is only the part from x = 2 to x = 3; the part from x = 0 to x = 2 is left out.
Option (2) 3 — Incorrect. This is the size of the single integral from 0 to 3, which is −4 + 1 = −3; the part below the axis is subtracted instead of added.
Option (4) 4 — Incorrect. This is only the part from x = 0 to x = 2; the part from x = 2 to x = 3 is left out.