The area (in sq. units) of the region bounded by the curve y = |x|, the x-axis and the lines x = −3 and x = 1 is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 5
Since \(|x| \ge 0\), the whole region is above the x-axis. The integral is split at x = 0, because |x| = −x for x < 0 and |x| = x for x ≥ 0:
\( \displaystyle A = \int_{-3}^{0}(-x)\,dx+\int_{0}^{1}x\,dx \)
\( \displaystyle = \left[-\frac{x^{2}}{2}\right]_{-3}^{0}+\left[\frac{x^{2}}{2}\right]_{0}^{1} \)
\( \displaystyle = \frac{9}{2}+\frac{1}{2} = 5 \)
Why the other options are incorrect
Option (1) 4 — Incorrect. This is the size of \(\int_{-3}^{1}x\,dx = \frac{1}{2}-\frac{9}{2} = -4\), which uses x in place of |x|, so the part on the left is subtracted instead of added.
Option (3) \(\displaystyle\frac{9}{2}\) — Incorrect. This is only the part from x = −3 to x = 0; the part from x = 0 to x = 1 is left out.
Option (4) 10 — Incorrect. The two parts add up to \(\frac{9}{2}+\frac{1}{2} = 5\), not 10.