The area (in sq. units) of the region bounded by the curve \(y = \sin x, -2\pi \le x \le 2\pi\) and x-axis is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 8
On \([-2\pi, 2\pi]\) the curve makes four arches, on \([-2\pi, -\pi]\), \([-\pi, 0]\), \([0, \pi]\) and \([\pi, 2\pi]\), alternately above and below the x-axis. Each arch has the same area:
\( \displaystyle \int_{0}^{\pi} \sin x\,dx = \left[-\cos x\right]_{0}^{\pi} = 1 + 1 = 2 \)
Area is always counted as positive, so
\(\text{Area} = 4 \times 2 = 8\)
Why the other options are incorrect
Option (1) 0 — Incorrect. This is \(\int_{-2\pi}^{2\pi} \sin x\,dx\), in which the arches below the axis cancel those above. The area adds them all as positive.
Option (2) 4 — Incorrect. This is the area of only two arches, for example over \([0, 2\pi]\).
Option (4) 16 — Incorrect. This would need eight arches; the interval has four.