The area (in square units) bounded by the curve \(y = \cos x\) between \(x = 0\) and \(x = 2\pi\) in first quadrant is equal to:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 2
The first quadrant is where \(x \ge 0\) and \(y \ge 0\). Between 0 and \(2\pi\), the curve \(y = \cos x\) is on or above the x-axis on two intervals:
\(\displaystyle 0 \le x \le \frac{\pi}{2}\quad\text{and}\quad \frac{3\pi}{2} \le x \le 2\pi\)
On \(\left(\frac{\pi}{2}, \frac{3\pi}{2}\right)\), \(\cos x \lt 0\), so that part of the region lies in the fourth quadrant and is left out.
\(\displaystyle \int_{0}^{\pi/2} \cos x\,dx = \Big[\sin x\Big]_{0}^{\pi/2} = 1 - 0 = 1\)
\(\displaystyle \int_{3\pi/2}^{2\pi} \cos x\,dx = \Big[\sin x\Big]_{3\pi/2}^{2\pi} = 0 - (-1) = 1\)
\(\text{Area} = 1 + 1 = 2\)
Why the other options are incorrect
Option (1) 4 — Incorrect. This is the whole area between the curve and the x-axis from 0 to \(2\pi\), including the part of area 2 below the axis, which is not in the first quadrant.
Option (3) 1 — Incorrect. This counts only the part from 0 to \(\frac{\pi}{2}\). The part from \(\frac{3\pi}{2}\) to \(2\pi\) also has \(x \gt 0\) and \(y \ge 0\), so it is also in the first quadrant.
Option (4) 3 — Incorrect. The two first-quadrant parts have area 1 each, so the total is 2, not 3.