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:

2

The correct answer is Option (2) → 2

$\text{Curve: } y=\cos x$

$\text{First quadrant: } y\ge 0,\; x\in[0,2\pi]$

In the first quadrant, $\cos x \ge 0$ only on:

$x\in[0,\frac{\pi}{2}]$ and $x\in[\frac{3\pi}{2},2\pi]$

So the required area is:

$\displaystyle \int_{0}^{\pi/2} \cos x\,dx \;+\; \int_{3\pi/2}^{2\pi} \cos x\,dx$

$= \left[\sin x\right]_{0}^{\pi/2} + \left[\sin x\right]_{3\pi/2}^{2\pi}$

$= (1-0) + (0 - (-1))$

$= 1 + 1$

$=2$