The area (in sq. units) bounded by the curve \(y = \cos x\) and x-axis between \(x = 0\) and \(x =\displaystyle\frac{3\pi}{2}\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 3
On \(\left[0, \frac{\pi}{2}\right]\), \(\cos x \ge 0\); on \(\left[\frac{\pi}{2}, \frac{3\pi}{2}\right]\), \(\cos x \le 0\), so that part lies below the x-axis and its area is the absolute value of the integral:
\(\displaystyle \text{Area} = \int_{0}^{\pi/2} \cos x\,dx + \left|\int_{\pi/2}^{3\pi/2} \cos x\,dx\right|\)
\(\displaystyle \int_{0}^{\pi/2} \cos x\,dx = \Big[\sin x\Big]_{0}^{\pi/2} = 1 - 0 = 1\)
\(\displaystyle \int_{\pi/2}^{3\pi/2} \cos x\,dx = \Big[\sin x\Big]_{\pi/2}^{3\pi/2} = -1 - 1 = -2\)
\(\text{Area} = 1 + 2 = 3\) sq. units
Tip: Integrating straight from 0 to \(\frac{3\pi}{2}\) gives \(\sin\frac{3\pi}{2} - \sin 0 = -1\), because the parts above and below the axis cancel. For an area, split the integral where the curve crosses the x-axis.