Area of the region bounded by the curve \(y = \sin x\) and x-axis between \(x =\displaystyle\frac{\pi}{2}\) and \(x=\displaystyle\frac{3\pi}{2}\) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 2 sq. units
On \(\left[\frac{\pi}{2}, \pi\right]\), \(\sin x \ge 0\) (curve above the x-axis); on \(\left[\pi, \frac{3\pi}{2}\right]\), \(\sin x \le 0\) (curve below it). The area is the sum of the absolute values of the two parts:
\(\displaystyle \int_{\pi/2}^{\pi} \sin x\,dx = \left[-\cos x\right]_{\pi/2}^{\pi} = 1 - 0 = 1\)
\(\displaystyle \int_{\pi}^{3\pi/2} \sin x\,dx = \left[-\cos x\right]_{\pi}^{3\pi/2} = 0 - 1 = -1\)
\(\text{Area} = 1 + |-1| = 2\) sq. units
Tip: Integrating straight from \(\frac{\pi}{2}\) to \(\frac{3\pi}{2}\) gives \(1 + (-1) = 0\), because the part below the x-axis counts as negative. Split the interval where the curve crosses the axis.