The area of region bounded by the x-axis, \(y=\cos x\) and \(y=\sin x\), \(0 \le x \le \displaystyle\frac{\pi}{2}\), is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(2-\sqrt{2}\) sq. units
The curves meet where \(\sin x = \cos x\), that is at \(x = \frac{\pi}{4}\), where both equal \(\frac{1}{\sqrt{2}}\).
The region bounded by the x-axis and both curves lies under the lower curve at each x: under \(y = \sin x\) for \(0 \le x \le \frac{\pi}{4}\) and under \(y = \cos x\) for \(\frac{\pi}{4} \le x \le \frac{\pi}{2}\).
\( \displaystyle \text{Area} = \int_{0}^{\pi/4}\sin x\,dx + \int_{\pi/4}^{\pi/2}\cos x\,dx \)
\( \displaystyle = \Big[-\cos x\Big]_{0}^{\pi/4} + \Big[\sin x\Big]_{\pi/4}^{\pi/2} \)
\( \displaystyle = \left(1 - \frac{1}{\sqrt{2}}\right) + \left(1 - \frac{1}{\sqrt{2}}\right) \)
\( \displaystyle = 2 - \frac{2}{\sqrt{2}} = 2 - \sqrt{2} \)
Why the other options are incorrect
Option (2) \((2\sqrt{2}-2)\) sq. units — Incorrect. This is the area between the two curves over the whole of \(\left[0, \frac{\pi}{2}\right]\), \(\int_{0}^{\pi/2}|\cos x - \sin x|\,dx\), a different region that is not bounded by the x-axis.
Option (3) \((\sqrt{2}-1)\) sq. units — Incorrect. This is the area between the two curves from x = 0 to \(x = \frac{\pi}{4}\) only, \(\int_{0}^{\pi/4}(\cos x - \sin x)\,dx\).
Option (4) \((\sqrt{2}+1)\) sq. units — Incorrect. This is about 2.41. The region lies under \(y = \sin x\) on \(\left[0, \frac{\pi}{2}\right]\), whose whole area is \(\int_{0}^{\pi/2}\sin x\,dx = 1\), so its area must be less than 1.