Area of the region bounded by the curve \( y=\sin x,y=\cos x \) and \( x \)-axis, \( \left(0\le x\le\displaystyle\frac{\pi}{2}\right) \) is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( (2-\sqrt{2}) \) Square units
The curves meet where \( \sin x=\cos x \), i.e. at \( x=\frac{\pi}{4} \). The region bounded by the two curves and the x-axis lies under sin x from 0 to \( \frac{\pi}{4} \) (there sin x is the lower curve) and under cos x from \( \frac{\pi}{4} \) to \( \frac{\pi}{2} \).
\( \displaystyle \text{Area}=\int_{0}^{\pi/4}\displaystyle\sin x\,dx+\int_{\pi/4}^{\pi/2}\displaystyle\cos x\,dx \)
\( \displaystyle =\Big[-\cos x\Big]_{0}^{\pi/4}+\Big[\sin x\Big]_{\pi/4}^{\pi/2} \)
\( \displaystyle =\left(1-\frac{1}{\sqrt{2}}\right)+\left(1-\frac{1}{\sqrt{2}}\right) \)
\( \displaystyle =2-\frac{2}{\sqrt{2}} \)
\( \displaystyle =2-\sqrt{2} \)
Shortcut: By symmetry about \( x=\frac{\pi}{4} \), the two pieces are equal, so the area is \( 2\left(1-\frac{1}{\sqrt{2}}\right)=2-\sqrt{2} \).
Why the other options are incorrect
Option (1) \( (2+\sqrt{2}) \) Square units — Incorrect. \( 2+\sqrt{2} \) is larger than 2, the whole area under both curves on \( \left[0,\frac{\pi}{2}\right] \) (1 each), so it cannot be a part of that area.
Option (2) \( (\sqrt{2}-1) \) Square units — Incorrect. \( \sqrt{2}-1=\int_{0}^{\pi/4}(\cos x-\sin x)\,dx \) is the area between the two curves from x = 0 to \( \frac{\pi}{4} \), which is bounded by the y-axis, not the x-axis.
Option (4) \( (\sqrt{2}+1) \) Square units — Incorrect. \( \sqrt{2}+1 \) is also larger than 2, so it cannot be a part of the area under the curves.