Question

The area bounded by the curves $y=\sin x,y=\cos x$ and $y-$axes in first quadrant is:

• A. $\sqrt{2}-1$
• B. $\sqrt{2}$
• C. $\sqrt{2}+1$
• D. None of the above

Answer 1

The correct option is A $\sqrt{2}-1$

The area bounded by the curves $y=\sin x,y=\cos x$ and $y-$ axis in the first quadrant is,

$A={\int }_0^{\pi /4}(\cos x-\sin x)dx$

$=[\sin x+\cos x{]}_0^{\pi /4}$

$=(\sin \frac{\pi }{4}+\cos \frac{\pi }{4})-(\sin 0+\cos 0)$

$=\sqrt{2}-1$