Question

The area of the figure bounded by the lines $x=0,x=\frac{\pi }{2},f\left(x\right)=\sin x$ and $g\left(x\right)=\cos x$ is

• A. $2(\sqrt{2}-1)$
• B. $\sqrt{3}-1$
• C. $2(\sqrt{3}-1)$
• D. $2(\sqrt{2}+1)$

Answer 1

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

For $x=0to\frac{\pi }{4}\cos x>\sin xandforx=\frac{\pi }{4}to\frac{\pi }{2}\sin x>\cos x$

So, Required area $={\int }_0^{\frac{\pi }{4}}(\cos x-\sin x)dx+{\int }_{\frac{\pi }{4}}^{\frac{\pi }{2}}(-\cos x+\sin x)dx$

$={[\sin x+\cos x]}_0^{\frac{\pi }{4}}+{[-\cos x-\sin x]}_{\frac{\pi }{4}}^{\frac{\pi }{2}}=2(\sqrt{2}-1)$