Question

Consider the curves $y=\sin x$ and $y=\cos x$.

What is the area of the region bounded by the above two curves and the lines $x=\frac{\pi }{4}$ and $x=\frac{\pi }{2}$?

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

Answer 1

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

The area enclosed by $y=\sin x,y=\cos x,x=\frac{\pi }{2},x=\frac{\pi }{4}$ is given by

${\int }_{\frac{\pi }{4}}^{\frac{\pi }{2}}(-\cos x+\sin x)dx$

$=|-\sin x-\cos x{|}_{\frac{\pi }{4}}^{\frac{\pi }{2}}$

$=-1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}$

$=\sqrt{2}-1$