Question

The area bounded by the curve $y=\sin x$ and $y=\cos x$ in between $x=0$ and $x=\frac{\pi }{2}$ is

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

Answer 1

The correct option is D $2-\sqrt{2}$


The area enclosed by the curves
$=\underset{0}{\overset{\frac{\pi }{4}}{\int }}\sin xdx+\underset{\frac{\pi }{4}}{\overset{\frac{\pi }{2}}{\int }}\cos xdx=-\cos x{|}_0^{\frac{\pi }{4}}+\sin x{|}_{\frac{\pi }{4}}^{\frac{\pi }{2}}=1-\frac{1}{\sqrt{2}}+1-\frac{1}{\sqrt{2}}=2-\sqrt{2}$