Question

Find the area of bounded by $y=\sin x$ from $x=\frac{\pi }{4}$ to $x=\frac{\pi }{2}$

• A. $\frac{\sqrt{2}-1}{\sqrt{2}}$
• B. $\frac{1}{2}$
• C. $\frac{1}{4}$
• D. None of these

Answer 1

Answer: (A)

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

The area bounded is given as ${\int }_{\pi /4}^{\pi /2}\sin xdx{-\cos x|}_{\pi /4}^{\pi /2}-\cos \frac{\pi }{2}+\cos \frac{\pi }{4}\frac{\sqrt{2}-1}{\sqrt{2}}$