Question

The area formed by triangular shaped region bounded by the curves $y=\sin x,y=\cos x$ and $x=0$ is

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

Answer 1

Answer: (A)

$(\sqrt{2}-1)$ sq unit

$(\because y=\sin x\text{and}y=\cos x\text{meet at}x=\pi /4)$

Then, the upper limit will be $\frac{\pi }{4}$.

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

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

$=(\sqrt{2}-1)$ sq unit