Question

The area of the figure bounded by the curves $y=\cos x$ and $y=\sin x$ and the co-ordinate $x=0$ and $x=\pi /4$ is

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

Answer 1

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

The given curves are $y=\cos x$ and $y=\sin x$ and $x$ lies between $0$ to $\frac{\pi }{4}$

$\therefore$ The required area is

$I={\int }_0^{\frac{\pi }{4}}\cos x-\sin x.dx$

$=|\sin x+\cos x{|}_0^{\frac{\pi }{4}}$

$=2\left(\frac{1}{\sqrt{2}}\right)-1$

$=\sqrt{2}-1$.