Question

The area of the figure bounded by $f\left(x\right)=\sin x,g\left(x\right)=\cos x$ in the first quadrant is:

• A. $2(\sqrt{2}-1)$ sq.unit
• B. $\sqrt{3}+1$ sq.unit
• C. $2(\sqrt{3}-1)$.sq.unit
• D. none of these.

Answer 1

The correct option is A $2(\sqrt{2}-1)$ sq.unit

We know that $\sin x\ge \cos x$ for $x\in [\frac{\pi }{4},\frac{\pi }{2}]$
and $\sin x\le \cos x$ for $x\in [0,\frac{\pi }{4}]$
$\therefore$ the required area is
$={\int }_0^{\frac{\pi }{4}}(\cos x-\sin x)dx+{\int }_{\frac{\pi }{4}}^{\frac{\pi }{2}}(\sin x-\cos x)dx$
$={(\sin x+\cos x)}_0^{\frac{\pi }{4}}+{(\sin x+\cos x)}_{\frac{\pi }{4}}^{\frac{\pi }{2}}$
$=(\sqrt{2}-1)+(\sqrt{2}-1)$
$=2(\sqrt{2}-1)$.sq.unit.