Question

The area of one curvilinear triangle formed by $y=\sin x,y=\cos x$ and x-axis is?

• A. $2$ sq. units
• B. $(2+\sqrt{2})$ sq units
• C. $(2-\sqrt{2})$ sq units
• D. $2\sqrt{3}$ sq units

Answer 1

Answer: (C)

$(2-\sqrt{2})$ sq units

$\begin{array}{c}Areaboundedbycurveis\\ {\int }_0^{\pi /4}\sin xdx+{\int }_{\pi /4}^{\pi /4}\cos xdx\\ -{\left[\cos x\right]}_0^{\pi /4}+{\left[\sin x\right]}_{\pi /4}^{\pi /4}\\ -\left[\cos \frac{\pi }{4}-\cos 0\right]+\left[\sin \frac{\pi }{2}-\sin \frac{\pi }{4}\right]\\ -\left[\frac{1}{\sqrt{2}}-1\right]+\left[1-\frac{1}{\sqrt{2}}\right]\\ 2\left[1-\frac{1}{\sqrt{2}}\right]=2-\sqrt{2}\end{array}$