Question

The area bounded by the curve $y=\cos x$ and $y=\sin x$ between the ordinates $x=0$ and $x=\frac{3\pi }{2}$ is

• A. $4\sqrt{2}+2$
• B. $4\sqrt{2}-1$
• C. $4\sqrt{2}+1$
• D. $4\sqrt{2}-2$

Answer 1

The correct option is D $4\sqrt{2}-2$

${\int }_0^{3\pi /2}|\sin x-\cos x|dx----\left(1\right)$

${\int }_0^{\pi /4}(\cos x-\sin x)dx+{\int }_{\pi /4}^{5\pi /4}(\sin x-\cos x)dx$

$\to {\int }_{5\pi /4}^{3\pi /2}(\cos x-\sin x)dx[\sin x+\cos x{]}_0^{\pi /4}$

$\to [\cos x-\sin x{]}_{\pi /4}^{5\pi /4}\to {\sin x+\cos x}_{5\pi /4}^{3\pi /4}$

$=\sqrt{2}-1+\sqrt{2}+\sqrt{2}+\sqrt{2}-1$

$4\sqrt{2}-2$