Question

The area bounded by the curves $y=$ cosx and $y=$ sinx between the ordinates $x=0$ and $x=\frac{3\pi }{2}$:

• 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 $4\sqrt{2}-2$

Required area$={\int }_0^{\frac{3\pi }{2}}|sinx-cosx|dx$
$={\int }_0^{\frac{\pi }{4}}(\cos x-\sin x)dx+{\int }_{\frac{\pi }{4}}^{\frac{5\pi }{4}}(\sin x-\cos x)dx+{\int }_{\frac{5\pi }{4}}^{\frac{3\pi }{2}}(\cos x-\sin x)dx$
$=\sqrt{2}-1+\sqrt{2}+\sqrt{2}+\sqrt{2}-1$
$=4\sqrt{2}-2$