Question

The area enclosed by the curves $y=|sinx+cosx|$ and $y=|cosx-sinx|$ in $[0,\frac{\pi }{2}]$ is (in sq. units)

• A. $4-2\sqrt{2}$
• B. $4-\sqrt{2}$
• C. $8-2\sqrt{2}$
• D. $8-\sqrt{2}$

Answer 1

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

$Area={\int }_0^{\frac{\pi }{4}}|2sinx|dx+{\int }_{\frac{\pi }{4}}^{\frac{\pi }{2}}|2cosx|dx$
$=2\sqrt{2}(\sqrt{2}-1)$