Q. The area bounded by the curves $y = \cos x\ and\ y = \sin x$ between the ordinance $x = 0\ and\ x = \frac{3\pi}{2}$ is
A) $\left( 4\sqrt{2} - 2 \right)sq\ units$
B) $\left( 4\sqrt{2} + 2 \right)sq\ units$
C) $\left( 4\sqrt{2} - 1 \right)sq\ units$
D) $\left( 4\sqrt{2} + 1 \right)sq\ units$
Explanation
(a)
[image:8_image29.jpeg]
Required area
$= \int_{0}^{x/4}{(\cos{x - \sin{x)dx + \int_{\pi/4}^{5\pi/4}{(\sin{x - \cos{x)dx + \int_{5\pi/4}^{3\pi/2}{(\cos{x - \sin{x)dx}}}}}}}}}$
$= (\sin{x + \cos{x)_{0}^{\pi/4} + ( - \cos{x - \sin{x)_{\pi/4}^{5\pi/4} + (\sin{x + \cos{x)_{5\pi/4}^{3\pi/2}}}}}}}$
$= \left( 4\sqrt{2} - 2 \right)sq\ units$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.