Question

The area bounded by the curves $y = cos\, x$ and $y = sin \,x$ between the ordinates $x = 0$ 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

Answer 1

Answer: (A)

$\left(4\sqrt{2}-2\right)$ sq units

Answer 2

Required area
=\int_\limits{0}^{\pi / 4}(\cos x-\sin x) d x+\int_{\pi / 4}^{5 \pi / 4}(\sin x-\cos x) d x
\int_\limits{5 \pi / 4}^{3 \pi / 2}(\cos x-\sin x) d x
$=[\sin x-\cos x]_{0}^{\pi / 4}+[-\cos x-\sin x]_{\pi / 4}^{5 \pi / 4}$
$+[\sin x-\cos x]_{5 x / 4}^{3 \pi / 4}$
$=(4 \sqrt{2}-2)$ sq unitsa