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

• Correct answer
  A

  $(4 \sqrt{2}-2)$ sq units
• B

  $(4 \sqrt{2}+2)$ sq units
• C

  $(4 \sqrt{2}-1)$ sq units
• D

  $(4 \sqrt{2}+1)$ sq units

Solution

Required area $\begin{aligned} =\int_{0}^{\pi / 4}(\cos x-\sin x) d x &+\int_{\pi / 4}^{5 \pi / 4}(\sin x-\cos x) d x \\ & \int_{5 \pi / 4}^{3 \pi / 2}(\cos x-\sin x) d x \\ =&[\sin x+\cos x]_{0}^{\pi / 4}+[-\cos x-\sin x]_{\pi / 4}^{5 \pi / 4} \\ &+[\sin x+\cos x]_{5 x / 4}^{3 \pi / 4} \\ =&(4 \sqrt{2}-2) \text { sq units } \end{aligned}$

Hence, the correct answer is option: 'A'.