question_answer

  The area bounded by the curves\[y=cos\text{ }x\]and\[y=sin\text{ }x\]between the ordinates \[x=0\]and\[x=\frac{3\pi }{2}\]is -       AIEEE  Solved  Paper-2010

A) \[4\sqrt{2}-2\]  

B)        \[4\sqrt{2}+2\] 

C)        \[4\sqrt{2}-1\]  

D)        \[4\sqrt{2}+1\]

Correct Answer: A

Solution :

\[Area=\int\limits_{0}^{\frac{\pi }{4}}{(\cos x-\sin x)dx}+\int\limits_{\frac{\pi }{4}}^{\frac{5\pi }{4}}{(\sin x-\cos x)dx}\] \[+\int\limits_{\frac{5\pi }{4}}^{\frac{3\pi }{2}}{(\cos x-\sin x)}dx\] \[=[\sin x+\cos x]_{0}^{\frac{\pi }{4}}+[-\cos x-\sin x]_{\frac{\pi }{4}}^{\frac{5\pi }{4}}\] \[+[\sin x+\cos x]_{\frac{5\pi }{4}}^{\frac{3\pi }{2}}\] \[=\left[ \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-(0+1) \right]-\left[ -\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}} \right]\] \[+\left[ 1+0-\left( -\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}} \right) \right]\] \[=\sqrt{2}-1+\frac{4}{\sqrt{2}}-1+\sqrt{2}=4\sqrt{2}-2\]