question_answer

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

A) \[4\sqrt{2}-1\]                  

B) \[4\sqrt{2}+1\]

C) \[4\sqrt{2}-2\]                  

D) \[4\sqrt{2}+2\]

Correct Answer: C

Solution :

Required area A is given by \[A=\int_{0}^{3\pi /2}{|\cos x-\sin x|}dx\] \[\Rightarrow \]\[A=\int_{0}^{\pi /4}{|\cos x-\sin x|}dx+\] \[\int_{\pi /4}^{5\pi /4}{|\cos x-\sin x|dx+}\]\[\int_{5\pi /4}^{3\pi /2}{|\cos x-\sin x|dx}\] \[\Rightarrow \] \[A=\int_{0}^{\pi }{(\cos \,x-\sin x)dx+}\] \[\int_{\pi /4}^{5\pi /4}{(\sin x-\cos x)dx+}\] \[\int_{5\pi /4}^{3\pi /2}{(\cos x-\sin x)dx}\] \[\Rightarrow \] \[[\sin x+\cos x]_{0}^{\pi /4}+[-\cos x-\sin x]_{\pi /4}^{5\pi /4}\] \[+[\sin x+\cos x]_{5\pi /4}^{3\pi /2}\] \[\Rightarrow \]\[A=(\sqrt{2}-1)+(\sqrt{2}+\sqrt{2})+(-1+\sqrt{2})\] \[\Rightarrow \]\[A=4\sqrt{2}-2\]