question_answer

The area of the figure bounded by\[f(x)=\sin \,x\] and \[g(x)=\cos x\] in the first quadrant is

A)  \[2(\sqrt{2}-1)\,sq\]units        

B)  \[(\sqrt{3}+1)\,sq\] units

C)  \[2(\sqrt{3}-1)\,sq\]units        

D)  None of these

Correct Answer: A

Solution :

 We know that, \[\sin \,x\ge \,\cos \,x\]for \[x\in \,\left[ \frac{x}{4},\,\frac{\pi }{2} \right]\] and \[\sin \,x\le \cos \,x\] for \[x\in \,\left[ 0,\,\frac{\pi }{4} \right]\]. Therefore, the required area \[=\,\int_{0}^{\pi /4}{(\cos \,x-\sin \,x)\,dx}\] \[+\,\int_{\pi /4}^{\pi /2}{(\sin \,x-\cos \,x)dx}\] \[=\,[\sin \,x+\cos \,x]_{0}^{\pi /4}+[-\cos \,x-\sin \,x]_{\pi /4}^{\pi /2}\] \[=(\sqrt{2}-1)+\sqrt{2}-1\] \[=2\,(\sqrt{2}-1)\,sq\] units