question_answer

The line\[x=\frac{\pi }{4}\]divides the area of the region bounded by\[y=\sin x,\,\,y=\cos x\]and x-axis\[\left( 0\le x\le \frac{\pi }{2} \right)\]into two regions of areas\[{{A}_{1}}\]and\[{{A}_{2}}\]. Then\[{{A}_{1}},\,\,{{A}_{2}}\]equals

A)  4 : 1                                      

B)  3 : 1

C)  2 : 1                      

D)         1 : 1

Correct Answer: D

Solution :

Area,\[{{A}_{1}}=\int_{0}^{\pi /4}{\sin x\,\,dx}\]                 \[=-[\cos x]_{0}^{\pi /4}=1-\frac{1}{\sqrt{2}}\]                 \[=\frac{\sqrt{2}-1}{\sqrt{2}}\] and area,             \[{{A}_{2}}=\int_{\pi /4}^{\pi /2}{\cos x\,\,dx}\]                 \[=[\sin x]_{\pi /4}^{\pi /2}=\left[ 1-\frac{1}{\sqrt{2}} \right]\]                 \[=\frac{\sqrt{2}-1}{\sqrt{2}}\] \[\therefore \]  \[{{A}_{1}}:{{A}_{2}}=\frac{\sqrt{2}-1}{\sqrt{2}}:\frac{\sqrt{2}-1}{\sqrt{2}}=1:1\]