question_answer

Let \[f\,(x)=\text{Min}.\]\[\left\{ x-2,\,\,\sqrt{4-x} \right\}\] then area bounded by \[y=f\,(x)\] and x-axis is

A) \[\frac{7}{6}sq.\,\,\text{units}\]              

B) \[\frac{11}{3}sq.\,\,\text{units}\]

C) \[\frac{19}{3}sq.\,\,\text{units}\]            

D) \[14\,\,sq.\,\,\text{units}\]

Correct Answer: A

Solution :

\[\text{Area}=\frac{1}{2}+\int\limits_{3}^{4}{\sqrt{4-x}\,dx}\]\[=\frac{1}{2}+\left( -\frac{{{(4-x)}^{3/2}}}{-\frac{3}{2}} \right)_{3}^{4}=\frac{1}{2}-\frac{2}{3}\,(0-1)=\frac{7}{6}\]