question_answer

The area (in square units) bounded by the curves\[y=\sqrt{x},\text{ }2y-x+3=0,\]x−axis, and lying in the first quadrant is:     AIEEE Solevd Paper-2013

A) 9                             

B) 36                          

C)        18                          

D)        \[\frac{27}{4}\]

Correct Answer: A

Solution :

                \[y=\sqrt{x}\] and \[2y-x+3=0\] \[\int\limits_{0}^{9}{\sqrt{x}}dx-\int\limits_{3}^{9}{\left( \frac{x-3}{2} \right)}dx\] \[\left( \frac{{{x}^{3/2}}}{3/2} \right)_{0}^{9}-\left[ \frac{\left( \frac{{{x}^{2}}}{2}-3x \right)}{2} \right]_{3}^{9}\] \[\Rightarrow \]9 square units