question_answer

The   area  bounded  by  the   curves \[y=\sqrt{x},\,\,2y+3=x\] and the x-axis lying in the first quadrant, is (in sq units)

A)  \[9\]              

B)  \[27\]           

C)  \[27/2\]            

D)  \[18\]

Correct Answer: A

Solution :

Given curves are \[y=\sqrt{x},\,2y+3=x.\] \[\therefore \]  point of intersection is \[2y+3={{y}^{2}}\] \[\Rightarrow \] \[{{y}^{2}}-2y-3=0\] \[\Rightarrow \] \[{{y}^{2}}-3y+y-3=0\] \[\Rightarrow \] \[y(y-3)+1(y-3)=0\] \[\Rightarrow \] \[(y-3)\,(y+1)=0\] \[\Rightarrow \] \[y=3,-1\] when \[y=3,\] then \[3=\sqrt{x}\Rightarrow x=9\] \[\therefore \]  The point of  intersection is \[(9,\,3)\]. \[\therefore \]  Required bounded area = Area of shaded region \[=\int_{0}^{3}{({{x}^{2}}-{{x}_{1}})\,dy}\] \[=\int_{0}^{3}{[(2y+3)-({{y}^{2}})]dy}\] \[=\left[ 2\frac{{{y}^{2}}}{2}+3y-\frac{{{y}^{3}}}{3} \right]_{0}^{3}\] \[=9+9-\frac{27}{3}=9\]