Question
The area (in square units) bounded by the curves $$y = \sqrt x ,\,2y - x + 3 = 0,$$ $$x$$-axis, and lying in the first quadrant is:
A.
$$9$$
B.
$$36$$
C.
$$18$$
D.
$$\frac{{27}}{4}$$
Answer :
$$9$$
Solution :
Given curves are $$y = \sqrt x .....(1)$$
and $$\,2y - x + 3 = 0.....(2)$$
On solving both we get $$y = - 1,\,3$$
Required area :
$$\eqalign{ & = \int\limits_0^3 {\left\{ {\left( {2y + 3} \right) - {y^2}} \right\}dy} \cr & = \left. {{y^2} + 3y - \frac{{{y^3}}}{3}} \right|_0^3 \cr & = 9 \cr} $$
Given curves are $$y = \sqrt x .....(1)$$
and $$\,2y - x + 3 = 0.....(2)$$
On solving both we get $$y = - 1,\,3$$
Required area :
$$\eqalign{ & = \int\limits_0^3 {\left\{ {\left( {2y + 3} \right) - {y^2}} \right\}dy} \cr & = \left. {{y^2} + 3y - \frac{{{y^3}}}{3}} \right|_0^3 \cr & = 9 \cr} $$