The area of the region (in sq. units), in the first quadrant bounded by the parabola $$y = 9{x^2}$$   and the lines $$x = 0,\, y = 1$$   and $$y = 4,$$  is :

Solution :
Required area
$$\eqalign{ & = \int\limits_{y = 1}^4 {\sqrt {\frac{y}{9}} dy} \cr & = \frac{1}{3}\int\limits_{y = 1}^4 {{y^{\frac{1}{2}}}dy} \cr & = \frac{1}{3} \times \frac{2}{3}\left| {\left( {{y^{\frac{3}{2}}}} \right)} \right|_1^4 \cr & = \frac{2}{9}\left[ {{{\left( {{4^{\frac{1}{2}}}} \right)}^3} - {{\left( {{1^{\frac{1}{2}}}} \right)}^3}} \right] \cr & = \frac{{14}}{9}{\text{ sq}}{\text{. units}} \cr} $$