# Find the area of the region included between $y^2=9x$ and $y=x$.

Toolbox:
• Let $f(x)$ be a continuous function defined on $[a,b]$.Then the area bounded by the curve $y=f(x)$,the $x$-axis and the ordinate $x=a$ and $x=b$ is given by
• $A=\int_a^bf(x) dx$ and $\int_a^b ydx.$
Step 1:
Given $y^2=9x$ and $y=x$
The curve $y^2=9x$ is a parabola .
The curve is open rightward.
$y=x$ is a straight line passing through the origin.
The points of intersection are :
$y=x\Rightarrow y^2=x^2$,substituting this in $y^2=9x$,we get
$9x=x^2\Rightarrow x^2-9x=0.$
$x(x-9)=0.$
The points of intersection are (0,0) and (9,9)
Step 2:
Area of the required portion is the shaded portion in the fig.
Now let us calculate the required area
$A=\int_0^9(y_2-y_1)dx.$
$y_1=\sqrt{9x}=3\sqrt x.$
$y_2=x.$
$\Rightarrow A=\int_0^9(3\sqrt x-x)dx.$
Step 3:
On integrating we get,
$A=\begin{bmatrix}3\big(\large\frac{x^{\Large\frac{3}{2}}}{\Large\frac{3}{2}}\big)-\frac{x^2}{2}\end{bmatrix}_0^9=\begin{bmatrix}2(x^{\Large\frac{3}{2}})-\large\frac{x^2}{2}\end{bmatrix}_0^9$
On applying limits we get
$\begin{bmatrix}2(9^{\Large\frac{3}{2}}\normalsize- 0)-(\large\frac{9^2}{2}-\normalsize 0)\end{bmatrix}=2(27)-\large\frac{81}{2}$
$\;\;\;=\large\frac{108-81}{2}=\frac{27}{2}$sq.units.
$A=\large\frac{27}{2}$sq.units.
edited Apr 28, 2013