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Home  >>  CBSE XII  >>  Math  >>  Application of Integrals
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Find the area of the region included between $y^2=9x$ and $y=x$.

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  • 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
$9x=x^2\Rightarrow x^2-9x=0.$
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
$y_1=\sqrt{9x}=3\sqrt x.$
$\Rightarrow A=\int_0^9(3\sqrt x-x)dx.$
Step 3:
On integrating we get,
On applying limits we get
$\begin{bmatrix}2(9^{\Large\frac{3}{2}}\normalsize- 0)-(\large\frac{9^2}{2}-\normalsize 0)\end{bmatrix}=2(27)-\large\frac{81}{2}$
answered Apr 26, 2013 by sreemathi.v
edited Apr 28, 2013 by sreemathi.v
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