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Home  >>  CBSE XII  >>  Math  >>  Application of Integrals
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Find the area of the region bounded by the curves $y^2=9x,y=3x.$

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  • Area of the region bounded by the curve $y=f(x)$,the $x$-axis and the ordinates $x=a$ and $x=b$,where $f(x)$ is a continuous function defined on $[a,b]$,is given by
  • $A=\int_a^bf(x)dx$ or $\int_a^b ydx.$
Step 1:
Given $y^2=9x$ and $y=3x$
Clearly $y^2=9x$ is a parabola,whose focus a can be calculated as shown below :
$y^2=4ax$ is the general equation .
In the given equation $4a=9\Rightarrow a=\large\frac{9}{4}$
The curve is open rightward with the focus as $\big(\large\frac{9}{4}$$,0\big)$
Step 2:
$y=3x$ is a straight line.
Next let us find the point of intersection of the line and the parabola.
$y=3x\Rightarrow y^2=9x^2$
Hence substituting this in equ(1) we get,
$\Rightarrow x=0$ and $1$
Hence the points of intersection are 0 and 1.
The area of the required region is as shown in the fig.
Now let us evaluate the area of the required region.
Step 3:
When $y_2=3\int_0^1\sqrt xdx$ and $y_1=\int_0^1 3xdx.$
$A=3\int_0^1\sqrt xdx-3\int_0^1 xdx.$
On integrating we get,
$\;\;=2\begin{bmatrix}x^{\large\frac{3}{2}}\end{bmatrix}_0^1-\large\frac{3}{2}\begin{bmatrix}\normalsize x^2\end{bmatrix}_0^1$
On applying limits we get,
Hence the required area is $\large\frac{1}{2}$sq.units.
answered Apr 25, 2013 by sreemathi.v
edited Apr 26, 2013 by sreemathi.v
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