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Home  >>  CBSE XII  >>  Math  >>  Application of Integrals
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Find the area of the region bounded by \(y^2 = 9x,\; x = 2,\; x = 4\) and the \(x\) - axis in the first quadrant.

$\begin{array}{1 1} 16-4\sqrt 2sq.units. \\8-4\sqrt 2sq.units. \\ 16-2\sqrt 2sq.units. \\ 8-4\sqrt 3sq.units.\end{array} $

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Toolbox:
  • To find the area bounded by the curve y=f(x),x-axis and the ordinate x=a,x=b then the required area is given by,
  • $A=\int_a^b y\;dx=\int_a^b f(x)\;dx.$
Given curve is $y^2=9x\Rightarrow y=3\sqrt x.$
Required area is bounded by the curve $y^2=9x,x=2,x=4 and the x-axis the first quadrant.
Hence $A=\int_2^4 3\sqrt x\;dx.$
This clearly shown in fig.
on integrating we get,
$A=3\begin{bmatrix}\frac{x^\frac{3}{2}}{\frac{3}{2}}\end{bmatrix}_2^4$
$\;\;\;\;=3\times\frac{2}{3}\begin{bmatrix}x^\frac{3}{2}\end{bmatrix}_2^4$
Applying limits,
$\;\;\;=2\begin{bmatrix}4^\frac{3}{2}-2^\frac{3}{2}\end{bmatrix}$
$\;\;\;=2\begin{bmatrix}8-2\sqrt 2\end{bmatrix}=\big(16-4\sqrt 2\big)sq.units.$
Hence the required area is $16-4\sqrt 2$sq.units.
answered Dec 20, 2013 by yamini.v
 

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