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Home  >>  CBSE XII  >>  Math  >>  Application of Integrals
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Find the area of the region bounded by the curve \(y^2 = x\) and the lines \(x = 1, x = 4\) and the \(x\) - axis.

$\begin{array}{1 1} \large \frac{14}{3}sq.units. \\ \large \frac{7}{3}sq.units. \\ \large \frac{28}{3}sq.units. \\ \large \frac{11}{3}sq.units.\end{array} $

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Toolbox:
  • To find the area bounded by the curve y=f(x),x-axis and the ordinates x=a and x=b,then the required area is given by,
  • $A=\int_a^by\;dx=\int_a^bf(x)\;dx$
Given curve is $y^2=x\Rightarrow y=\sqrt x$
Required area A is bounded by the curve and the lines x=1 and x=4 and the x-axis.
Hence $A=\int_1^4\sqrt x\;dx$.
This is clearly shown in the fig.
$\int_1^4x^\frac{1}{2}\;dx$.
on integrating we get,
$A=\begin{bmatrix}\frac{x^\frac{3}{2}}{\frac{3}{2}}\end{bmatrix}_1^4$
$\;\;\;=\frac{2}{3}\begin{bmatrix}x^\frac{3}{2}\end{bmatrix}_1^4$
on applying limits we get,
$A=\frac{2}{3}\begin{bmatrix}4^\frac{3}{2}-1^\frac{3}{2}\end{bmatrix}$
$\;\;\;=\frac{2}{3}\begin{bmatrix}2^3-1\end{bmatrix}=\frac{14}{3}sq.units.$
hence the required area =$\frac{14}{3}$sq.units.
answered Dec 20, 2013 by yamini.v
 

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