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Home  >>  CBSE XII  >>  Math  >>  Application of Integrals
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The area between \(x = y^2\) and \(x = 4\) is divided into two equal parts by the line \(x = a\), find the value of \(a.\)

$\begin{array}{1 1} \large\frac{2}{3} \\ \large\frac{4}{3} \\ \large\frac{1}{3} \\ \large\frac{2}{5} \end{array} $

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  • whenever the area bounded by a curve and a line is divided into two equal parts,then the limits of the curve can also be split into two.
Here it is given in the area between $x=y^2$ and x=4 is divided into two equal parts by the line x=a.
From the fig we can see that the area is symmetrical about x-axis.
hence the required area can be divided as,$y_1$ and $y_2$.
 
Given $y_1=y_2$.
 
Area of $y_1=\int_0^a \sqrt x\;dx$--------(1)
Integrating this we get,
Area of $y_1=\begin{bmatrix}\frac{x^\frac{3}{2}}{\frac{3}{2}}\end{bmatrix}_0^9$
 
$\;\;\;\;\quad\qquad=\frac{2}{3}[x^\frac{3}{2}]$
on applying limits we get,
Area of $y_1=\frac{2}{3}[a^\frac{3}{2}]$--------(2)
 
Area of $y_2=\int_a^4\sqrt x\;dx$----------(3)
on integrating we get,
Area of $y_2=\begin{bmatrix}\frac{x^\frac{3}{2}}{\frac{3}{2}}\end{bmatrix}_a^4$
 
$\;\;\;\;\quad\qquad=\frac{2}{3}[x^\frac{3}{2}]_a^4$
on applying limits we get,
Area of $y_2=\frac{2}{3}[4^\frac{3}{2}-a^\frac{3}{2}]$
 
$\;\;\;\;\quad\qquad=\frac{2}{3}[8-a^\frac{3}{2}]$---------(4)
Combining (3) and (4) gives the required area.
But since they are equal in area let us equate both the equations.
$\quad\frac{2}{3}a^\frac{3}{2}=\frac{2}{3}[8-a^\frac{3}{2}]$
 
$\;\;\Rightarrow a^\frac{3}{2}=[8-a^\frac{3}{2}]$
 
$\;\Rightarrow 2a^\frac{3}{2}=8$
 
$\;\;\Rightarrow a=(4)^\frac{2}{3}$
 
The value of a is $(4)^\frac{2}{3}$.

 

answered Jan 24, 2013 by sreemathi.v
 

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