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Home  >>  CBSE XII  >>  Math  >>  Application of Integrals
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Choose the correct answer in the the area bounded by the curve \(y = x |\; x\; |\) , \(x\) - axis and the ordinates \(x = -1\) and \(x = 1\) is given by [Hint : \(y = x^2\) if \(x > 0\) and \(y = -x^2\) if \(x < 0\)].

$(a)\;0\qquad(b)\;\large\frac{1}{3}\qquad(c)\;\large\frac{2}{3}\qquad(d)\;\large\frac{4}{3}$

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Toolbox:
  • Whenever a function is represented by y=|x| two cases arises.
  • (i) y=x if $x\geq 0;$
  • (ii) y=-x if $x< 0;$
  • where $x\geq 0$ represents the portion that lies to the right side of the curve and x<0 lies to the left side of the curve.
Here the given curve is y=x|x|.
Hence we can take $y=x^2$ if x>0 and $y=-x^2$ if x<0
The required area is shown in the fig.
clearly the ordinate given are x=-1 and x=1.
Hence $A=\int_{-1}^1x|x|dx.$
$\Rightarrow A=\int_{-1}^0x^2dx+\int_0^1x^2dx$
on integrating we get,
$\Rightarrow A=\begin{bmatrix}\frac{x^3}{3}\end{bmatrix}_{-1}^0+\begin{bmatrix}\frac{x^3}{3}\end{bmatrix}_0^1$
on applying limits we get,
$A=\begin{bmatrix}0-\frac{(-1)^3}{3}\end{bmatrix}+\frac{(1)^3}{3}$
$\;\;\;=\frac{1}{3}+\frac{1}{3}=\frac{2}{3}sq.units.$
Hence the required area is $\frac{2}{3}sq. units.$
The correct answer is C.
answered Jan 23, 2013 by sreemathi.v
 

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