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Find the area bounded by curves {(x, y)$ : y\: \geq\: x^2\: and \: y = |\; x\; |$}.

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Suppose we are given two curves represented by y=f(x),y=g(x) where $f(x)\geq g(x)$ in [a,b] the points of intersection of these two curves are given by x=a and x=b,by taking common values of y from the given equation of two curves.
If the given function is y=|x| these two cases arise:
case(i) y=x where $x\geq 0$
case(ii) y=-x where x<0.
$x\geq 0$ represents the portion that lies to the right side of the curve and x<0 represents the portion that lies to the left side of the curve .
The area bounded by the curves $\{(x,y):y\geq x^2$ and $y=|x|\},$ is represented as shown in the fig.
It can be said that the required area is symmetrical about y-axis.
Hence the required area can also be written as $2\times$ area bounded between the straight line y=x and parabola $y=x^2.$
Now to obtain the limit let us find the point of intersection.
Given y=x and $y=x^2$
Equating both the equation we get,
x(1-x)=0 $\Rightarrow$ x=0;x=1.
Hence the limits are 0 and 1.
The required area $A=2x\int_a^b[f(x)-g(x)]dx$
Here a=0,b=1.f(x)=1x and $g(x)=x^2$
$A=2\left\{\int_0^1 xdx -\int_0^1 x^2dx\right \}$
on integrating ,
$\:=2\left\{\begin{bmatrix}\frac{x^2}{2}\end{bmatrix}_0^1-\begin{bmatrix}\frac{x^3}{3}\end{bmatrix}_0^1\right \}$
on applying the limits we get ,
$A=2\left\{\frac{1^2}{2}-\frac{1^3}{3}\right \}$
$\;\;\;=2\times \frac{1}{6}=\frac{1}{3}sq.units.$
Hence the required area is $ \frac{1}{3}$sq.units.
answered Jan 22, 2013 by sreemathi.v

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