$\begin{array}{1 1} \frac{1}{3}sq.units. \\ \frac{2}{3}sq.units. \\ \frac{4}{3}sq.units. \\ \frac{\pi}{3}sq.units.\end{array} $

- Whenever a function is represented by y=|x|,two cases arise:
- (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.
- If we are given two curves represented by y= f(x);y=g(x) where $f(x)\geq g(x)$ in [a,b]the point of intersection of two curves are given by x=a and x=b by taking common values of y from the equation of the two curves.

Here $x^2=y$ represents a parabola with vertex at (0,0),positive direction of y-axis as its axis and it opens upwards.Let the interior of the parabola be required region $R_1$.

y=x;$x\geq 0$ is a line passing through the origin (0,0) and on the positive direction of x-axis.Hence let $R_2$ be the required region which lies below y=x.

y=-x ,x<0 is a line passing through the origin with the positive direction of x-axis.Let the region $R_3$ be the required region which lies below y=-x.

Hence the required region is as shown in the fig.

Clearly both the curves are symmetrical about y-axis.

Hence the required area A is given by,

A=2(shaded area in Ist quadrant)

$A=2\int_0^1(x-x^2)dx.$

On integrating we get,

$A=2\begin{bmatrix}\frac{x^2}{2}-\frac{x^3}{3}\end{bmatrix}_0^1$

On applying limits we get,

$2[\frac{1}{2}-\frac{1}{3}]=2\times \frac{1}{6}=\frac{1}{3}$sq.units.

The required area is $\frac{1}{3}$sq.units.

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