Browse Questions

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 $x < 0$].

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

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 $\frac{2}{3}sq. units.$