Question

Find the area bounded by the curve $y=x\left|x\right|,$$x$-axis and the ordinates $x=1,x=-1$.

• A. $0$
• B. $\frac{1}{3}$
• C. $\frac{2}{3}$
• D. None of these

Answer 1

The correct option is C

$\frac{2}{3}$

Explanation of correct answer:

Finding area within the curves

Given, $y=x\left|x\right|=\left\{\begin{array}{l}-x^2-1\le x<0\\ x^{2}0\le x\le 1\end{array}\right.$

In the diagram: blue line is the curve of $y=x\left|x\right|$, orange line is of $x=1$and violet line is of $x=-1$

The area bounded by these three curves and the $x$-axis is the area within the curves.

Required area $={\int }_{-1}^{1}x\left|x\right|dx$

Using properties of definite integral

$$\begin{gathered} ={\int }_{-1}^{0}x\left|x\right|dx+{\int }_0^{1}x\left|x\right|dx \\ ={\int }_{-1}^0-x^{2}dx+{\int }_0^1{x}^{2}dx \\ =-{\int }_{-1}^0{x}^{2}dx+{\int }_0^1{x}^{2}dx \\ ={\int }_0^1{x}^{2}dx+{\int }_0^1{x}^{2}dx \end{gathered}$$

$=2{\int }_0^1{x}^{2}dx$

$$\begin{gathered} =2{\left[\frac{x^3}{3}\right]}_0^1 \\ =\frac{2}{3}\text{sq units} \end{gathered}$$

Hence, the area within the curve is $\frac{2}{3}\mathrm{sq}\mathrm{units}$

Hence, option (C) is the correct answer.