Question

Using integration, find the area of the region bounded by the curve \( y = x|x| \), the x-axis, and the vertical lines \( x = -2 \) and \( x = 2 \).

Hint:

Since \( y = x|x| \) is an odd function, the geometric shapes on either side of the origin are completely identical in area. You can find the area for the positive half and multiply it by 2.

Answer 1

Concept:
The function \[ y=x|x| \] is defined differently for positive and negative values of \(x\). Since \[ |x|= \begin{cases} x, & x\ge 0,\\ -x, & x<0, \end{cases} \] we have \[ y=x|x|= \begin{cases} x^2, & x\ge 0,\\ -x^2, & x<0. \end{cases} \] The area enclosed by the curve and the \(x\)-axis from \(x=a\) to \(x=b\) is \[ \text{Area}=\int_a^b |y|\,dx. \] 

Step 1: Split the integral at \(x=0\).
Since the expression changes at \(x=0\), \[ \text{Area} = \int_{-2}^{0}|-x^2|\,dx + \int_{0}^{2}|x^2|\,dx. \] As \[ |-x^2|=x^2, \] we get \[ \text{Area} = \int_{-2}^{0}x^2\,dx + \int_{0}^{2}x^2\,dx. \] 

Step 2: Evaluate each integral.
Using \[ \int x^2\,dx=\frac{x^3}{3}, \] First integral: \[ \int_{-2}^{0}x^2\,dx = \left[\frac{x^3}{3}\right]_{-2}^{0} = 0-\left(-\frac{8}{3}\right) = \frac{8}{3}. \] Second integral: \[ \int_{0}^{2}x^2\,dx = \left[\frac{x^3}{3}\right]_{0}^{2} = \frac{8}{3}. \] 

Step 3: Find the total area.
\[ \text{Area} = \frac{8}{3} + \frac{8}{3} = \frac{16}{3}\text{ sq. units}. \] Therefore, \[ \boxed{\text{Area}=\frac{16}{3}\text{ sq. units}.} \]