The evaluation of the definite integral ${\int }_{-1}^{1.4}x\left|x\right|dx$ by using Simpson’s 1/3^rd (one –third) rule with step size h = 0.6 yields

The correct option is 0.592.

To evaluate the definite integral by Simpson's 1/3^rd rule, let us first identify the given function, the integration limits, and the step size:
The integrand is:
$f(x)=x\left|x\right|$
The lower limit of integration is $a=-1$, and the upper limit is $b=1.4$.
The step size is given as $h=0.6$.

First, we calculate the number of intervals, $n$:
$n=\frac{b-a}{h}=\frac{1.4-(-1)}{0.6}=\frac{2.4}{0.6}=4$
Since $n=4$ is an even number, we can apply Simpson's 1/3^rd rule.

Now, let us find the grid points $x_i$ and the corresponding function values $y_i=f(x_i)$:
For $x_0=-1$:
$y_0=-1\cdot |-1|=-1$
For $x_1=-1+0.6=-0.4$:
$y_1=-0.4\cdot |-0.4|=-0.16$
For $x_2=-0.4+0.6=0.2$:
$y_2=0.2\cdot \left|0.2\right|=0.04$
For $x_3=0.2+0.6=0.8$:
$y_3=0.8\cdot \left|0.8\right|=0.64$
For $x_4=0.8+0.6=1.4$:
$y_4=1.4\cdot \left|1.4\right|=1.96$

Simpson's 1/3^rd rule formula for 4 intervals is given by:
$I\approx \frac{h}{3}[(y_0+y_4)+4(y_1+y_3)+2(y_2)]$

Substituting the values into the formula:
$I\approx \frac{0.6}{3}[(-1+1.96)+4(-0.16+0.64)+2(0.04)]$
$I\approx 0.2[0.96+4(0.48)+0.08]$
$I\approx 0.2[0.96+1.92+0.08]$
$I\approx 0.2\left[2.96\right]$
$I\approx 0.592$

Thus, evaluating the definite integral yields approximately 0.592.