Answer :

Solution :

Given: f(x) = x|x|

a = -1, b = 1.4, h = 0.6

$\mathrm{N}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{o}\mathrm{f}\mathrm{i}\mathrm{n}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{v}\mathrm{a}\mathrm{l}=\frac{1.4-\left(-1\right)}{0.6}=4$

According to Simpson's 1/3^rd rule

$\underset{-1}{\overset{1.4}{\int }}x\left|x\right|dx=\frac{\mathrm{h}}{3}\left[{\mathrm{y}}_{\mathrm{o}}+{\mathrm{y}}_4+2\left({\mathrm{y}}_2\right)+4\left({\mathrm{y}}_1+{\mathrm{y}}_3\right)\right]$

$\underset{-1}{\overset{1.4}{\int }}x\left|x\right|dx=\frac{0.6}{3}\left[-1+1.96+2\left(0.04\right)+4\left(-0.16+0.64\right)\right]$

$\therefore \underset{-1}{\overset{1.4}{\int }}x\left|x\right|dx=0.592$