Solution:

$\begin{array}{ll}{\text { Let }} & {I=\int_{0}^{1} \frac{x^{2}}{1+x^{3}} d x} \\ {a=0,} & {b=1}\end{array}$

Dividing limits into 4 parts i.e $n=4 \quad \therefore h=\frac{b-a}{n}=\frac{1}{4}=0.25$

$\begin{array}{|c|c|c|c|c|}\hline x_{0}=0 & {x_{1}=0.25} & {x_{2}=0.50} & {x_{3}=0.75} & {x_{4}=1.0} \\ \hline y_{0=0} & {y_{1}=0.06153} & {y_{2}=0.2222} & {y_{3}=0.39560} & {y_{4}=0.5} \\ \hline\end{array}$

(i) Trapezoidal rule:

$\quad I=\frac{h}{2}[X+2 R]------(1)$

$\begin{aligned} X=& \text {sum of extreme ordinates}=y_{0}+y_{4}=0+0.5=0.5 \\ R=& \text {sum of remaining ordinates}=y_{1}+y_{2}+y_{3} \\ &=0.06153+0.2222+0.39560=0.67933----From(1) \end{aligned}$

(ii) Simpson's $(1 / 3)^{r d}$ rule:

$\mathrm{I}=\frac{h}{3}[X+2 E+40]------------(2)$

$\begin{aligned} X =\text {sum of extreme ordinates}=y_{0}+y_{4}=0+0.5=0.5 \\E=\text {sum of even base} \text { ordinates }=y_{2}=0.2222 \end{aligned}$

$\boldsymbol{o}=$ sum of odd base ordinates $=y_{1}+y_{3}=0.06153+0.39560=$ 0.45713

$\begin{aligned} I &=\frac{0.25}{3}(0.5+2 \times 0.2222+4 \times 0.45713) \\ \therefore I &=0.23108------From(2) \end{aligned}$

(iii) Simpson's $(3 / 8)^{\text { th }}rule :$

$I=\frac{3 h}{8}[X+2 T+3 R]--------(3)$

$X=$sum of extreme ordinates$=y_{0}+y_{4}=0+0.5=0.5$

$T=$ sum of multiple of three base ordinates $=y_{3}=0.39560$

$R=$ sum of remaining ordinates $=y_{1}+y_{2}=0.06153+0.2222=0.28373$

$I=\frac{3 \times 0.25}{8}(0.5+2 \times 0.39560+3 \times 0.28373)$

$\therefore I=0.2008----------From (3)$