Solution:

$\begin{array}{|c|c|c|c|c|c|c|}\hline X & {0} & {1} & {2} & {3} & {4} & {5} & {6} \\ \hline Y & {1} & {0.25} & {0.1428} & {0.1} & {0.0769} & {0.0625} & {0.0526} \\ \hline \text { Ordinate } & {y_{0}} & {y_{1}} & {y_{2}} & {y_{3}} & {y_{4}} & {y_{5}} & {y_{6}} \\ \hline\end{array}$

1) Trapezoidal Rule:

$1=\frac{h}{2}(X+2 R)$

$x=$ Sum of extreme value $=1+0.0526=1.0526$

$\mathrm{R}=$ Sum of Remaining values $=0.25+0.1428+0.1+0.0769+0.0625$

$=0.6322$

$I=\frac{1}{2}(1.0526+2(0.6322))$

$I=1.1585$

$\mathrm{~ 2 ) ~ S i m p s o n s ~}(1 / 3)$ rd rule

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

$X=$ Sum of Extreme values $=1+0.0526=1.0526$

$E=$ Sum of even ordinates $=0.1428+0.0769=0.2197$

$\begin{aligned} \mathrm{O} &=\text { Sum of odd ordinates }=0.25+0.1+0.0625=0.4125 \\ I &=\frac{1}{3}(1.0526+2(0.2197)+4(0.4125)) \\ I &=0.5616 \end{aligned}$

3) Simpsons $(3 / 8)$ Th rule.

$I=\frac{3 h}{8}(X+2 T+4 R)$

$x=$ Sum of extreme value $=1+0.0526=1.0526$

$\mathrm{T}=$ Sum of multiple of three $=0.1$

$\mathrm{R}=$ Sum of Remaining values $=0.25+0.1428+0.0769+0.0625=0.5322$

$I=\frac{3 * 1}{8}(1.0526+2(0.1)+4(0.5322))$

$I=1.06845$