Hey all, i asked a question on this the other day, just wanted to check my answer & it might be useful to someone else as a worked example

Evaluate the area correct to 4 significant figures, using trapezoidal,mid-ordinate&simpsons rule each using 4 intervals  <br />
\begin{gathered}<br />
  \int_1^2 {\sqrt[3]{x}} .dx \hfill \\<br />
\end{gathered} <br />
These are the 4 ordinates with 4 mid-ordinates inbetween
<br />
\left[ {\begin{array}{*{20}c}<br />
   x & y  \\<br />
   1 & 1  \\<br />
   {1.125} & {1.040}  \\<br />
   {1.250} & {1.077}  \\<br />
   {1.375} & {1.112}  \\<br />
   {1.5} & {1.145}  \\<br />
   {1.625} & {1.176}  \\<br />
   {1.750} & {1.205}  \\<br />
   {1.875} & {1.233}  \\<br />
   2 & {1.260}  \\<br />
 \end{array} } \right]<br />
Trapezoidal Rule:
<br />
\begin{gathered}<br />
  A = \frac{{0.25}}<br />
{2}\left[ {(1 + 1.26) + 2(1.077 + 1.145 + 1.205)} \right] \hfill \\<br />
  A = 1.139 \hfill \\ <br />
\end{gathered} <br />
Mid-Ordinate:
<br />
\begin{gathered}<br />
  A = 0.25(1.040 + 1.112 + 1.176 + 1.233) \hfill \\<br />
  A = 1.140 \hfill \\ <br />
\end{gathered} <br />
Simpsons:
<br />
\begin{gathered}<br />
  A = \frac{{0.25}}<br />
{3}\left[ {(1 + 1.26) + 4(1.077 + 1.205) + 2(1.145)} \right] \hfill \\<br />
  A = 1.140 \hfill \\ <br />
\end{gathered} <br />
The theoretical value by integration:
<br />
\begin{gathered}<br />
  \int_1^2 {(x)^{1/3} .dx}  \hfill \\<br />
  \left[ {\frac{{x^{4/3} }}<br />
{{4/3}}} \right]_1^2  = \left[ {\frac{{3(2)^{4/3} }}<br />
{4} - \frac{{3(1)^{4/3} }}<br />
{4}} \right] = 1.140 \hfill \\<br />
  \% {\text{error}}\left( {{\text{trapezoid}}} \right) = 0.0{\text{88}}\%  \hfill \\<br />
  \% error(mid - ordinate) = 0\%  \hfill \\<br />
  \% error(simpsons) = 0\%  \hfill \\ <br />
\end{gathered} <br />