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