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 />