## Approximation Integral

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:
$\displaystyle \begin{gathered} \int_1^2 {(x)^{1/3} .dx} \hfill \\ \left[ {\frac{{x^{4/3} }} {{4/3}}} \right]_1^2 = \left[ {\frac{{3(2)^{4/3} }} {4} - \frac{{3(1)^{4/3} }} {4}} \right] = 1.140 \hfill \\ \% {\text{error}}\left( {{\text{trapezoid}}} \right) = 0.0{\text{88}}\% \hfill \\ \% error(mid - ordinate) = 0\% \hfill \\ \% error(simpsons) = 0\% \hfill \\ \end{gathered}$