Simpson’s 1/3 Rule

Simpson’s 1/3 Rule

In Simpson's Rule, we will use parabolas to approximate each part of the curve. This proves to be very efficient since it's generally more accurate than the other numerical methods we've seen.

We divide the area into $\displaystyle{n}$$n$ equal segments of width $\displaystyle\Delta{x}$$\Delta x$. The approximate area is given by the following.

Area $\displaystyle={\int_{{a}}^{{b}}} f{{\left({x}\right)}}{\left.{d}{x}\right.}$

$\displaystyle\approx\frac{{\Delta{x}}}{{3}}{\left({y}_{{0}}+{4}{y}_{{1}}+{2}{y}_{{2}}+{4}{y}_{{3}}+{2}{y}_{{4}}+\right.}$ $\displaystyle{\left.\ldots+{4}{y}_{{{n}-{1}}}+{y}_{{n}}\right)}$​+yn​)

where $$\displaystyle\Delta{x}=\frac{{{b}-{a}}}{{n}}

We can re-write Simpson's Rule by grouping it as follows:

$\displaystyle{\int_{{a}}^{{b}}} f{{\left({x}\right)}}{\left.{d}{x}\right.}$ $\displaystyle\approx\frac{{\Delta{x}}}{{3}}{\left[{y}_{{0}}+{4}{\left({y}_{{1}}+{y}_{{3}}+{y}_{{5}}+\ldots\right)}\right.}$ $\displaystyle{\left.+{2}{\left({y}_{{2}}+{y}_{{4}}+{y}_{{6}}+\ldots\right)}+{y}_{{n}}\right]}$

This gives us an easy way to remember Simpson's Rule:

$\displaystyle{\int_{{a}}^{{b}}} f{{\left({x}\right)}}{\left.{d}{x}\right.}$ $\displaystyle\approx\frac{{\Delta{x}}}{{3}}{\left[\text{FIRST}+{4}{\left(\text{sum of ODDs}\right)}\right.}$ $\displaystyle{\left.+{2}{\left(\text{sum of EVENs}\right)}+\text{LAST}\right]}$

➽  Example using Simpson's 1/3 Rule

Approximate $\displaystyle{\int_{{2}}^{{3}}}\frac{{{\left.{d}{x}\right.}}}{{{x}+{1}}}$ using Simpson's Rule with $\displaystyle{n}={4}$.

We haven't seen how to integrate this using algebraic processes yet, but we can use Simpson's Rule to get a good approximation for the value.

Solution :

                                                                                     

$\Delta x=\frac{b-a}{n}=\frac{3-2}{4}$= 0.25

$\displaystyle{y}_{{0}}= f{{\left({a}\right)}}$

$\displaystyle= f{{\left({2}\right)}}$

$\displaystyle=\frac{1}{{{2}+{1}}}={0.3333333}$= 0.3333333

$\displaystyle{y}_{{1}}= f{{\left({a}+\Delta{x}\right)}}= f{{\left({2.25}\right)}}$ $\displaystyle=\frac{1}{{{2.25}+{1}}}={0.3076923}$= 0.3076923

$\displaystyle{y}_{{2}}= f{{\left({a}+{2}\Delta{x}\right)}}= f{{\left({2.5}\right)}}$=      f(2.5)$\displaystyle=\frac{1}{{{2.5}+{1}}}={0.2857142}$= 0.2857142

$\displaystyle{y}_{{3}}= f{{\left({a}+{3}\Delta{x}\right)}}= f{{\left({2.75}\right)}}$=f       (2.75) $\displaystyle=\frac{1}{{{2.75}+{1}}}={0.2666667}$= 0.2666667

$\displaystyle{y}_{{4}}= f{{\left({b}\right)}}= f{{\left({3}\right)}}$ $\displaystyle=\frac{1}{{{3}+{1}}}={0.25}$= 0.25

So

Area $\displaystyle={\int_{{a}}^{{b}}} f{{\left({x}\right)}}\text{d}{x}$

$\displaystyle\approx\frac{0.25}{{3}}{\left({0.333333}+{4}{\left({0.3076923}\right)}\right.}$ $\displaystyle+{2}{\left({0.2857142}\right)}+{4}{\left({0.2666667}\right)}$ $\displaystyle{\left.+{0.25}\right)}$ 

$\displaystyle={0.2876831}$

Notes

1. The actual answer to this problem is 0.287682 (to 6 decimal places) so our Simpson's Rule approximation has an error of only 0.00036%. 

$\displaystyle\approx\frac{{\Delta{x}}}{{3}}{\left({y}_{{0}}+{4}{y}_{{1}}+{2}{y}_{{2}}+{4}{y}_{{3}}+{2}{y}_{{4}}+\right.$