
Simpson's Rule

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The Trapezoidal and Midpoint estimates provided better accuracy than the Left and Right endpoint estimates. It turns out that a certain combination of the Trapezoid and Midpoint estimates is even better.

Definitions: Trapezoidal and Midpoint Estimates

Let $$f(x)$$ be a function defined on $$[a,b]$$. Then

$S(n) = \dfrac{1}{3} T(n) + \dfrac{2}{3} M(n)$

where $$T(n)$$ and $$M(n)$$ are the Trapezoidal and Midpoint Estimates.

Geometrically, if $$n$$ is an even number then Simpson's Estimate gives the area under the parabolas defined by connecting three adjacent points.

Let $$n$$ be even then using the even subscripted $$x$$ values for the trapezoidal estimate and the midpoint estimate, gives

\begin{align} S(n) &= \dfrac{1}{3}\Big[\dfrac{(b-a)}{2n} \big( f(x_0) +2f(x_2) +...+ f(x_{2n-2}) +f(x_{2n}) \big)\Big] \\ &\;\;\; + \dfrac{2}{3}\Big[\dfrac{b-a}{n}\big( f(x_1)f(x_3)+f(x_5)+...+f(x_{2n-1}) \big) \Big] \\ &= \dfrac{b-a}{3n}\Big(f(x_0)+4f(x_1)+2f(x_2)+4f(x_3)+2f(x_4)+4f(x_5)+...+2f(x_{n-2})+4f(x_{n-1})+f(x_n) \Big) \end{align}

Notice the "1 2 4 2 4 ... 2 4 2 4 1" pattern in the coefficients.

Example 1

Use Simpson's Estimate to approximate

$\int_{0}^{2} e^{x^2} dx \nonumber$

Using $$n = 6$$

Solution

We partition

$$0 < 1/3 < 2/3 < 1 < 4/3 < 5/3 < 2 \nonumber$$

and calculate

$e^{0^2}=1, e^{(\frac{1}{3})^2}=1.12, e^{(\frac{2}{3})^2}=1.56, e^{(1)^2}=2.72 \\ e^{(\frac{4}{3})^2}=5.92, e^{(\frac{5}{3})^2}=16.08, e^{(2)^2}=54.60 \nonumber$

and put these into the formula for Simpson's Estimate

$\dfrac{2-0}{6} [ 1 + 2 \cdot 1.12 + 4 \cdot 1.56 + 2 \cdot 2.72 + 4 \cdot 5.92 + 2 \cdot 16.08 + 54.60 ] =41.79 \nonumber$

Exercise

Approximate

$\int_{2}^{6} \dfrac{1}{1+x^3} dx$

Error in Simpson's Estimate

• Without proof, we state:

Let $$M = max |f''''(x)|$$ and let $$E_s$$ be the error in using Simpson's estimate then $$|E_s| \leq \dfrac{M(b - a)^5}{180n^4}$$