6.1: Elementary Formulas
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We first consider integration from 0 to h, with h small, to serve as the building blocks for integration over larger domains. We here define Ih as the following integral:
Ih=∫h0f(x)dx.
To perform this integral, we consider a Taylor series expansion of f(x) about the value x=h/2 :
f(x)=f(h/2)+(x−h/2)f′(h/2)+(x−h/2)22f′′(h/2)+(x−h/2)36f′′′(h/2)+(x−h/2)424f′′′′(h/2)+…
6.1.1. Midpoint rule
The midpoint rule makes use of only the first term in the Taylor series expansion. Here, we will determine the error in this approximation. Integrating,
Ih=hf(h/2)+∫h0((x−h/2)f′(h/2)+(x−h/2)22f′′(h/2)+(x−h/2)36f′′′(h/2)+(x−h/2)424f′′′′(h/2)+…)dx.
Changing variables by letting y=x−h/2 and dy=dx, and simplifying the integral depending on whether the integrand is even or odd, we have
Ih=hf(h/2)+∫h/2−h/2(yf′(h/2)+y22f′′(h/2)+y36f′′′(h/2)+y424f′′′′(h/2)+…)dy=hf(h/2)+∫h/20(y2f′′(h/2)+y412f′′′′(h/2)+…)dy
The integrals that we need here are
∫h20y2dy=h324,∫h20y4dy=h5160
Therefore,
Ih=hf(h/2)+h324f′′(h/2)+h51920f′′′′(h/2)+…
6.1.2. Trapezoidal rule
From the Taylor series expansion of f(x) about x=h/2, we have
f(0)=f(h/2)−h2f′(h/2)+h28f′′(h/2)−h348f′′′(h/2)+h4384f′′′′(h/2)+…,
and
f(h)=f(h/2)+h2f′(h/2)+h28f′′(h/2)+h348f′′′(h/2)+h4384f′′′′(h/2)+……
Adding and multiplying by h/2 we obtain
h2(f(0)+f(h))=hf(h/2)+h38f′′(h/2)+h5384f′′′′(h/2)+….
We now substitute for the first term on the right-hand-side using the midpoint rule formula:
h2(f(0)+f(h))=(Ih−h324f′′(h/2)−h51920f′′′′(h/2))+h38f′′(h/2)+h5384f′′′′(h/2)+…,
and solving for Ih, we find
Ih=h2(f(0)+f(h))−h312f′′(h/2)−h5480f′′′′(h/2)+…
6.1.3. Simpson’s rule
To obtain Simpson’s rule, we combine the midpoint and trapezoidal rule to eliminate the error term proportional to h3. Multiplying (6.3) by two and adding to (6.4), we obtain
3Ih=h(2f(h/2)+12(f(0)+f(h)))+h5(21920−1480)f′′′′(h/2)+…,
or
Ih=h6(f(0)+4f(h/2)+f(h))−h52880f′′′′(h/2)+….
Usually, Simpson’s rule is written by considering the three consecutive points 0,h and 2h. Substituting h→2h, we obtain the standard result
I2h=h3(f(0)+4f(h)+f(2h))−h590f′′′′(h)+…