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The error for the trapezoidal rule may be found by integrating the corresponding Taylor series and applying integration by parts. Taylor's series gives us that:

$f\left(a\right)=f\left(x\right)+f^{\left(1\right)}\left(x\right)\left(a-x\right)+\frac{1}{2}{f}^{\left(2\right)}\left(\xi \right){\left(a-x\right)}^2$

where ξ is in the interval [a, x].

Integrating each side with respect to the variable x from a to b, we get the expression:

$f\left(a\right)\left(b-a\right)=\underset{a}{\overset{b}{\int }}f\left(x\right)dx+af\left(b\right)-af\left(a\right)-\underset{a}{\overset{b}{\int }}x{f}^{\left(1\right)}\left(x\right)dx+\frac{1}{6}{f}^{\left(2\right)}\left(\xi \right){\left(b-a\right)}^3$

Recall from calculus the technique of integration by parts. In this case, may rewrite the integral of x f⁽¹⁾(x) as:

$\underset{a}{\overset{b}{\int }}xf\left(x\right)dx=\left(xf\left(x\right)\right|\underset{x=a}{\overset{b}{}}-\underset{a}{\overset{b}{\int }}f\left(x\right)dx$

Substituting this into our expression, we get:

$f\left(a\right)\left(b-a\right)=2\underset{a}{\overset{b}{\int }}f\left(x\right)dx-f\left(b\right)\left(b-a\right)+\frac{1}{6}{f}^{\left(2\right)}\left(\xi \right){\left(b-a\right)}^3$ ">

Exchanging the integral and the left-hand side, dividing each side by −2, and collecting appropriately, we get:

$\underset{a}{\overset{b}{\int }}f\left(x\right)dx=\frac{1}{2}\left(f\left(a\right)+f\left(b\right)\right)\left(b-a\right)-\frac{1}{12}{f}^{\left(2\right)}\left(\xi \right){\left(b-a\right)}^3$

Note, we sort of cheated here, because ξ depends on x, however we may replace f⁽²⁾(ξ) by the average value:

$\underset{a}{\overset{b}{\int }}f\left(x\right)dx\approx \frac{1}{2}\left(f\left(a\right)+f\left(b\right)\right)\left(b-a\right)-\frac{1}{12}\overline{f^{\left(2\right)}}{\left(b-a\right)}^3$

Example of Error

Consider the integral of cos(x) from 0.2 to 0.4. The correct value of this integral is 0.1907490115, 0.5⋅(cos(0.2) + cos(0.4))⋅0.2 = 0.1901127572, and the difference between these is −0.0006362543. If we calculate the average value of the 2nd derivative and multiply this by −0.2³/12, we get −0.0006358300384, and thus our approximation of the error is very close (less than 0.1% error in our estimation of the error).

Copyright ©2005 by Douglas Wilhelm Harder. All rights reserved.

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