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I am well aware of five approximate integration techniques:

1) Left endpoint approximation

2) Right endpoint approximation

3) Midpoint rule

4) Trapezoidal Rule

5) Simpson's Rule

It is very natural to dervive all of these rules using Taylor Series (all you have to do is choose your centre point , integrate term by term, shave of a few terms, and use approximate differentiaton place of the derivatives to obtain them!)

However, no matter what I do, I can't seem to obtain the trapezoidal rule. Just wondering if anyone has any ideas!

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    $\begingroup$ the Trapezoidal rule is the average between the left-endpoint and the right-endpoint approximation. $\endgroup$ – Doug M Jul 31 '17 at 22:25
  • $\begingroup$ I know but can I obtain it independently? $\endgroup$ – user463792 Jul 31 '17 at 22:26
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An easy way to derive the trapezoidal rule (and estimate the error) uses an integration-by-parts.

For a subinterval $[x_{n-1},x_n]$ with $h = x_n - x_{n-1}$ we derive the trapezoidal approximation to the integral and get a local error estimate as follows.

Let $c = (x_{n+1}+x_n)/2$ be the midpoint and note that

$$x_{n+1} - c = c - x_n = \frac{x_{n+1} - x_n}{2} = \frac{h}{2}.$$

Using integration by parts with $u = (x-c)$ and $dv = f'(x)dx$ we get

$$\int_{x_n}^{x_{n+1}}(x-c)f'(x) \, dx = (x-c)f(x)|_{x_n}^{x_{n+1}} - \int_{x_n}^{x_{n+1}} f(x) \, dx \\ = \frac{h}{2} \left(f(x_{n}) + f(x_{n+1}) \right) - \int_{x_n}^{x_{n+1}} f(x) \, dx. $$

Consequently, the local error of the trapezoidal approximation is

$$E = \int_{x_n}^{x_{n+1}}(x-c)f'(x) \, dx. $$

Depending on the smoothness of $f$ and bounds on derivatives this can be used to get more refined estimates.

If the derivative is bounded with $|f'| \leqslant M$ we get the well-known estimate

$$|E| = \left|\frac{h}{2} \left(f(x_{n}) + f(x_{n+1}) \right) - \int_{x_n}^{x_{n+1}} f(x) \, dx \right| \leqslant \frac{M}{4}h^2$$

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