Wikipedia says the midpoint formula for numerical integration has error of order $h^3 f''(\xi)$. I am trying to replicate this result.
I'm guessing that I want to use Lagrange's formulation of the remainder for Taylor series. Let $x_0=\frac{a+b}{2}$ (i.e. the midpoint).
The midpoint method says $\int_a^b f(x)dx \approx (b-a)f(\frac{a+b}{2})$, so to get the error I find $(b-a) f(\frac{a+b}{2}) – \int_a^bf(x)dx$. If I expand this using Taylor's theorem I get:
$ \begin{aligned}
error & =(b-a) f(x_0) – \int_a^bf(x_0)+\frac{f'(\xi)(x-x_0)}{2}dx \\
& =\frac{f'(\xi)}{2}\int_a^b(x-x_0)dx \\
& =\frac{f'(\xi)}{2}\int_a^b(x-\frac{a+b}{2})dx \\
& = 0
\end{aligned}$
So apparently I have just proven that it has zero error? Any hints as to what I did wrong? (I realize that since wikipedia gives it in terms of $f''$ I probably want to take the expansion one level further to match them, but I don't understand why this doesn't work.)
Best Answer
You are right that you need one more term in your Taylor series. If you write $f(x)=f(x_0)+(x-x_0)f'(x_0)+(x-x_0)^2f''(\xi)/2$ and plug that into the midpoint method, the term in $f''$ will survive. What you have done essentially shows that if $f$ is linear, the method is exact.