midpoint ruleMidpointRule2006-06-08 16:30:312008-03-12 08:03:25TheoremRiemann integral\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{amsfonts}The \emph{midpoint rule} for computing the Riemann integral $\displaystyle \int\limits_a^b f(x) \, dx$ is
\[
\int\limits_a^b f(x) \, dx = \lim_{n \to \infty} \sum_{j=1}^n f \left( a + \left( j-\frac{1}{2} \right) \left( \frac{b-a}{n} \right) \right) \left( \frac{b-a}{n} \right).
\]
If the Riemann integral is considered as a measure of area under a curve, then the expressions $\displaystyle f \left( a + \left( j-\frac{1}{2} \right) \left( \frac{b-a}{n} \right) \right)$ \PMlinkescapetext{represent} the \PMlinkescapetext{heights} of the rectangles, and $\displaystyle \frac{b-a}{n}$ is the common \PMlinkescapetext{width} of the rectangles.
The Riemann integral can be approximated by using a definite value for $n$ rather than taking a limit. In this case, the partition is $\displaystyle \left\{ \left[ a, a+\frac{b-a}{n} \right) , \dots , \left[ a+\frac{(b-a)(n-1)}{n}, b \right] \right\}$, and the function is evaluated at the midpoints of each of these intervals. Note that this is a special case of a Riemann sum in which the $x_j$'s are evenly spaced and the $c_j$'s chosen are the midpoints.
If $f$ is Riemann integrable on $[a,b]$ such that $|f''(x)| \le M$ for every $x \in [a,b]$, then
\[
\left| \int\limits_a^b f(x) \, dx - \sum_{j=1}^n f \left( a + \left( j-\frac{1}{2} \right) \left( \frac{b-a}{n} \right) \right) \left( \frac{b-a}{n} \right) \right| \le \frac{M(b-a)^3}{24n^2}.
\]