The 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 $\displaystyle f \left( a + \left( j-\frac{1}{2} \right) \left( \frac{b-a}{n} \right) \right)$ represent the heights of the rectangles, and $\displaystyle \frac{b-a}{n}$ is the common 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$$ \left| \int\limits_a^b f(x) \, dx - \sum_{j=1}^n f \left( a + \left( j-\frac{1}{2} \right) \left( \frac{b-a}{n} \right) \right) \left( \frac{b-a}{n} \right) \right| \le \frac{M(b-a)^3}{24n^2}.$$