# midpoint rule

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 $\displaystyle f\left(a+\left(j-\frac{1}{2}\right)\left(\frac{b-a}{n}\right)\right)$ the of the rectangles, and $\displaystyle\frac{b-a}{n}$ is the common 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 $|f^{\prime\prime}(x)|\leq M$ for every $x\in[a,b]$, then

 $\left|\int\limits_{a}^{b}f(x)\,dx-\sum_{j=1}^{n}f\left(a+\left(j-\frac{1}{2}% \right)\left(\frac{b-a}{n}\right)\right)\left(\frac{b-a}{n}\right)\right|\leq% \frac{M(b-a)^{3}}{24n^{2}}.$
 Title midpoint rule Canonical name MidpointRule Date of creation 2013-03-22 15:57:44 Last modified on 2013-03-22 15:57:44 Owner Wkbj79 (1863) Last modified by Wkbj79 (1863) Numerical id 16 Author Wkbj79 (1863) Entry type Theorem Classification msc 41-01 Classification msc 28-00 Classification msc 26A42 Related topic LeftHandRule Related topic RightHandRule Related topic RiemannSum Related topic ExampleOfEstimatingARiemannIntegral