midpoint rule

The midpoint rule for computing the Riemann integral ${\displaystyle \underset{a}{\overset{b}{\int }}}f(x)𝑑x$ is

$$\underset{a}{\overset{b}{\int }}f(x)𝑑x=\underset{n\to \mathrm{\infty }}{lim}\sum _{j=1}^{n}f\left(a+\left(j-\frac{1}{2}\right)\left(\frac{b-a}{n}\right)\right)\left(\frac{b-a}{n}\right).$$

If the Riemann integral is considered as a measure of area under a curve, then the expressions $f\left(a+\left(j-{\displaystyle \frac{1}{2}}\right)\left({\displaystyle \frac{b-a}{n}}\right)\right)$ 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 $\{[a,a+{\displaystyle \frac{b-a}{n}}),\mathrm{\dots },[a+{\displaystyle \frac{(b-a)(n-1)}{n}},b]\}$, 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 $|f^{\prime \prime }(x)|\le M$ for every $x\in [a,b]$, then

$$\left|\underset{a}{\overset{b}{\int }}f(x)𝑑x-\sum _{j=1}^{n}f\left(a+\left(j-\frac{1}{2}\right)\left(\frac{b-a}{n}\right)\right)\left(\frac{b-a}{n}\right)\right|\le \frac{M{(b-a)}^3}{24n^2}.$$

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