left hand rule LeftHandRule 2006-06-08 16:23:45 2008-03-12 07:56:25 Theorem Riemann integral \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} The \emph{left hand rule} for computing the Riemann integral $\displaystyle \int\limits_a^b f(x) \, dx$ is \[ \int\limits_a^b f(x) \, dx = \lim_{n \to \infty} \sum_{j=1}^n f \left( a+(j-1) \left( \frac{b-a}{n} \right) \right) \left( \frac{b-a}{n} \right). \] If the Riemann integral is considered as a measure of area under a curve, then the expressions $\displaystyle f \left( a+(j-1) \left( \frac{b-a}{n} \right) \right)$ \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 left endpoints of each of these intervals. Note that this is a special case of a left Riemann sum in which the $x_j$'s are evenly spaced.