| Version 3 |
Version 2 |
| The {\sl midpoint rule\/} for computing the Riemann integral $\int_a^b f(x) \, dx$ is |
The {\sl midpoint rule\/} for computing the Riemann integral $\int_a^b f(x) \, dx$ is |
|
|
| $$\int_a^b f(x) \, dx = \lim_{n \to \infty} \sum_{j=1}^n f \left( a + \frac{(b-a)(j-\frac{1}{2})}{n} \right) \left( \frac{b-a}{n} \right).$$ |
$$\int_a^b f(x) \, dx = \lim_{n \to \infty} \sum_{j=1}^n f \left( a + \frac{(b-a)(j-\frac{1}{2})}{n} \right) \left( \frac{b-a}{n} \right).$$ |
|
|
| The Riemann integral can be approximated by using a definite value for $n$ rather than taking a limit. |
|
|
|
| If the Riemann integral is considered as a measure of area under a curve, then the expressions $f \left( a + \frac{(b-a)(j-\frac{1}{2})}{n} \right)$ represent the heights of the rectangles, and $\frac{b-a}{n}$ is the common width of the rectangles. |
If the Riemann integral is considered as a measure of area under a curve, then the expressions $f \left( a + \frac{(b-a)(j-\frac{1}{2})}{n} \right)$ represent the heights of the rectangles, and $\frac{b-a}{n}$ is the common width of the rectangles. |
