18.1 Midpoint Rule

In the midpoint rule you approximate the area under the curve as a rectangle with the height as the function value at the midpoint of the interval:

\[ \int_a^b f(x)~ dx \approx f\left(\frac{a + b}{2}\right) (b - a) \]

Composite Midpoint Rule

For a more accurate solution we can subdivide the interval further, constructing rectangles for each subinterval, with the function value of the midpoint used as the height:

For \(n\) subdivisions:

\[ \int_a^b f(x)~ dx \approx \sum_{i=1}^n (x_i - x_{i-1}) f\left(\frac{x_i + x_{i-1}}{2}\right) \]

If these divisions are equal, then

\[ x_i - x_{i-1} = \frac{b - a}{n} \]

which makes the approximation:

\[ \int_a^b f(x) ~dx \approx \frac{b - a}{n} \sum_{i=1}^n f\left(\frac{x_i + x_{i-1}}{2}\right) \]

Assuming that \(n\) is chosen so that \(0 < \tfrac{b - a}{n} < 1\), the error for this method is \(O\left(\tfrac{1}{n}^3\right)\) [IntMid1].

Composite Midpoint Rule with a Discrete Data Set

Let’s consider the case where we have a discrete set of data points \((x_i, y_i)\) for \(i = 0, \dots, n\), where:

\[ y_i = f(x_i) \]

We want to approximate the integral of \(f(x)\) using this data and the midpoint rule. We can treat each \(x_i\) as the midpoint (except for \(x_0\) and \(x_n\) at the boundaries) and determine the size of the interval around it using the adjacent values.

For equally spaced data points, where \(\Delta x = x_i - x_{i-1}\) is constant, we can approximate the integral as:

\[ \int_{x_0}^{x_n} f(x) ~dx \approx \Delta x \left[ \frac{1}{2} y_0 + \left\{\sum_{i = 1}^{n-1} y_i \right\} + \frac{1}{2} y_n \right] \]

where the first and final contributions are halved as the intervals they represent are halved (note that these aren’t at the midpoints of their intervals, rather at the boundaries).

References

[IntMid1]

James F. Epperson. An Introduction to Numerical Methods and Analysis. John Wiley & Sons, Inc., Hoboken, New Jersey, second edition edition, 2013.