18.2 Trapezoidal Rule

For the trapezoidal rule we approximate the integral of \(f(x)\) on the interval \([a-b]\) by constructing the trapezium below:

and calculating it’s area.

The area of the trapezium is given by:

\[ A_\text{trapezoid} = A_{\text{rectangle}} + A_{\text{triangle}} \]

In the case where \(f(a) < f(b)\), this area is given by:

\[\begin{align*} A_\text{trapezoid} &= (b - a)f(a) + \tfrac{1}{2} (b - a) \left[f(b) - f(a)\right]\\ &= \tfrac{1}{2}(b - a) \left[f(a) + f(b)\right] \end{align*}\]

In the case where \(f(a) > f(b)\), the area is give by:

\[\begin{align*} A_\text{trapezoid} &= (b - a)f(b) + \tfrac{1}{2} (b - a) \left[f(a) - f(b)\right]\\ &= \tfrac{1}{2}(b - a) \left[f(a) + f(b)\right] \end{align*}\]

which is the same as above, so in general we can approximate the integral as:

\[ \int_a^b f(x)~ dx \approx T(f) = \tfrac{1}{2} (b - a) \left[f(a) + f(b)\right] \]

This can be shown [IntTrap1] to have an error of:

\[ I(f) - T(f) = - \frac{1}{12} (b - a)^3 f''(\xi) \]

for some \(\xi \in [a, b]\). Note that this is double the error of the midpoint rule. As before, a large interval will have a large error, to reduce this we can sub-divide the interval.

Composite Trapezoid Rule

Now, if we were to break up this interval into \(n\) equal sub-intervals, and approximate the integral on each of these, we arrive at the composite trapezoidal rule (illustrated in the diagrams that follow).

To calculate this we use the sum:

\[ \int_a^b f(x)~ dx \approx T_n(f) = \sum_{i=0}^{n-1} \tfrac{1}{2} (x_{i+1} - x_{i}) \left[f(x_{i}) + f(x_{i+1})\right] \]

where \(x_0 = a\) and \(x_n = b\). As we have specified that each of the \(n\) sub-intervals are of equal sizes, we have that:

\[ x_{i+1} - x_{i} = \frac{b- a}{n} \equiv h \]

we can therefore write \(T_n(f)\) as:

\[ T_n(f) = \frac{h}{2} \sum_{i=0}^{n-1} \left[f(x_i) + f(x_{i+1})\right] \]

note how each \(f(x_i)\) in the sum above is repeated twice, except for \(f(x_0)\) and \(f(x_n)\), which only feature once each. We can now write the \(T_n(f)\) as:

(19)\[ T_n(f) = \frac{h}{2} \left[f(a) + 2 \left\{\sum_{i=1}^{n-1} f(a + ih)\right\} + f(b)\right] \]

This can be shown [IntTrap1] to have a (global) error of:

(20)\[ I(f) - T_n(f) = - \frac{b-a}{12} h^2 f''(\xi) = O(h^2) \]

for some different \(\xi \in [a, b]\). Again, this is double the error of the composite midpoint rule.

Composite Trapezoidal Rule with a Discrete Data Set

The composite trapezoidal rule can be used to integrate a discrete set of data. Consider 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 \(f(x)\) on the interval \([x_0, x_n]\) using this data.

If we wanted to approximate the integral of this data set using the trapezoidal rule, we can apply this to each interval individually:

\[ \int_{x_0}^{x_n} f(x)~ dx \approx \tfrac{1}{2} \sum_{i = 1}^n (x_i - x_{i-1}) \left[y_{i-1} + y_i\right] \]

If the \(x_i\) values are evenly spaced, with \(x_i - x_{i-1} = \Delta x\) constant, then we can use the composite formula from the section above:

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

References

[IntTrap1] (1,2)

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