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Numerical Integration

The Newton-Leibniz formula requires us to “be able to find an antiderivative”. But in reality many important integrals — such as $\int e^{-x^2} dx$ (the normal distribution) and $\int \frac{\sin x}{x} dx$ — have no elementary antiderivative. At that point you have to return to the essence of integration: computing the area approximately. That is what numerical integration is.

Starting from the Riemann Sum

The definition of a definite integral is itself an approximation: divide the interval up and sum the areas of the rectangles.

Divide the interval $[a, b]$ into $n$ equal parts, each of width $\Delta x = \frac{b-a}{n}$ , and take the division points $x_i = a + i\Delta x$ . Then

$R_n = \sum_{i=1}^{n} f(\xi_i)\, \Delta x$

When $n \to \infty$ , if the limit exists, it is the definite integral $\int_a^b f(x)\, dx$ .

In actual computation, $R_n$ with a finite $n$ is an approximation to the integral. According to how the sample point $\xi_i$ is chosen, there are:

The left-endpoint Riemann sum: $\xi_i = x_{i-1}$ , usually on the small side (for an increasing function)
The right-endpoint Riemann sum: $\xi_i = x_i$ , usually on the large side (for an increasing function)
The midpoint Riemann sum: $\xi_i = \frac{x_{i-1} + x_i}{2}$ , which is more accurate

The idea behind the rectangular method is “approximate the area under the curve by rectangles whose tops are straight segments”; the error comes from the gaps between the rectangles and the curve.

The Trapezoidal Rule: Approximating with Straight Lines

A slightly cleverer approach: approximate each small interval by a trapezoid rather than a rectangle.

$\int_a^b f(x)\, dx \approx T_n = \frac{\Delta x}{2} \left[ f(x_0) + 2f(x_1) + 2f(x_2) + \cdots + 2f(x_{n-1}) + f(x_n) \right]$

Here $\Delta x = \frac{b-a}{n}$ . The values at the two ends are multiplied by one only, while each value in the middle appears twice (because two neighbouring trapezoids share an endpoint).

Simpson’s Rule: Approximating with Parabolas

The trapezoidal rule joins adjacent points with straight lines; Simpson’s rule goes a step further: every three points (two small intervals) are approximated by a single parabola. This raises the accuracy considerably.

$\int_a^b f(x)\, dx \approx S_n = \frac{\Delta x}{3} \left[ f(x_0) + 4f(x_1) + 2f(x_2) + 4f(x_3) + \cdots + 2f(x_{n-2}) + 4f(x_{n-1}) + f(x_n) \right]$

Note: Simpson’s rule requires $n$ to be even. The coefficients alternate $4, 2, 4, 2, \ldots, 4$ , with a $1$ at each end.

Comparing the Accuracy of the Three Methods

The error of the rectangular method is $O(1/n)$ , of the trapezoidal rule $O(1/n^2)$ , and of Simpson’s rule $O(1/n^4)$ . So for the same number of divisions, Simpson’s rule is usually the most accurate.

Where Numerical Integration Is Used

Statistics: the probability $P(a \le X \le b) = \frac{1}{\sqrt{2\pi}}\int_a^b e^{-x^2/2} dx$ of the normal distribution has no elementary antiderivative, so numerical integration is the only way.
Physics: many potential-energy, electric-field and radiation integrals have no analytic solution.
Engineering computation: areas, volumes and centroids bounded by curves whose antiderivatives cannot be found analytically.

A Typical Application

Application: Approximating $\int_0^1 e^{x^2} dx$ by Three Methods ( $n = 4$ )

Reference Answer

numerical integrationthe trapezoidal ruleSimpson's rule

Approach:

With $n = 4$ , $\Delta x = 0.25$ , and the division points $x_0 = 0, x_1 = 0.25, x_2 = 0.5, x_3 = 0.75, x_4 = 1$ . Compute the value of the function at each point and substitute into the formulas.

Steps:

Function values: $f(0) = 1$ , $f(0.25) = e^{0.0625} \approx 1.0645$ , $f(0.5) = e^{0.25} \approx 1.2840$ , $f(0.75) = e^{0.5625} \approx 1.7551$ , $f(1) = e \approx 2.7183$
Trapezoidal rule: $T_4 = \frac{0.25}{2}[1 + 2(1.0645 + 1.2840 + 1.7551) + 2.7183] \approx 0.125 \times 11.7265 \approx 1.4658$
Simpson’s rule: $S_4 = \frac{0.25}{3}[1 + 4(1.0645 + 1.7551) + 2(1.2840) + 2.7183] \approx 0.08333 \times 17.4291 \approx 1.4524$
The exact value is about $1.4627$ , so Simpson’s result is already very close

Answer: $T_4 \approx 1.4658$ , $S_4 \approx 1.4524$ (the exact value is about $1.4627$ )

Summary

Symbols Used in This Article

Symbol	Type	Pronunciation/Explanation	Meaning in This Article
$\Delta x$	mathematical symbol	delta x	The width of a small interval (the step size)
$n$	mathematical symbol	n	The number of equal divisions of the interval
$x_i$	mathematical symbol	x-sub-i	The $i$ -th division point
$\xi_i$	Greek letter	Xi-sub-i	The sample point on the $i$ -th small interval
$R_n$	mathematical symbol	R-sub-n	The Riemann sum
$T_n$	mathematical symbol	T-sub-n	The approximation given by the trapezoidal rule
$S_n$	mathematical symbol	S-sub-n	The approximation given by Simpson’s rule
$\sum_{i=1}^{n}$	mathematical symbol	sum from i=1 to n	The summation sign
$e$	mathematical symbol	Euler’s number	The base of the natural logarithm
$\pi$	Greek letter	Pi	The ratio of a circle’s circumference to its diameter
$P(a \le X \le b)$	mathematical symbol	probability	The probability that a random variable falls in the interval

The Fundamental Theorem of Calculus

Comprehensive Exercises in the Integral Calculus of One Variable

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Numerical Integration

Starting from the Riemann Sum

The Trapezoidal Rule: Approximating with Straight Lines

Simpson’s Rule: Approximating with Parabolas

Comparing the Accuracy of the Three Methods

Where Numerical Integration Is Used

A Typical Application

Application: Approximating $\int_0^1 e^{x^2} dx$ by Three Methods ( $n = 4$ )

Summary

Symbols Used in This Article

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Numerical Integration

Starting from the Riemann Sum

The Trapezoidal Rule: Approximating with Straight Lines

Simpson’s Rule: Approximating with Parabolas

Comparing the Accuracy of the Three Methods

Where Numerical Integration Is Used

A Typical Application

Application: Approximating ∫01ex2dx\int_0^1 e^{x^2} dx∫01​ex2dx by Three Methods (n=4n = 4n=4)

Summary

Symbols Used in This Article

Application: Approximating $\int_0^1 e^{x^2} dx$ by Three Methods ( $n = 4$ )