Integral Approximation Rules, calculus lesson

Most integrals have no antiderivative in elementary functions. The area under $e^{-x^2}$ is the standard example, and data from an experiment has no formula at all. Yet a definite integral is a limit of sums, so a well chosen finite sum gets close. The rules below are the everyday ones, and the error bounds tell you in advance how close they get.

Key terms

Composite midpoint rule

Definition Let $f$ be defined on $[a, b]$ , let $n$ be a positive integer, $h = \frac{b - a}{n}$ and $x_i = a + ih$ for $i = 0, 1, \dots, n$ . Let $\bar{x}_i = \frac{x_{i-1} + x_i}{2}$ be the midpoint of the $i$ th subinterval. The composite midpoint rule is

M_n = h \left[ f(\bar{x}_1) + f(\bar{x}_2) + \dots + f(\bar{x}_n) \right].

It is the Riemann sum that samples each subinterval at its midpoint.

In plain words Cut the interval into $n$ equal strips and build each rectangle at the height of $f$ in the middle of the strip.

Example Approximate

\int_1^2 \frac{1}{x} \, dx = \ln 2 \approx 0.693147

with $n = 4$ . Then $h = 0.25$ , the midpoints are $1.125, 1.375, 1.625, 1.875$ , and

M_4 = 0.25 \left( \frac{1}{1.125} + \frac{1}{1.375} + \frac{1}{1.625} + \frac{1}{1.875} \right) \approx 0.691220,

with error $M_4 - \ln 2 \approx -0.00193$ .

Composite trapezoidal rule

Definition With the same $h$ and $x_i$ , the composite trapezoidal rule is

T_n = \frac{h}{2} \left[ f(x_0) + 2f(x_1) + 2f(x_2) + \dots + 2f(x_{n-1}) + f(x_n) \right].

It adds the areas of the trapezoids under the line segments joining consecutive points $(x_{i-1}, f(x_{i-1}))$ and $(x_i, f(x_i))$ , and it equals the average of the left and right Riemann sums.

In plain words Replace the curve on each strip by the straight chord between its endpoints and add up the trapezoids.

Example For the same integral of $\frac{1}{x}$ over $[1, 2]$ with $n = 4$ ,

T_4 = 0.125 \left( 1 + 2 \cdot \frac{1}{1.25} + 2 \cdot \frac{1}{1.5} + 2 \cdot \frac{1}{1.75} + \frac{1}{2} \right) \approx 0.697024,

with error about $+0.00388$ . The midpoint error was about half as large and of the opposite sign, which is typical when $f''$ does not change sign.

Simpson’s rule and the even-subinterval condition

Definition Let $n$ be even, $h = \frac{b - a}{n}$ and $x_i = a + ih$ . Simpson’s rule is

S_n = \frac{h}{3} \left[ f(x_0) + 4f(x_1) + 2f(x_2) + 4f(x_3) + \dots + 2f(x_{n-2}) + 4f(x_{n-1}) + f(x_n) \right],

with coefficients $1, 4, 2, 4, \dots, 2, 4, 1$ . On each pair of subintervals $[x_{2j}, x_{2j+2}]$ it integrates the parabola through the three points at $x_{2j}$ , $x_{2j+1}$ , $x_{2j+2}$ . That is why $n$ must be even, since the strips are used two at a time. For every even $n$ Simpson’s rule is a weighted average of the trapezoidal and midpoint rules with half as many strips,

S_{n} = \frac{T_{n/2} + 2M_{n/2}}{3}.

In plain words Fit a parabola through every three neighboring points instead of a line through every two. With an odd number of strips one strip has no partner, and the pattern of weights breaks.

Example For the integral of $\frac{1}{x}$ over $[1, 2]$ with $n = 4$ , the points are $1, 1.25, 1.5, 1.75, 2$ and

S_4 = \frac{0.25}{3} \left( 1 + \frac{4}{1.25} + \frac{2}{1.5} + \frac{4}{1.75} + \frac{1}{2} \right) \approx 0.693254,

with error about $0.000107$ , more than thirty times smaller than the trapezoidal error with the same five function values. As a check of the identity,

\frac{T_4 + 2M_4}{3} \approx \frac{0.697024 + 2 \cdot 0.691220}{3} \approx 0.693155,

which is $S_8$ .

To compare the three rules yourself, choose $\frac{1}{x}$ on $[1, 2]$ with $n = 4$ in the Riemann sum explorer, then raise $n$ and watch the errors shrink.

Derivative-based error bounds

Definition Let $E_M = \int_a^b f(x) \, dx - M_n$ and define $E_T$ and $E_S$ the same way. If $f''$ is continuous on $[a, b]$ and $\lvert f''(x) \rvert \le K$ for all $x$ in $[a, b]$ , then

\lvert E_T \rvert \le \frac{K (b - a)^3}{12 n^2}, \qquad \lvert E_M \rvert \le \frac{K (b - a)^3}{24 n^2}.

If $f^{(4)}$ is continuous on $[a, b]$ , $\lvert f^{(4)}(x) \rvert \le K_4$ for all $x$ in $[a, b]$ , and $n$ is even, then

\lvert E_S \rvert \le \frac{K_4 (b - a)^5}{180 n^4}.

In plain words Doubling $n$ cuts the trapezoidal and midpoint errors by about four and Simpson’s error by about sixteen. The bounds also let you pick $n$ before computing anything.

Example For $f(x) = \frac{1}{x}$ on $[1, 2]$ the derivatives satisfy

\lvert f''(x) \rvert = \frac{2}{x^3} \le 2, \qquad \lvert f^{(4)}(x) \rvert = \frac{24}{x^5} \le 24.

With $n = 4$ the bounds are $\frac{2}{12 \cdot 16} \approx 0.0104$ for $T_4$ , $\frac{2}{24 \cdot 16} \approx 0.0052$ for $M_4$ and $\frac{24}{180 \cdot 256} \approx 0.00052$ for $S_4$ . All three actual errors lie within them. To guarantee an error of at most $10^{-6}$ , the trapezoidal rule needs

\frac{2}{12 n^2} \le 10^{-6}, \quad \text{so} \quad n \ge 409.

Simpson’s rule needs

\frac{24}{180 n^4} \le 10^{-6}, \quad \text{so} \quad n \ge 19.1,

and since $n$ must be even, $n = 20$ is enough.

Adaptive subdivision

Definition An adaptive method chooses subintervals of different lengths according to how hard $f$ is to integrate on each piece. In adaptive Simpson integration with tolerance $\tau$ on a piece $[\alpha, \beta]$ , compute $S_{(1)}$ , Simpson’s rule with $n = 2$ on $[\alpha, \beta]$ , and $S_{(2)}$ , the sum of Simpson’s rule with $n = 2$ on each half. If

\lvert S_{(2)} - S_{(1)} \rvert \le 15 \tau,

accept $S_{(2)}$ for that piece. Otherwise apply the same test to each half with tolerance $\frac{\tau}{2}$ . The factor $15$ comes from the error formula. On one piece the error of Simpson’s rule is proportional to $h^4 f^{(4)}$ at some point. Write $E_1$ and $E_2$ for the errors of $S_{(1)}$ and $S_{(2)}$ . If $f^{(4)}$ is nearly constant on the piece, halving $h$ gives $E_1 \approx 16 E_2$ , so

S_{(2)} - S_{(1)} = E_1 - E_2 \approx 15 E_2.

The test therefore asks for $\lvert E_2 \rvert \lesssim \tau$ . This estimate is a heuristic, not a guaranteed bound, since $f^{(4)}$ may change a lot on the piece.

In plain words Do not waste effort where $f$ is gentle. Compare a coarse and a finer answer on each piece and split only the pieces where they disagree.

Example Take

\int_0^1 \sqrt{x} \, dx = \frac{2}{3},

where $f$ has an unbounded derivative near $0$ , so the error bounds above do not apply on $[0, 1]$ . Comparing coarse and split Simpson values on each half gives this.

Piece	Coarse	Split	Difference	Exact
$[0, 0.5]$	$0.225592$	$0.232117$	$6.5 \times 10^{-3}$	$0.235702$
$[0.5, 1]$	$0.430934$	$0.430962$	$2.8 \times 10^{-5}$	$0.430964$

The right half is nearly settled after one split, while the left half, which touches $0$ , needs many more. An adaptive method keeps splitting near $0$ and leaves the right half alone.

Check yourself

Compute $T_2$ and $M_2$ for $\int_0^2 x^2 \, dx$ and compare with the exact value.

Show the answer

Here $h = 1$ , so

T_2 = \frac{1}{2}(0 + 2 \cdot 1 + 4) = 3, \qquad M_2 = 1 \cdot (0.25 + 2.25) = 2.5.

The exact value is $\frac{8}{3} \approx 2.667$ , so $T_2$ is $\frac{1}{3}$ too big and $M_2$ is $\frac{1}{6}$ too small.

Show that Simpson’s rule with $n = 2$ gives the exact value of $\int_0^2 x^3 \, dx$ .

Show the answer

With $h = 1$ ,

S_2 = \frac{1}{3}(0 + 4 \cdot 1 + 8) = 4,

and $\int_0^2 x^3 \, dx = 4$ . This agrees with the error bound, since $f^{(4)} = 0$ for a cubic.

How large must $n$ be for the midpoint rule to approximate $\int_0^1 e^{x} \, dx$ within $10^{-4}$ ?

Show the answer

On $[0, 1]$ we have $\lvert f'' \rvert = e^x \le e$ . You need

\frac{e}{24 n^2} \le 10^{-4}, \quad \text{so} \quad n^2 \ge \frac{e}{24} \cdot 10^4 \approx 1132.6,

which means $n \ge 34$ .

Integral Approximation Rules

Key terms

Composite midpoint rule

Composite trapezoidal rule

Simpson’s rule and the even-subinterval condition

Derivative-based error bounds

Adaptive subdivision

Check yourself

How this lesson connects

Builds on

Used later in

Resources for this lesson

Riemann sum explorer