The ApproximateInt(f(x), x = a..b, method = boole, opts) command approximates the integral of f(x) from a to b by using Boole's rule. The first two arguments (function expression and range) can be replaced by a definite integral.

If the independent variable can be uniquely determined from the expression, the parameter x need not be included in the calling sequence.

Given a partition $P=\left(a=x_0,x_1,...,x_N=b\right)$ of the interval $\left(a\,b\right)$, Boole's rule approximates the integral on each subinterval $\left(x_{i-1}\,x_i\right)$ by integrating the quartic function that interpolates five equally spaced points in that subinterval.

In the case that the widths of the subintervals are equal, the approximation can be written as

Traditionally, Boole's rule is written as: given N, where N is a positive multiple of 3, and given equally spaced points $a=x_0,x_1,x_2,...,x_N=b$, an approximation to the integral ${\int }_a^{b}f\left(x\right)\ⅆx$ is

By default, the interval is divided into $10$ equal-sized subintervals.

For the options opts, see the ApproximateInt help page.

This rule can be applied interactively, through the ApproximateInt Tutor.

This rule is also sometimes known as Bode's Rule due to a misattribution in the literature.  The ApproximateInt command will accept either method=boole or method=bode.

$\mathrm{polynomial}≔\mathrm{CurveFitting}\left[\mathrm{PolynomialInterpolation}\right]\left(\left[x\left[0\right]\,\frac{3x\left[0\right]+x\left[1\right]}{4}\,\frac{x\left[0\right]+x\left[2\right]}{2}\,\frac{x\left[0\right]+3x\left[1\right]}{4}\,x\left[1\right]\right]\,\left[f\left(0\right)\,f\left(\frac{1}{4}\right)\,f\left(\frac{1}{2}\right)\,f\left(\frac{3}{4}\right)\,f\left(1\right)\right]\,z\right)\:$

$\mathrm{integrated}≔\mathrm{int}\left(\mathrm{polynomial}\,z=x\left[0\right]..x\left[1\right]\right)\:$

$\mathrm{factor}\left(\mathrm{integrated}\right)$

$-\frac{\left(x_0-x_1\right)\left(-72f\left(0\right)x_1{x}_2^3-88f\left(1\right)x_1{x}_2^3+72f\left(\frac{1}{2}\right)x_0^2{x}_1^2-656f\left(\frac{1}{4}\right)x_1{x}_2^3-624f\left(\frac{3}{4}\right)x_1{x}_2^3+60f\left(0\right)x_0^2{x}_1{x}_2-150f\left(0\right)x_0{x}_1^2{x}_2+90f\left(0\right)x_0{x}_1{x}_2^2+432f\left(\frac{1}{4}\right)x_0^2{x}_1{x}_2-768f\left(\frac{1}{4}\right)x_0{x}_1^2{x}_2+336f\left(\frac{1}{4}\right)x_0{x}_1{x}_2^2+336f\left(\frac{3}{4}\right)x_0^2{x}_1{x}_2-768f\left(\frac{3}{4}\right)x_0{x}_1^2{x}_2+432f\left(\frac{3}{4}\right)x_0{x}_1{x}_2^2+72f\left(1\right)x_0^2{x}_1{x}_2-114f\left(1\right)x_0{x}_1^2{x}_2+42f\left(1\right)x_0{x}_1{x}_2^2+20f\left(1\right)x_2^4+160f\left(\frac{3}{4}\right)x_2^4+12f\left(\frac{1}{2}\right)x_0^4+12f\left(\frac{1}{2}\right)x_1^4+7f\left(1\right)x_0^4+7f\left(0\right)x_0^4-12f\left(0\right)x_1^4+20f\left(0\right)x_2^4+32f\left(\frac{1}{4}\right)x_0^4+160f\left(\frac{1}{4}\right)x_2^4+32f\left(\frac{3}{4}\right)x_0^4-36f\left(1\right)x_1^3{x}_2+111f\left(1\right)x_1^2{x}_2^2+36f\left(1\right)x_0{x}_1^3+8f\left(1\right)x_0{x}_2^3-30f\left(0\right)x_0^3{x}_1+2f\left(0\right)x_0^3{x}_2+15f\left(0\right)x_0^2{x}_1^2-33f\left(0\right)x_0^2{x}_2^2+40f\left(0\right)x_0{x}_1^3-8f\left(0\right)x_0{x}_2^3+8f\left(0\right)x_1^3{x}_2+63f\left(0\right)x_1^2{x}_2^2-112f\left(\frac{1}{4}\right)x_0^3{x}_1-16f\left(\frac{1}{4}\right)x_0^3{x}_2-48f\left(\frac{1}{4}\right)x_0^2{x}_1^2-192f\left(\frac{1}{4}\right)x_0^2{x}_2^2+288f\left(\frac{1}{4}\right)x_0{x}_1^3+16f\left(\frac{1}{4}\right)x_0{x}_2^3-288f\left(\frac{1}{4}\right)x_1^3{x}_2+816f\left(\frac{1}{4}\right)x_1^2{x}_2^2-144f\left(\frac{3}{4}\right)x_0^3{x}_1+16f\left(\frac{3}{4}\right)x_0^3{x}_2+48f\left(\frac{3}{4}\right)x_0^2{x}_1^2-192f\left(\frac{3}{4}\right)x_0^2{x}_2^2+224f\left(\frac{3}{4}\right)x_0{x}_1^3-16f\left(\frac{3}{4}\right)x_0{x}_2^3-224f\left(\frac{3}{4}\right)x_1^3{x}_2+720f\left(\frac{3}{4}\right)x_1^2{x}_2^2-48f\left(\frac{1}{2}\right)x_0^3{x}_1-48f\left(\frac{1}{2}\right)x_0{x}_1^3-26f\left(1\right)x_0^3{x}_1-2f\left(1\right)x_0^3{x}_2+3f\left(1\right)x_0^2{x}_1^2-33f\left(1\right)x_0^2{x}_2^2\right)}{90\left(x_0-x_2\right)\left(x_0-2x_2+x_1\right)\left(x_0-3x_1+2x_2\right)\left(-2x_1+x_0+x_2\right)}$

$\mathrm{with}\left(\mathrm{Student}\left[\mathrm{Calculus1}\right]\right)\:$

$\mathrm{ApproximateInt}\left(\sin \left(x\right)\,x=0..5\,\mathrm{method}=\mathrm{boole}\right)$

$\frac{8\sin \left(\frac{37}{8}\right)}{45}+\frac{\sin \left(\frac{19}{4}\right)}{15}+\frac{8\sin \left(\frac{39}{8}\right)}{45}+\frac{7\sin \left(5\right)}{180}+\frac{\sin \left(\frac{15}{4}\right)}{15}+\frac{8\sin \left(\frac{31}{8}\right)}{45}+\frac{7\sin \left(4\right)}{90}+\frac{8\sin \left(\frac{33}{8}\right)}{45}+\frac{\sin \left(\frac{17}{4}\right)}{15}+\frac{8\sin \left(\frac{35}{8}\right)}{45}+\frac{7\sin \left(\frac{9}{2}\right)}{90}+\frac{8\sin \left(\frac{23}{8}\right)}{45}+\frac{7\sin \left(3\right)}{90}+\frac{8\sin \left(\frac{25}{8}\right)}{45}+\frac{\sin \left(\frac{13}{4}\right)}{15}+\frac{8\sin \left(\frac{27}{8}\right)}{45}+\frac{7\sin \left(\frac{7}{2}\right)}{90}+\frac{8\sin \left(\frac{29}{8}\right)}{45}+\frac{8\sin \left(\frac{17}{8}\right)}{45}+\frac{\sin \left(\frac{9}{4}\right)}{15}+\frac{8\sin \left(\frac{19}{8}\right)}{45}+\frac{7\sin \left(\frac{5}{2}\right)}{90}+\frac{8\sin \left(\frac{21}{8}\right)}{45}+\frac{\sin \left(\frac{11}{4}\right)}{15}+\frac{\sin \left(\frac{5}{4}\right)}{15}+\frac{8\sin \left(\frac{11}{8}\right)}{45}+\frac{7\sin \left(\frac{3}{2}\right)}{90}+\frac{8\sin \left(\frac{13}{8}\right)}{45}+\frac{\sin \left(\frac{7}{4}\right)}{15}+\frac{8\sin \left(\frac{15}{8}\right)}{45}+\frac{7\sin \left(2\right)}{90}+\frac{8\sin \left(\frac{3}{8}\right)}{45}+\frac{7\sin \left(\frac{1}{2}\right)}{90}+\frac{8\sin \left(\frac{5}{8}\right)}{45}+\frac{\sin \left(\frac{3}{4}\right)}{15}+\frac{8\sin \left(\frac{7}{8}\right)}{45}+\frac{8\sin \left(\frac{9}{8}\right)}{45}+\frac{8\sin \left(\frac{1}{8}\right)}{45}+\frac{\sin \left(\frac{1}{4}\right)}{15}+\frac{7\sin \left(1\right)}{90}$

$\mathrm{ApproximateInt}\left(x\left(x-2\right)\left(x-3\right)\,x=0..5\,\mathrm{method}=\mathrm{boole}\,\mathrm{output}=\mathrm{plot}\right)$

$\mathrm{ApproximateInt}\left(\tan \left(x\right)-2x\,-1..1\,\mathrm{method}=\mathrm{boole}\,\mathrm{output}=\mathrm{plot}\,\mathrm{partition}=50\right)$

$\mathrm{ApproximateInt}\left(\ln \left(x\right)\,x=1..100\,\mathrm{method}=\mathrm{boole}\,\mathrm{output}=\mathrm{animation}\right)$