Deriving the first and second order Newton-Cotes formulas

[MarkFL]

Overview​

Much like Simpson's Rule (also known as the prismoidal method) improves upon the trapezoidal rule by fitting parabolic arcs across discrete points of a function $f(x)$ rather than line segments, Newton–Cotes formulas of order 1 and 2 fit cubic and quartic polynomials, respectively, across points on the curve.

Newton–Cotes Formula: Order 1​

We begin by computing the definite integral

$$
\int_{-h}^h Ax^3+Bx^2+Cx+D\,dx
$$

Applying the Fundamental Theorem of Calculus, we obtain

$$
\int_{-h}^h Ax^3+Bx^2+Cx+D\,dx
=
\left[\frac{Ax^4}{4}+\frac{Bx^3}{3}+\frac{Cx^2}{2}+Dx \right]_{-h}^h
$$

$$
=
\left(\frac{Ah^4}{4}+\frac{Bh^3}{3}+\frac{Ch^2}{2}+Dh\right)
-
\left(\frac{A(-h)^4}{4}+\frac{B(-h)^3}{3}+\frac{C(-h)^2}{2}+D(-h)\right)
$$

$$
=
\frac{2Bh^3}{3}+2Dh
=
\frac{2h}{3}\left(Bh^2+3D \right)
$$

We require that the cubic interpolant pass through the points

$$
\left(-h,y_0 \right),\,
\left(-\frac{h}{3},y_1 \right),\,
\left(\frac{h}{3},y_2 \right),\,
\left(h,y_3 \right)
$$

This yields the system

$$
y_0=-Ah^3+Bh^2-Ch+D
$$

$$
y_1=-\frac{Ah^3}{27}+\frac{Bh^2}{9}-\frac{Ch}{3}+D
$$

$$
y_2=\frac{Ah^3}{27}+\frac{Bh^2}{9}+\frac{Ch}{3}+D
$$

$$
y_3=Ah^3+Bh^2+Ch+D
$$

From these equations we find

$$
y_0-y_1-y_2+y_3=\frac{16}{9}Bh^2
$$

$$
Bh^2=\frac{9}{16}\left(y_0-y_1-y_2+y_3 \right)
$$

and

$$
-y_0+9y_1+9y_2-y_3=16D
$$

$$
3D=\frac{3}{16}\left(-y_0+9y_1+9y_2-y_3 \right)
$$

Substituting these results gives

$$
\int_{-h}^h Ax^3+Bx^2+Cx+D\,dx
=
\frac{2h}{3}\left(Bh^2+3D \right)
=
\frac{h}{4}\left(y_0+3y_1+3y_2+y_3 \right)
$$

Now suppose the interval $[a,b]$ is partitioned into $n$ subintervals of equal width

$$
\Delta x=x_{k+1}-x_k=\frac{b-a}{n}
$$

where $n$ is a multiple of 3. On each subinterval $[x_{k-3},x_k]$, with

$$
h=\frac{3\Delta x}{2}
$$

the integral may be approximated by

$$
\int_{x_{k-3}}^{x_k} f(x)\,dx
\approx
\frac{3\Delta x}{8}
\left(
f\left(x_{k-3}\right)
+3f\left(x_{k-2}\right)
+3f\left(x_{k-1}\right)
+f\left(x_k\right)
\right)
$$

Hence,

$$
\int_a^b f(x)\,dx
\approx
\frac{3(b-a)}{8n}
\left(
f\left(x_0\right)
+3f\left(x_1\right)
+3f\left(x_2\right)
+2f\left(x_3\right)
+\cdots
\right.
$$

$$
\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left.
+2f\left(x_{n-3}\right)
+3f\left(x_{n-2}\right)
+3f\left(x_{n-1}\right)
+f\left(x_n\right)
\right)
$$

This is also known as Simpson's $3/8$ rule.

Newton–Cotes Formula: Order 2​

We now proceed similarly to derive the second-order Newton–Cotes formula. First, compute

$$
\int_{-h}^h Ax^4+Bx^3+Cx^2+Dx+E\,dx
=
\frac{2h}{15}\left(3Ah^4+5Ch^2+15E \right)
$$

We require that the quartic interpolant pass through the points

$$
\left(-h,y_0 \right),\,
\left(-\frac{h}{2},y_1 \right),\,
\left(0,y_2 \right),\,
\left(\frac{h}{2},y_3 \right),\,
\left(h,y_4 \right)
$$

This gives the system

$$
y_0=Ah^4-Bh^3+Ch^2-Dh+E
$$

$$
y_1=\frac{Ah^4}{16}-\frac{Bh^3}{8}+\frac{Ch^2}{4}-\frac{Dh}{2}+E
$$

$$
y_2=E
$$

$$
y_3=\frac{Ah^4}{16}+\frac{Bh^3}{8}+\frac{Ch^2}{4}+\frac{Dh}{2}+E
$$

$$
y_4=Ah^4+Bh^3+Ch^2+Dh+E
$$

From these equations we obtain

$$
y_0-4y_1+6y_2-4y_3+y_4=\frac{3Ah^4}{2}
$$

$$
Ah^4=\frac{2}{3}\left(y_0-4y_1+6y_2-4y_3+y_4 \right)
$$

and

$$
-y_0+16y_1-30y_2+16y_3-y_4=6Ch^2
$$

$$
Ch^2=\frac{1}{6}\left(-y_0+16y_1-30y_2+16y_3-y_4 \right)
$$

Substitution yields

$$
\int_{-h}^h Ax^4+Bx^3+Cx^2+Dx+E\,dx
=
\frac{2h}{15}\left(3Ah^4+5Ch^2+15E \right)
=
\frac{h}{45}\left(7y_0+32y_1+12y_2+32y_3+7y_4 \right)
$$

Finally,

$$
\int_a^b f(x)\,dx
\approx
\frac{2(b-a)}{45n}
\left(
7f\left(x_0\right)
+32f\left(x_1\right)
+12f\left(x_2\right)
+32f\left(x_3\right)
+14f\left(x_4\right)
+\cdots
\right.
$$

$$
\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left.
+14f\left(x_{n-4}\right)
+32f\left(x_{n-3}\right)
+12f\left(x_{n-2}\right)
+32f\left(x_{n-1}\right)
+7f\left(x_n\right)
\right)
$$

This is also known as Boole's rule.

 

[Greg Bernhardt]

Thanks @MarkFL! What math forum can we move this to?