21 Jun 2025

Trapezoidal, Simpson, and Gaussian Quadrature

numerical-methods quadrature trapezoidal-rule simpson-rule gaussian-quadrature

Numerical quadrature replaces

\[I=\int_a^b f(x)\,dx\]

by a weighted sum of sampled values. The weights follow by integrating an interpolating polynomial, rather than by guessing the area.

The construction is available as editable TikZ.

Trapezoidal rule

On $[x_i,x_{i+1}]$, where $h=x_{i+1}-x_i$, the linear interpolant is

\[P_1(x)=f_i\frac{x_{i+1}-x}{h} +f_{i+1}\frac{x-x_i}{h}.\]

Integrating both basis functions gives $h/2$, so

\[\int_{x_i}^{x_{i+1}}f(x)\,dx \approx\frac h2(f_i+f_{i+1}).\]

For $n$ equal subintervals, $x_i=a+ih$ and $h=(b-a)/n$. Adding adjacent panels yields the composite rule

\[\boxed{I_T=\frac h2\left[ f_0+2\sum_{i=1}^{n-1}f_i+f_n\right]}.\]

For a sufficiently smooth function,

\[I-I_T=-\frac{b-a}{12}h^2 f^{\prime\prime}(\xi),\]

for some $\xi\in(a,b)$. Halving $h$ therefore reduces the leading error by about a factor of four.

Simpson’s one-third rule

Use three equally spaced nodes $x_0,x_1,x_2$, with $x_1-x_0=x_2-x_1=h$, and put $t=(x-x_1)/h$. The nodes become $-1,0,1$, whose Lagrange basis polynomials are

\[L_{-1}(t)=\frac{t(t-1)}2,\qquad L_0(t)=1-t^2,\qquad L_1(t)=\frac{t(t+1)}2.\]

Their integrals from $-1$ to $1$ are $1/3,4/3,1/3$. Since $dx=h\,dt$,

\[\int_{x_0}^{x_2}f(x)\,dx \approx\frac h3(f_0+4f_1+f_2).\]

Adding pairs of panels gives, for even $n$,

\[\boxed{I_S=\frac h3\left[ f_0+f_n +4\sum_{\substack{i=1\\i\ {\rm odd}}}^{n-1}f_i +2\sum_{\substack{i=2\\i\ {\rm even}}}^{n-2}f_i \right]}.\]

The composite error is

\[I-I_S=-\frac{b-a}{180}h^4f^{(4)}(\xi).\]

Simpson’s rule is exact for every polynomial of degree at most three.

Two-point Gaussian quadrature

On $[-1,1]$, seek a symmetric two-point rule

\[\int_{-1}^{1}f(t)\,dt\approx w[f(-c)+f(c)].\]

Exactness for $f(t)=1$ requires $2w=2$, so $w=1$. Exactness for $f(t)=t^2$ then requires

\[2c^2=\int_{-1}^{1}t^2\,dt=\frac23,\]

which gives $c=1/\sqrt3$. Odd moments vanish by symmetry. Hence

\[\boxed{\int_{-1}^{1}f(t)\,dt \approx f\!\left(-\frac1{\sqrt3}\right) +f\!\left(\frac1{\sqrt3}\right)}\]

is exact through degree three. To integrate over $[a,b]$, use

\[x=\frac{a+b}{2}+\frac{b-a}{2}t,\qquad dx=\frac{b-a}{2}\,dt,\]

so

\[\boxed{\int_a^b f(x)\,dx\approx\frac{b-a}{2} \left[ f\!\left(\frac{a+b}{2}-\frac{b-a}{2\sqrt3}\right) +f\!\left(\frac{a+b}{2}+\frac{b-a}{2\sqrt3}\right) \right]}.\]

Exactness checks

For $f(x)=x^3-2x+4$ on $[0,2]$, Simpson’s rule with $h=1$ gives

\[I_S=\frac13[f(0)+4f(1)+f(2)] =\frac13(4+12+8)=8,\]

equal to the analytic integral. For $g(t)=t^2+2t+3$ on $[-1,1]$, the two-point Gaussian rule gives $20/3$, also exactly.

The Maxima worksheet displays

simpson_residual = 0
gauss_residual   = 0

Solved Problems

1. Composite trapezoidal rule for a convex function

Approximate

\[I=\int_0^1x^2\,dx\]

with four equal subintervals. Here $h=1/4$ and

\[f_0=0,\quad f_1=\frac1{16},\quad f_2=\frac14, \quad f_3=\frac9{16},\quad f_4=1.\]

The composite trapezoidal rule gives

\[\begin{aligned} I_T&=\frac{1/4}{2} \left[0+2\left(\frac1{16}+\frac14+\frac9{16}\right)+1\right]\\ &=\frac18\left(\frac{11}{4}\right)\\ &=\boxed{\frac{11}{32}=0.34375}. \end{aligned}\]

The exact integral is $1/3$, so

\[I_T-I=\frac{11}{32}-\frac13=\frac1{96} \approx0.0104167.\]

The positive signed error is physically consistent with the chords lying above the convex curve $x^2$.

2. Two-point Gaussian quadrature on a general interval

Evaluate

\[I=\int_0^2(x^3+x^2+1)\,dx\]

with the two-point Gaussian rule. The map $x=1+t$ sends $[0,2]$ to $[-1,1]$ and has $dx=dt$. Therefore

\[I_G=f\!\left(1-\frac1{\sqrt3}\right) +f\!\left(1+\frac1{\sqrt3}\right).\]

Put $u=1/\sqrt3$. By symmetry,

\[\begin{aligned} f(1-u)+f(1+u) &=[(1-u)^3+(1+u)^3]\\ &\quad+[(1-u)^2+(1+u)^2]+2\\ &=(2+6u^2)+(2+2u^2)+2\\ &=6+8\left(\frac13\right)\\ &=\boxed{\frac{26}{3}}. \end{aligned}\]

Direct integration also gives $4+8/3+2=26/3$. Exact agreement is expected because the integrand is cubic.

Descriptive Questions

1. Derive the trapezoidal rule by integrating a linear interpolant.
2. Derive Simpson’s one-third weights and explain why the composite rule requires an even number of subintervals.
3. Derive the nodes and weights of two-point Gaussian quadrature from polynomial exactness.
4. Compare the degree of exactness and leading composite error of the trapezoidal and Simpson rules.

Numerical Problems

1. Apply one Simpson panel to $\int_0^2x^4\,dx$ and compare with the exact result. Answer: $I_S=20/3\approx6.66667$; exact $I=32/5=6.4$; error $I_S-I=4/15\approx0.26667$.

2. Apply one trapezoidal panel to $\int_1^3(3x+2)\,dx$. Answer: $I_T=(2/2)[5+11]=16$, exactly.

3. On $[0,3]$ with step $h=0.5$, decide whether the composite Simpson one-third rule is admissible. Answer: $n=(3-0)/0.5=6$, which is even, so it is admissible.

4. Apply two-point Gaussian quadrature to $\int_{-1}^{1}t^4\,dt$ and find the signed error $I_G-I$. Answer: $I_G=2/9$, exact $I=2/5$, and $I_G-I=-8/45$.

The panel sums, mapped nodes, and exactness comparisons are checked in the Unit II Maxima worksheet; every displayed residual is zero.

References

1. Numerical integration — Wikipedia.
2. Richard L. Burden, J. Douglas Faires, and Annette M. Burden, Numerical Analysis, 10th ed., Chapter 4, “Numerical Differentiation and Integration.”
3. Steven C. Chapra and Raymond P. Canale, Numerical Methods for Engineers, 8th ed., Chapter 21, “Newton-Cotes Integration Formulas.”