Gaussian quadrature

At a Glance

• Frequency: 3 sub-parts across 3 of 13 years (2016, 2018, 2020)
• Priority tier: T3
• Marks (count): 15 (2), 20 (1)
• Average solve time: ~14 min
• Difficulty mix: medium 3
• Section: B | Dominant type: derivation

Why This Chapter Matters

Gaussian quadrature and undetermined-coefficients quadrature are among the highest-mark derivation problems in Paper 2’s numerical methods section. The 2016 question (15 marks) derives the two-point Gauss-Legendre rule from scratch and applies it; the 2018 question (15 marks) finds a three-parameter formula by matching monomials; the 2020 question (20 marks) constructs a weighted quadrature rule using Beta-integral moments. All three use the same core technique — set up a linear system by requiring the formula to be exact on $1, x, x^2, \ldots$ — and then evaluate. Getting the moment integrals right is the only genuinely hard part.

Minimum Theory

Undetermined coefficients method. A quadrature formula $\int_a^b w(x)f(x)\,dx \approx \sum_i c_i f(x_i)$ with $k$ free parameters is determined by requiring exactness on the $k$ monomials $1, x, x^2, \ldots, x^{k-1}$. This gives a $k\times k$ linear system for the parameters.

Gauss-Legendre rule (2 points). On $[-1,1]$ with $w(x)=1$ and two free nodes $x_1,x_2$ and weights $w_1,w_2$ (four parameters), impose exactness on $1,x,x^2,x^3$. By symmetry: $w_1=w_2=1$, $x_1=-x_2=1/\sqrt{3}$. The rule $\int_{-1}^1 f\,dx \approx f(1/\sqrt3)+f(-1/\sqrt3)$ is exact for all polynomials of degree $\le 3$.

Rescaling. For $\int_a^b f(x)\,dx$, use $x=\tfrac{a+b}{2}+\tfrac{b-a}{2}t$ ($t\in[-1,1]$, $dx=\tfrac{b-a}{2}\,dt$) to convert to the standard $[-1,1]$ interval. The Jacobian factor $\tfrac{b-a}{2}$ must be included unless $b-a=2$.

Question Archetypes

gauss-quadrature (1 question(s); 2016)

Recognition Cues

• “Show that the two-point Gauss quadrature rule on $[-1,1]$ is $f(1/\sqrt3)+f(-1/\sqrt3)$.”
• “Using this rule, estimate $\int_a^b f(x)\,dx$” where $[a,b]\ne[-1,1]$.
• Free nodes and weights; exactness through degree $2n-1$ for $n$ points.

Solution Template

1. Set up the system. Write $w_1f(x_1)+w_2f(x_2)=\int_{-1}^1 x^k\,dx$ for $k=0,1,2,3$.
2. Invoke symmetry. Take $w_1=w_2=w$, $x_2=-x_1$ to halve the unknowns.
3. Solve. $k=0$ gives $w=1$; $k=2$ gives $x_1^2=1/3$, so $x_1=1/\sqrt3$.
4. Rescale to $[a,b]$. Map $t\in[-1,1]$ to $x\in[a,b]$ via $x=\tfrac{a+b}{2}+\tfrac{b-a}{2}t$; include Jacobian.
5. Evaluate the rescaled function at $t=\pm1/\sqrt3$.

Worked Example

2016 Paper 2, 2016-P2-Q7c (15 marks)

Show that the two-point Gauss quadrature rule is $\int_{-1}^1 f(x)\,dx = f(1/\sqrt3)+f(-1/\sqrt3)$ and use it to estimate $\int_2^4 2xe^x\,dx$.

Derivation. Four unknowns $w_1,w_2,x_1,x_2$; impose exactness on $1,x,x^2,x^3$ over $[-1,1]$:

$w_1+w_2=2,\quad w_1x_1+w_2x_2=0,\quad w_1x_1^2+w_2x_2^2=\tfrac{2}{3},\quad w_1x_1^3+w_2x_2^3=0.$

By symmetry $w_1=w_2=1$, $x_2=-x_1$. Equation (3): $2x_1^2=\tfrac{2}{3}\Rightarrow x_1=\tfrac{1}{\sqrt3}$.

$\boxed{\int_{-1}^1 f(x)\,dx = f\!\left(\tfrac{1}{\sqrt3}\right)+f\!\left(-\tfrac{1}{\sqrt3}\right)}$

Application. Map $[2,4]$ to $[-1,1]$: $x=3+t$ ($dx=dt$). Define $g(t)=2(3+t)e^{3+t}$.

$g(0.5774)=2(3.5774)\,e^{3.5774}=7.1548\times 35.779\approx255.98$

$g(-0.5774)=2(2.4226)\,e^{2.4226}=4.8452\times 11.277\approx54.64$

$\boxed{\int_2^4 2xe^x\,dx \approx 255.98+54.64 = 310.62}$

(Exact: $2(3e^4-e^2)\approx312.81$; error $\approx2.19$, expected since $2xe^x$ is not a polynomial.)

undetermined-coefficients-quadrature (2 question(s); 2018, 2020)

Recognition Cues

• “Find $a,b,c$ such that $\int_0^h f\,dx = h[af(0)+bf(h/3)+cf(h)]$ is exact to highest degree.”
• “Find $\alpha_1,\alpha_2,\alpha_3$ for $\int_0^1 f(x)\,dx/\sqrt{x(1-x)} = \alpha_1 f(0)+\alpha_2 f(1/2)+\alpha_3 f(1)$ exact to highest degree.”
• Fixed nodes given; weights are the unknowns; matched against moments.

Solution Template

1. Evaluate the integral of each monomial $x^k$ (possibly with weight $w(x)$). These are the moments $\mu_k=\int x^k w(x)\,dx$.
2. Set up the exactness conditions: $\sum_i c_i x_i^k = \mu_k$ for $k=0,1,\ldots,m-1$ where $m$ = number of free parameters.
3. Solve the linear system for the weights.
4. Check the next monomial ($x^m$) to identify the degree of exactness; it will fail.
5. Compute the truncation error from the first-failing monomial: $E\propto h^{m+1}f^{(m)}(\xi)/m!$.

Worked Example

2018 Paper 2, 2018-P2-Q7b (15 marks)

Find $a,b,c$ such that $\int_0^h f(x)\,dx = h[af(0)+bf(h/3)+cf(h)]$ is exact for polynomials of highest degree. Find the order of the truncation error.

Three parameters $\Rightarrow$ impose exactness on $f=1,x,x^2$.

For $f=1$: $h = h(a+b+c)$, so $a+b+c=1$.

For $f=x$: $h^2/2 = h(b\cdot h/3 + c\cdot h) = h^2(b/3+c)$, so $b/3+c=1/2$.

For $f=x^2$: $h^3/3 = h(b\cdot h^2/9+c\cdot h^2) = h^3(b/9+c)$, so $b/9+c=1/3$.

Subtract: $\tfrac{b}{3}-\tfrac{b}{9}=\tfrac{1}{2}-\tfrac{1}{3} \Rightarrow \tfrac{2b}{9}=\tfrac{1}{6} \Rightarrow b=\tfrac{3}{4}$. Then $c=\tfrac{1}{2}-\tfrac{1}{4}=\tfrac{1}{4}$, and $a=1-\tfrac{3}{4}-\tfrac{1}{4}=0$.

$\boxed{a=0,\quad b=\frac{3}{4},\quad c=\frac{1}{4}}$

Check $f=x^3$: LHS $=h^4/4$; RHS $=h^4[\tfrac{3}{4}\cdot\tfrac{1}{27}+\tfrac{1}{4}]=h^4[\tfrac{1}{36}+\tfrac{1}{4}]=\tfrac{5h^4}{18}\ne\tfrac{h^4}{4}$. Not exact for degree 3.

Truncation error. Error for $x^3$: $\tfrac{h^4}{4}-\tfrac{5h^4}{18}=-\tfrac{h^4}{36}$. With $f^{(3)}=6$: $E=Ch^4\cdot6=-\tfrac{h^4}{36}\Rightarrow C=-\tfrac{1}{216}$.

$\boxed{E = -\frac{h^4}{216}\,f'''(\xi),\quad O(h^4)}$

2020 Paper 2, 2020-P2-Q7b (20 marks)

Find $\alpha_1,\alpha_2,\alpha_3$ such that $\int_0^1 \tfrac{f(x)\,dx}{\sqrt{x(1-x)}}=\alpha_1 f(0)+\alpha_2 f(\tfrac12)+\alpha_3 f(1)$ is exact to highest degree; evaluate $\int_0^1\tfrac{dx}{\sqrt{x-x^3}}$.

Moments of $w(x)=1/\sqrt{x(1-x)}$ (Beta integrals):

$\mu_0=\pi,\quad \mu_1=\frac{\pi}{2},\quad \mu_2=\frac{3\pi}{8},\quad \mu_3=\frac{5\pi}{16}$

Exactness conditions (nodes $0,\tfrac12,1$; $k=0,1,2$):

$\alpha_1+\alpha_2+\alpha_3=\pi$; $\tfrac{1}{2}\alpha_2+\alpha_3=\tfrac{\pi}{2}$; $\tfrac{1}{4}\alpha_2+\alpha_3=\tfrac{3\pi}{8}$.

Subtract: $\tfrac{1}{4}\alpha_2=\tfrac{\pi}{8}\Rightarrow\alpha_2=\tfrac{\pi}{2}$. Then $\alpha_3=\tfrac{\pi}{4}$, $\alpha_1=\tfrac{\pi}{4}$.

$\boxed{\alpha_1=\frac{\pi}{4},\quad \alpha_2=\frac{\pi}{2},\quad \alpha_3=\frac{\pi}{4}}$

Check: rule is exact for $x^3$ (the weight’s symmetry gives a bonus) but fails for $x^4$. Exact to degree 3.

Evaluate $\int_0^1 dx/\sqrt{x-x^3}$: factor $x-x^3=x(1-x)(1+x)$, so take $f(x)=1/\sqrt{1+x}$.

$f(0)=1$, $f(1/2)=\sqrt{2/3}\approx0.8165$, $f(1)=1/\sqrt2\approx0.7071$.

$I\approx\frac{\pi}{4}(1)+\frac{\pi}{2}(0.8165)+\frac{\pi}{4}(0.7071)=0.7854+1.2826+0.5554=2.6234$

$\boxed{\int_0^1\frac{dx}{\sqrt{x-x^3}}\approx 2.623}$

Common Traps

• Result $a=0$ is correct. In the 2018 question, the weight on $f(0)$ genuinely vanishes — the node drops out entirely. Do not try to force $a\ne0$.
• Three conditions, degree 2 or 3. With 3 free weights and fixed nodes, the rule is normally exact only through degree 2. The 2020 rule reaches degree 3 by coincidence of the weight’s symmetry — but do not assume this in general.
• Get the error constant, not just the order. The examiner expects $E=-\tfrac{h^4}{216}f'''(\xi)$, not just "$O(h^4)$".
• Rescaling factor. For Gauss-Legendre on $[a,b]$, the Jacobian is $\tfrac{b-a}{2}$. In 2016 it equals 1 (since $b-a=2$) but this is coincidental; on any other interval forgetting the factor loses marks.
• Identify $f$ in the weighted integral. In 2020, split $1/\sqrt{x-x^3}=f(x)/\sqrt{x(1-x)}$ by factoring the denominator; then $f(x)=1/\sqrt{1+x}$.

Marks-Aware Writing

A 15-mark derivation answer must show: explicit exactness conditions for each monomial (written as equations, not just stated), the solution of the linear system step by step, a check on the next failing monomial, and the error constant computed from the first failing case. A 20-mark answer additionally requires: the moment integrals evaluated explicitly, the identification of the highest degree of exactness, and the application to the specific integral with all numerical values shown.

Practice Set

• 2016-P2-Q7c (15 m) — — Hint: symmetry gives $w_1=w_2=1$; rescale $[2,4]\to[-1,1]$ via $x=3+t$; Jacobian = 1 here.
• 2018-P2-Q7b (15 m) — — Hint: $a=0$ is the correct answer; error constant is $-1/216$, not just $O(h^4)$.
• 2020-P2-Q7b (20 m) — — Hint: use Beta moments $\mu_0=\pi$, $\mu_1=\pi/2$, $\mu_2=3\pi/8$; factor denominator to identify $f(x)=1/\sqrt{1+x}$.