Romberg Integration of a Tabulated Experimental Curve: 04-BS-5, Undated paper, Q5

Question 5: Romberg Integration of a Tabulated Experimental Curve (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Nine equally spaced ordinates of an unknown experimental curve on $[0,8]$ at unit spacing, plus the two Romberg formulas quoted by the question.

Given data — nine tabulated ordinates
$x$	$0.0$	$1.0$	$2.0$	$3.0$	$4.0$	$5.0$	$6.0$	$7.0$	$8.0$
$f(x)$	$19.0$	$12.0$	$10.0$	$8.0$	$9.0$	$12.0$	$18.0$	$26.0$	$31.0$

Find. The complete Romberg triangle available from nine points, and the best estimate of $\int_{0}^{8}f(x)\,dx$ it yields.

The nine tabulated points joined by straight segments, with the region bounded by the curve, the lines $x=0$ and $x=8$ and the x-axis shaded. Romberg extrapolation of the trapezoidal estimates gives this area as $118.55$.

Approach. Nine points admit trapezoidal rules with $n=1,2,4,8$ sub-intervals, i.e. $h=8,4,2,1$; these fill the first Romberg column. Each further column is one Richardson extrapolation step that cancels the leading term of the remaining error.

Column one, $k=1$ ($h_{1}=8$, one panel). Only the end ordinates enter: $$R(1,1) = \frac{8}{2}\big[f(0)+f(8)\big] = 4\,(19.0+31.0) = 200.0 .$$
Column one, $k=2$ ($h_{2}=4$, two panels). The midpoint $x=4$ joins in: $$R(2,1) = \frac{4}{2}\big[19.0+31.0+2(9.0)\big] = 2\,(68.0) = 136.0 .$$
Column one, $k=3$ ($h_{3}=2$, four panels). Interior ordinates at $x=2,4,6$: $$R(3,1) = \frac{2}{2}\big[19.0+31.0+2(10.0+9.0+18.0)\big] = 50.0 + 74.0 = 124.0 .$$
Column one, $k=4$ ($h_{4}=1$, eight panels). All seven interior ordinates now contribute, and they sum to $12.0+10.0+8.0+9.0+12.0+18.0+26.0 = 95.0$: $$R(4,1) = \frac{1}{2}\big[50.0 + 2(95.0)\big] = \frac{240.0}{2} = 120.0 .$$ The four trapezoidal values are converging from above, as expected for a curve that is convex over most of the range.
Second column (divisor $4^{1}-1=3$). Each entry cancels the $O(h^{2})$ error of the trapezoidal rule and is equivalent to Simpson's rule: $$R(2,2) = 136.0 + \frac{136.0-200.0}{3} = 114.6667,\qquad R(3,2) = 124.0 + \frac{124.0-136.0}{3} = 120.0000,$$ and likewise $R(4,2) = 120.0 + (120.0-124.0)/3 = 118.6667$.
Third column (divisor $4^{2}-1=15$) and fourth (divisor $4^{3}-1=63$). Continuing the same extrapolation, $$R(3,3) = 120.0 + \frac{120.0-114.6667}{15} = 120.3556,\qquad R(4,3) = 118.6667 + \frac{118.6667-120.0}{15} = 118.5778,$$ and finally $$R(4,4) = 118.5778 + \frac{118.5778-120.3556}{63} = 118.5496 .$$

Collect the triangle and quote the answer.

Romberg triangle for $\int_{0}^{8}f(x)\,dx$
$k$	$h_{k}$	$R(k,1)$	$R(k,2)$	$R(k,3)$	$R(k,4)$
$1$	$8$	$200.0000$
$2$	$4$	$136.0000$	$114.6667$
$3$	$2$	$124.0000$	$120.0000$	$120.3556$
$4$	$1$	$120.0000$	$118.6667$	$118.5778$	$118.5496$

The bottom-right entry is the most accurate estimate the data support, so $$\boxed{\;\int_{0}^{8}f(x)\,dx \approx 118.55\;}$$ The last two entries of the bottom row agree to about $0.03$, which is a reasonable indication of the accuracy actually achieved. Since the tabulated ordinates carry only three significant figures, quoting the area as $118.5$ would be defensible and anything beyond $118.55$ is spurious precision.

Final results — Question 5
Quantity	Result
$R(1,1)$, $R(2,1)$, $R(3,1)$, $R(4,1)$	$200.0000$, $136.0000$, $124.0000$, $120.0000$
$R(2,2)$, $R(3,2)$, $R(4,2)$	$114.6667$, $120.0000$, $118.6667$
$R(3,3)$, $R(4,3)$	$120.3556$, $118.5778$
$R(4,4)$ — best estimate	$118.5496$
Area bounded by the curve, $x=0$, $x=8$, x-axis	$\approx 118.55$

Topic. Romberg integration: repeated Richardson extrapolation of the composite trapezoidal rule.

Key relations. The composite trapezoidal rule $R(k,1)=\tfrac{h_{k}}{2}\big[f_{0}+f_{n}+2\sum f_{i}\big]$ with $h_{k}=(b-a)/2^{\,k-1}$; the extrapolation $R(k,j)=R(k,j-1)+\big[R(k,j-1)-R(k-1,j-1)\big]/(4^{\,j-1}-1)$; and the Euler–Maclaurin error expansion $T(h)=I+c_{1}h^{2}+c_{2}h^{4}+\cdots$.

Why this works. For a smooth integrand the trapezoidal error contains only even powers of $h$. Halving $h$ therefore divides the leading error by four, so the particular combination $\big[4T(h/2)-T(h)\big]/3$ — exactly what the $j=2$ column computes — eliminates the $h^{2}$ term and leaves an $O(h^{4})$ estimate, which is why that column reproduces Simpson's rule. Repeating the argument with divisors $4^{2}-1$ and $4^{3}-1$ removes the $h^{4}$ and $h^{6}$ terms in turn, so each new column gains two orders for the price of one extra halving. Nine tabulated points allow exactly four halvings, hence a $4\times4$ triangle and no more; asking for a fifth row would require data at half-integer $x$ that the experiment did not provide.

Common pitfall. Starting column one at the finest spacing and working outward, which inverts the extrapolation and produces nonsense; and using the divisor $4^{j}-1$ instead of $4^{\,j-1}-1$, which divides the third column by $63$ rather than $15$. It is also wrong to read Romberg accuracy as unlimited: the extrapolation assumes a smooth integrand, and experimental ordinates rounded to three figures cap the useful precision.

Source. Kreyszig, Advanced Engineering Mathematics, 10th ed., §19.5 (numeric integration; Romberg); Burden and Faires, Numerical Analysis, 9th ed., §4.5.

Question 5 of 7: Romberg Integration of a Tabulated Experimental Curve

Question 5: Romberg Integration of a Tabulated Experimental Curve (20 marks)