Romberg Integration of Tabulated Data: 04-BS-5, December 2014, Q5

Question 5: Romberg Integration of Tabulated Data (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. Tabulated $y=f(x)$ at $x=1,1.5,2,\dots,5$ (step $0.5$), integration limits $a=1$, $b=5$.

Tabulated data for the Romberg estimate
x	1	1.5	2	2.5	3	3.5	4	4.5	5
y	230	345	436	527	500	773	864	955	1,070

Tabulated y(x) used for the Romberg estimate of ∫₁⁵y dx (linear segments shown between the 9 sample points)

Find. The Romberg triangle $R(k,j)$, $k=1,\dots,4$, and the best available estimate of $\int_1^5 y\,dx$.

Approach. Build the trapezoidal column $R(k,1)$ by successively halving the panel width $H_k=4/2^{k-1}$ (using every other tabulated point at each level, since the table already supplies $h=0.5=H_4$), then apply Richardson extrapolation across the columns via $R(k,j)=R(k,j-1)+\dfrac{R(k,j-1)-R(k-1,j-1)}{4^{j-1}-1}$.

Level 1 ($H_1=4$): single trapezoid using the endpoints only. $$R(1,1)=\dfrac{4}{2}\big[f(1)+f(5)\big]=2(230+1070)=\boxed{2600}$$
Level 2 ($H_2=2$): add the midpoint $x=3$. $$R(2,1)=\dfrac12\big[R(1,1)+H_1f(3)\big]=\dfrac12\big[2600+4(500)\big]=\boxed{2300}$$
Level 3 ($H_3=1$): add $x=2,4$. $$R(3,1)=\dfrac12\big[R(2,1)+H_2\big(f(2)+f(4)\big)\big]=\dfrac12\big[2300+2(436+864)\big]=\boxed{2450}$$
Level 4 ($H_4=0.5$): add $x=1.5,2.5,3.5,4.5$ — this uses every tabulated point. $$R(4,1)=\dfrac12\big[R(3,1)+H_3\big(f(1.5)+f(2.5)+f(3.5)+f(4.5)\big)\big]=\dfrac12\big[2450+1(345+527+773+955)\big]=\boxed{2525}$$
Richardson-extrapolate across each column. Using $R(k,j)=R(k,j-1)+\dfrac{R(k,j-1)-R(k-1,j-1)}{4^{j-1}-1}$: $$R(2,2)=2300+\tfrac{2300-2600}{3}=2200,\qquad R(3,2)=2450+\tfrac{2450-2300}{3}=2500,\qquad R(4,2)=2525+\tfrac{2525-2450}{3}=2550$$ $$R(3,3)=2500+\tfrac{2500-2200}{15}=2520,\qquad R(4,3)=2550+\tfrac{2550-2500}{15}=2553.333$$ $$R(4,4)=2553.333+\tfrac{2553.333-2520}{63}=\boxed{2553.86}$$

Full Romberg triangle
k	R(k,1)	R(k,2)	R(k,3)	R(k,4)
1	2600	—	—	—
2	2300	2200	—	—
3	2450	2500	2520	—
4	2525	2550	2553.33	2553.86

Topic: Romberg integration (Richardson extrapolation applied to the composite trapezoidal rule).

Key relations: the trapezoidal error is $O(h^2)$, so combining two trapezoidal estimates at $h$ and $h/2$ via $R(k,j)=R(k,j-1)+\dfrac{R(k,j-1)-R(k-1,j-1)}{4^{j-1}-1}$ cancels the leading error term and produces an estimate accurate to $O(h^{2j})$; $R(k,2)$ is equivalent to Simpson’s rule, $R(k,3)$ to Boole’s rule, and so on down the diagonal.

Why this works: the trapezoidal rule's error expansion (Euler–Maclaurin) contains only even powers of $h$, so a single linear combination of two step sizes eliminates the $h^2$ term exactly, without needing to know the integrand's derivatives.

Common pitfall: reusing the wrong subset of tabulated points when halving $H_k$ (each new level adds only the newly-needed midpoints, it does not resample everything), and mixing up $H_{k-1}$ vs. $H_k$ in the trapezoidal recursion.

Source: Kreyszig, Advanced Engineering Mathematics, Numerical Integration (Ch. 19).

Question 5 of 7: Romberg Integration of Tabulated Data

Question 5: Romberg Integration of Tabulated Data (20 marks)