Romberg Integration from Tabulated Data: 04-BS-5, May 2014, Q5

Question 5: Romberg Integration from 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 data
x	1.00	1.50	2.00	2.50	3.00	3.50	4.00	4.50	5.00
y	10.60	11.25	15.20	17.50	21.85	26.25	32.30	39.15	46.90

Find. The area $\int_1^5 y(x)\,dx$, approximated by Romberg's algorithm through $R(4,4)$ using the tabulated values at $H_4=0.5$.

The nine tabulated (x, y) points; the shaded region is the area Romberg's algorithm estimates.

Approach. Build the trapezoidal column $R(k,1)$ at successively halved step sizes $H_k=(b-a)/2^{k-1}$ (using coarser-to-finer subsets of the table), then apply Richardson extrapolation column by column to reach $R(4,4)$.

Level 1: coarsest trapezoid, $H_1=4$ (endpoints only). $$R(1,1)=\frac{H_1}{2}\big[f(1)+f(5)\big]=\frac{4}{2}(10.60+46.90)=\boxed{115.000}$$
Level 2: $H_2=2$, add the midpoint $x=3$. $$R(2,1)=\frac12\big[R(1,1)+H_1f(3)\big]=\frac12\big[115.000+4(21.85)\big]=101.200$$ $$R(2,2)=R(2,1)+\frac{R(2,1)-R(1,1)}{4^1-1}=101.200+\frac{101.200-115.000}{3}=96.600$$
Level 3: $H_3=1$, add $x=2,4$. $$R(3,1)=\frac12\big[R(2,1)+H_2\big(f(2)+f(4)\big)\big]=\frac12\big[101.200+2(15.20+32.30)\big]=98.100$$ $$R(3,2)=98.100+\frac{98.100-101.200}{3}=97.067,\qquad R(3,3)=97.067+\frac{97.067-96.600}{4^2-1}=97.098$$
Level 4: $H_4=0.5$, add $x=1.5,2.5,3.5,4.5$ (the remaining table rows). $$R(4,1)=\frac12\big[R(3,1)+H_3\big(f(1.5)+f(2.5)+f(3.5)+f(4.5)\big)\big]=\frac12\big[98.100+1(11.25+17.50+26.25+39.15)\big]=96.125$$ $$R(4,2)=95.467,\qquad R(4,3)=95.360,\qquad R(4,4)=95.360+\frac{95.360-97.098}{4^3-1}=\boxed{95.332}$$

Romberg array R(k,j)
k	R(k,1)	R(k,2)	R(k,3)	R(k,4)
1	115.000
2	101.200	96.600
3	98.100	97.067	97.098
4	96.125	95.467	95.360	95.332

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

Key relations: $R(k,1)$ — composite trapezoid at step $H_k$, built recursively from $R(k-1,1)$ plus the new midpoints; $R(k,j)=R(k,j-1)+\dfrac{R(k,j-1)-R(k-1,j-1)}{4^{j-1}-1}$.

Why this works: the trapezoidal rule's error has an asymptotic expansion in even powers of $h$ (Euler–Maclaurin). Each Richardson extrapolation column cancels the next leading error term, so $R(k,k)$ converges as $O(h^{2k})$ — far faster than any single trapezoid column, without evaluating the integrand at any new points beyond level $k$.

Common pitfall: at each new level, reusing the previous trapezoid estimate (halving the step, not restarting from scratch) and adding only the newly-required function values; also using the wrong denominator $4^{j-1}-1$ per column (it changes with $j$, not $k$).

Source: Kreyszig, Advanced Engineering Mathematics, Romberg integration (Ch. 19).

Question 5 of 7: Romberg Integration from Tabulated Data

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