Romberg Integration of Tabulated Data: 04-BS-5, December 2013, Q6

Question 6: 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 data
$x$	1	2	3	4	5	6	7	8	9
$y$	12	17	21	24	27	25	23	20	16

Find. $\int_1^9 y(x)\,dx$ approximated by the Romberg array through $R(4,4)$.

Approach. Build successively finer composite-trapezoid estimates $R(k,1)$ with $H_k=(b-a)/2^{k-1}$ using only the tabulated points at each level, then Richardson-extrapolate across the array using the supplied $R(k,j)$ formula.

Tabulated experimental data $y(x)$, $x=1$ to $9$.

Level 1 ($H_1=8$, nodes $x=1,9$). $$R(1,1)=\dfrac{H_1}{2}[y(1)+y(9)]=\dfrac82[12+16]=\boxed{112.00000}$$
Level 2 ($H_2=4$, new node $x=5$). $$R(2,1)=\dfrac12\big[R(1,1)+H_1\,y(5)\big]=\dfrac12[112+8(27)]=164.00000$$ $$R(2,2)=R(2,1)+\dfrac{R(2,1)-R(1,1)}{4^1-1}=164+\dfrac{164-112}{3}=\boxed{181.33333}$$
Level 3 ($H_3=2$, new nodes $x=3,7$). $$R(3,1)=\dfrac12\big[R(2,1)+H_2(y(3)+y(7))\big]=\dfrac12[164+4(21+23)]=170.00000$$ $$R(3,2)=R(3,1)+\dfrac{R(3,1)-R(2,1)}{3}=172.00000,\qquad R(3,3)=R(3,2)+\dfrac{R(3,2)-R(2,2)}{15}=\boxed{171.37778}$$
Level 4 ($H_4=1$, new nodes $x=2,4,6,8$ — every remaining table entry). $$R(4,1)=\dfrac12\big[R(3,1)+H_3(y(2)+y(4)+y(6)+y(8))\big]=\dfrac12[170+2(17+24+25+20)]=171.00000$$ $$R(4,2)=R(4,1)+\dfrac{R(4,1)-R(3,1)}{3}=171.33333$$ $$R(4,3)=R(4,2)+\dfrac{R(4,2)-R(3,2)}{15}=171.28889$$ $$\boxed{R(4,4)=R(4,3)+\dfrac{R(4,3)-R(3,3)}{63}=171.28748}$$

$k$	$R(k,1)$	$R(k,2)$	$R(k,3)$	$R(k,4)$
1	112.00000
2	164.00000	181.33333
3	170.00000	172.00000	171.37778
4	171.00000	171.33333	171.28889	171.28748

Concept & Theory

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

Key relations. $R(1,1)$ is the one-panel trapezoid; each $R(k,1)$ adds the midpoints of the previous level's panels; $R(k,j)=R(k,j-1)+\dfrac{R(k,j-1)-R(k-1,j-1)}{4^{j-1}-1}$ cancels the leading error term at each extrapolation order.

Why this works. The trapezoidal rule's error expands in even powers of $H$ (Euler–Maclaurin); Richardson extrapolation combines two estimates at different $H$ to cancel the leading-order term, and repeating the process up the triangular array cancels successively higher-order terms, converging much faster than the raw trapezoid values.

Common pitfall. At each new level, only the newly introduced midpoints are summed and added to the previous $R(k-1,1)$ — re-summing every point from scratch (instead of reusing $R(k-1,1)$) is unnecessary extra work and a common source of arithmetic slips.

Source. Kreyszig, Advanced Engineering Mathematics, Ch. 19 (Numerics for Integration; Romberg Integration).

Question 6 of 7: Romberg Integration of Tabulated Data

Question 6: Romberg Integration of Tabulated Data (20 marks)