Romberg Integration and the Area Between Two Curves: 04-BS-5, December 2018, Q5

Question 5: Romberg Integration and the Area Between Two Curves (a: 17 marks; b: 3 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. $G(x)$ tabulated at nine equally spaced points, $x=0$ to $x=4$ in steps of $h=0.5$; separately, $R(4,4)_F=274.476180$ for $\int_0^4F(x)\,dx$ (from an identical table of $F(x)$ values, all lying above $G(x)$ on $[0,4]$).

Find. (A) The full Romberg triangular array $R(k,j)$ for $G$, $1\le j\le k\le4$, giving $\int_0^4 G(x)\,dx$. (B) The area between $F(x)$ and $G(x)$ on $[0,4]$.

Approach. Build the composite-trapezoid estimates $R(k,1)$ at $H_k=4/2^{k-1}$ (using every $2^{4-k}$-th tabulated point), then Richardson-extrapolate across the table using $R(k,j)=R(k,j-1)+\dfrac{R(k,j-1)-R(k-1,j-1)}{4^{j-1}-1}$.

$R(1,1)$: one panel, $H_1=4$. Using only $x=0,4$: $R(1,1)=\dfrac{H_1}{2}[G(0)+G(4)]=\dfrac42[6.00+64.00]=140.000$.
$R(2,1)$: two panels, $H_2=2$. Using $x=0,2,4$: $R(2,1)=\dfrac{H_2}{2}[G(0)+2G(2)+G(4)]=\dfrac22[6.00+2(66.00)+64.00]=202.000$.
$R(3,1)$: four panels, $H_3=1$. Using $x=0,1,2,3,4$: $R(3,1)=\dfrac{H_3}2[G(0)+2(G(1)+G(2)+G(3))+G(4)]=\dfrac12[6.00+2(53.00+66.00+41.00)+64.00]=195.000$.
$R(4,1)$: eight panels, $H_4=0.5$ (all nine tabulated points). $$R(4,1)=\frac{0.5}{2}\Big[G(0)+2\!\!\sum_{k=1}^{7}G(0.5k)+G(4)\Big]=202.500$$
Richardson-extrapolate across the table using $R(k,j)=R(k,j-1)+\dfrac{R(k,j-1)-R(k-1,j-1)}{4^{j-1}-1}$: $$\begin{array}{l}R(2,2)=202.000+\dfrac{202.000-140.000}{3}=222.667\\[2pt] R(3,2)=195.000+\dfrac{195.000-202.000}{3}=192.667,\quad R(3,3)=192.667+\dfrac{192.667-222.667}{15}=190.667\\[2pt] R(4,2)=202.500+\dfrac{202.500-195.000}{3}=205.000,\quad R(4,3)=205.000+\dfrac{205.000-192.667}{15}=205.822\\[2pt] R(4,4)=205.822+\dfrac{205.822-190.667}{63}=\boxed{206.063}\end{array}$$
(B) Area between $F$ and $G$. Since the tabulated $F$ values exceed $G$ at every one of the nine points, $F(x)\ge G(x)$ throughout $[0,4]$, so the enclosed area is simply the difference of the two Romberg integrals: $$\boxed{\text{Area}=R(4,4)_F-R(4,4)_G=274.476180-206.062787\approx68.413}$$

$k$	Romberg row (H_k = 4/2^(k-1))
1	$140.000$
2	$202.000,\ 222.667$
3	$195.000,\ 192.667,\ 190.667$
4	$202.500,\ 205.000,\ 205.822,\ \mathbf{206.063}$
Area between F and G	$\approx68.413$

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

Key relations: composite trapezoid $T(H)=\frac H2[f_0+2(f_1+\dots+f_{n-1})+f_n]$; Richardson extrapolation $R(k,j)=R(k,j-1)+\dfrac{R(k,j-1)-R(k-1,j-1)}{4^{j-1}-1}$, which cancels the leading $O(H^{2j-2})$ error term at each column.

Why this works: the trapezoidal rule's error expands in even powers of $H$ (Euler–Maclaurin), so successively halving $H$ and combining estimates with weight $4^{j-1}$ eliminates one more error order per column, converging to the true integral far faster than refining the trapezoid rule alone would. Because both $F$ and $G$ share the exact same node spacing, their Romberg estimates can be subtracted directly to estimate $\int(F-G)$ without recomputing a joint table.

Common pitfall: re-tabulating panels from scratch at each level instead of reusing already-computed function values (only the previous row's points are needed, halved); also subtracting the wrong-order rows (must compare the two functions' final, most-refined $R(4,4)$ estimates, not intermediate rows).

Source: Kreyszig, Advanced Engineering Mathematics, 10th ed., Ch. 19.5 (Numerical Integration and Differentiation: Romberg); Chapra & Canale, Numerical Methods for Engineers, 7th ed., Ch. 22.

Question 5 of 7: Romberg Integration and the Area Between Two Curves

Question 5: Romberg Integration and the Area Between Two Curves (a: 17 marks; b: 3 marks)