Question 5 of 7: Romberg Integration of Tabulated Data

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.

Find. A four-level Romberg estimate $R(4,4)$ of $\int_{-4}^{4} y(x)\,dx$.

Approach. Build the trapezoidal column $R(k,1)$ at successively halved step sizes $H_k=(b-a)/2^{k-1}$ using the tabulated $y$-values, then apply Richardson extrapolation column by column ($R(k,j)=R(k,j-1)+\dfrac{R(k,j-1)-R(k-1,j-1)}{4^{j-1}-1}$) to reach $R(4,4)$.

1. Trapezoidal column $R(k,1)$. With $a=-4$, $b=4$ ($b-a=8$), $H_1=8,H_2=4,H_3=2,H_4=1$: $$R(1,1)=\frac{H_1}{2}\big[f(-4)+f(4)\big]=4(113+239)=1408.$$ $$R(2,1)=\tfrac12\big[R(1,1)+H_1 f(0)\big]=\tfrac12(1408+8\cdot124)=1200.$$ $$R(3,1)=\tfrac12\big[R(2,1)+H_2\big(f(-2)+f(2)\big)\big]=\tfrac12(1200+4\cdot282)=1164.$$ $$R(4,1)=\tfrac12\big[R(3,1)+H_3\big(f(-3)+f(-1)+f(1)+f(3)\big)\big]=\tfrac12(1164+2\cdot572)=1154.$$ Each $R(k,1)$ is exactly the composite trapezoidal rule at step $H_k$.
2. First Richardson column, $R(k,2)=R(k,1)+\dfrac{R(k,1)-R(k-1,1)}{3}$. $$R(2,2)=1200+\frac{1200-1408}{3}=1130.667,\quad R(3,2)=1164+\frac{1164-1200}{3}=1152,$$ $$R(4,2)=1154+\frac{1154-1164}{3}=1150.667.$$
3. Second column, $R(k,3)=R(k,2)+\dfrac{R(k,2)-R(k-1,2)}{15}$. $$R(3,3)=1152+\frac{1152-1130.667}{15}=1153.422,\qquad R(4,3)=1150.667+\frac{1150.667-1152}{15}=1150.578.$$
4. Final extrapolation, $R(4,4)=R(4,3)+\dfrac{R(4,3)-R(3,3)}{63}$. $$R(4,4) = 1150.578 + \frac{1150.578-1153.422}{63} = 1150.578 - 0.0452.$$ $$\boxed{R(4,4) \approx 1150.53}$$