The Composite Trapezoidal Rule

• We subdivide [a,b] into $n$ smaller intervals with $\Delta x=h$, apply the rule to each subinterval, and add.
• This gives the composite trapezoidal rule;
  
  $$\int_a^b\approx\sum_{i=0}^{n-1} \frac{h}{2} (f_i+f_{i+1})=\frac{h}{2}(f_0+2f_1+2f_2+\ldots+2f_{n-1}+f_n)$$
  
  

• The error is not the local error $O(h^3)$ but the global error, the sum of $n$ local errors;
  
  $$Global~error=(-1/12)h^3[f''(\xi_1) +f''(\xi_2) +\ldots+ f''(\xi_n)]$$
  
  

• In this equation, each of the $\xi_i$ is somewhere within each subinterval.
• If $f''(x)$ is continuous in [a, b], there is some point within [a,b] at which the sum of the $f''(\xi_i)$ is equal to $nf''(\xi)$, where $\xi$ in [a, b].
• We then see that, because $nh=(b - a)$,
  
  $$Global~error=(-1/12)h^3nf''(\xi)= \frac{-(b-a)}{12}h^2f''(\xi)=O(h^2)$$
  
  

• Example: Given the values for $x$ and $f(x)$ in Table3.
  

  Table 3: Example for the trapezoidal rule.

  
  
  
• Use the trapezoidal rule to estimate the integral from $x=1.8$ to $x=3.4$.
• Applying the trapezoidal rule:
  
  $$\begin{array}{rl} \int_{1.8}^{3.4}f(x)dx& \approx \frac{0.2}... ...+ 2(20.086) + 2(24.533)\\ & + 29.964] = 23.9944\\ \end{array}$$
  
  

• The data in Table 3 are for $f(x) = e^x$ and the true value is $e^{3.4}-e^{1.8}=23.9144$.
• The trapezoidal rule value is off by $0.08$; there are three digits of accuracy.
• How does this compare to the estimated error?

$$\begin{array}{rl} Error&=-\frac{1}{12}h^3nf''(\xi),~1.8\leq\... ...0323&(max)\\ -0.1598&(min)\\ \end{array}\right \} \end{array}$$

Alternatively,

$$\begin{array}{rl} Error&=-\frac{1}{12}(0.2)^2(3.4-1.8)*\left... ....0323&(max)\\ -0.1598&(min)\\ \end{array}\right\} \end{array}$$

• The actual error was $-0.080$. It is reasonable since the value is in the error bounds.

 
 
Thanks for attending and listening.