The Composite Trapezoidal Rule

• If we are getting the integral of a known function over a larger span of $x$-values, say, from $x=a$ to $x=b$, 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;
  
  $$\int_a^b\approx\sum_{i=0}^{n-1} (h/2)(f_i+f_{i+1})=(h/2)(f_0+2f_1+2f_2+\ldots+2f_{n-1}+f_n)$$
  
  
  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 $f''(\xi)$, where $\xi$ in [a, b]. We then see that, because $nh=(b - a)$,
  
  $$Global~error=(-1/12)h^3f''(\xi)= \frac{-(b-a)}{12}h^2f''(\xi)=O(h^2)$$
  
  
  An Algorithm for Integration by the Composite Trapezoidal Rule:
  
• Example: Given the values for $x$ and $f(x)$ in Table3, use the trapezoidal rule to estimate the integral from $x=1.8$ to $x=3.4$. Applying the trapezoidal rule:
  
  $$\begin{array}{rl} \int_{1.8}^{3.4}f(x)dx& \approx \frac{0.2}... ...+ 2(20.086) + 2(24.533) + 29.964]\\ & = 23.9944\\ \end{array}$$
  
  
  
  

  Table 3: Example for the trapezoidal rule.

  
  
  The data in Table 3 are for $f(x) = e^x$ and the true value is $e^{3.4}-e^{l.8}=23.9144$. 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 $-0.080$.