To derive the composite formula for the trapezoidal rule and show that it takes the form:

[ T_n[f](a, b) = \sum_{i=0}^{n-1} T[f](x_i, x_{i+1}) = \frac{h}{2}(f(x_0) + f(x_n)) + h \sum_{i=1}^{n-1} f(x_i), ]

we need to consider how the trapezoidal rule is applied to the interval $[a, b]$ which is divided into $n$ segments.

Step-by-step Solution:

1. Divide the Interval into Subintervals:

   The interval $[a, b]$ is divided into $n$ segments, each of length $h = \frac{b-a}{n}$. The points of division are denoted by:

   $x_i = a + ih, \quad \text{for } i = 0, 1, \ldots, n.$

   The first point is $x_0 = a$ and the last point is $x_n = b$.

2. Apply the Trapezoidal Rule on Each Subinterval:

   The trapezoidal rule for approximating the integral of a function $f(x)$ over a small subinterval $[x_i, x_{i+1}]$ is given by:

   $T[f](x_i, x_{i+1}) = \frac{h}{2} \left( f(x_i) + f(x_{i+1}) \right).$

   This formula uses the fact that the area under the curve $f(x)$ on the interval $[x_i, x_{i+1}]$ can be approximated by the area of a trapezoid with base $h$ and heights $f(x_i)$ and $f(x_{i+1})$.

3. Sum the Contributions Over All Subintervals:

   The composite trapezoidal rule sums up the contributions from all subintervals:

   [ T_n[f](a, b) = \sum_{i=0}^{n-1} T[f](x_i, x_{i+1}). ]

   Substituting the formula for $T[f](x_i, x_{i+1})$ gives:

   $T_n[f](a, b) = \sum_{i=0}^{n-1} \frac{h}{2} \left( f(x_i) + f(x_{i+1}) \right).$

   This can be rewritten as:

   $T_n[f](a, b) = \frac{h}{2} \left( \sum_{i=0}^{n-1} f(x_i) + \sum_{i=0}^{n-1} f(x_{i+1}) \right).$

4. Combine Terms to Simplify the Expression:

   Notice that each $f(x_i)$ appears twice in the sums above, except for $f(x_0)$ and $f(x_n)$ which only appear once (at the beginning and end). Therefore, we can combine terms as follows:

   $T_n[f](a, b) = \frac{h}{2} \left( f(x_0) + 2 \sum_{i=1}^{n-1} f(x_i) + f(x_n) \right).$

5. Factor Out the Constants to Get the Desired Form:

   This simplifies to:

   $T_n[f](a, b) = \frac{h}{2} \left( f(x_0) + f(x_n) \right) + h \sum_{i=1}^{n-1} f(x_i).$

   And this is the desired formula for the composite trapezoidal rule.

Conclusion:

The composite trapezoidal rule approximates the integral of $f(x)$ over the interval $[a, b]$ by summing up the areas of trapezoids over each subinterval. The derived formula is:

$T_n[f](a, b) = \frac{h}{2}(f(x_0) + f(x_n)) + h \sum_{i=1}^{n-1} f(x_i).$

This formula is used to estimate the definite integral of $f(x)$ over $[a, b]$ using $n$ subintervals.

Would you like more details or have any specific questions?

Related Questions:

1. What is the error term for the composite trapezoidal rule?
2. How does the composite trapezoidal rule differ from Simpson's rule?
3. What happens to the accuracy of the trapezoidal rule as $n$ increases?
4. How can we derive the trapezoidal rule for unequal intervals?
5. What are the advantages and disadvantages of using the trapezoidal rule compared to other numerical integration methods?

Tip: The error in the trapezoidal rule is proportional to the square of the width of the intervals, $h^2$. Reducing the interval width by increasing $n$ improves the accuracy significantly.