Composite Trapezoidal Rule

The basic Trapezoidal Rule fits a single straight line over the entire interval $[a, b]$. When $f(x)$ curves significantly, this single trapezoid can be a poor approximation. The Composite Trapezoidal Rule fixes this by dividing $[a, b]$ into $m$ equal sub-intervals and applying the basic Trapezoidal Rule on each one. The total integral is then the sum of all the small trapezoids.

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Partitioning the Interval

Divide $[a, b]$ into $m$ equal sub-intervals of width:

$h = \frac{b - a}{m}$

This defines $m + 1$ equally spaced nodes:

$x_0 = a, \quad x_1 = a + h, \quad x_2 = a + 2h, \quad \ldots, \quad x_m = a + mh = b$

Applying the Trapezoidal Rule to the $j$-th sub-interval $[x_{j-1}, x_j]$ gives:

$\frac{h}{2}\left[f(x_{j-1}) + f(x_j)\right]$

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Assembling the Sum

The composite approximation is the sum over all $m$ sub-intervals:

$C_{1,m} = \frac{h}{2}\left[f(x_0) + f(x_1)\right] + \frac{h}{2}\left[f(x_1) + f(x_2)\right] + \cdots + \frac{h}{2}\left[f(x_{m-1}) + f(x_m)\right]$

Every interior node $x_1, x_2, \ldots, x_{m-1}$ appears once as the right endpoint of one sub-interval and once as the left endpoint of the next. Each therefore contributes twice. The two endpoints $x_0 = a$ and $x_m = b$ appear only once. Collecting terms:

$\boxed{C_{1,m} = \frac{h}{2}\left[f(x_0) + 2f(x_1) + 2f(x_2) + \cdots + 2f(x_{m-1}) + f(x_m)\right]}$

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Example: $m = 4$ Sub-intervals

Writing out all four trapezoids explicitly shows exactly how the interior nodes combine:

$C_{1,4} = \frac{h}{2}\left[f(x_0)+f(x_1)\right] + \frac{h}{2}\left[f(x_1)+f(x_2)\right] + \frac{h}{2}\left[f(x_2)+f(x_3)\right] + \frac{h}{2}\left[f(x_3)+f(x_4)\right]$

$= \frac{h}{2}\left[f(x_0) + 2f(x_1) + 2f(x_2) + 2f(x_3) + f(x_4)\right]$

The pattern is clear: coefficients of 1 at the two endpoints and 2 at every interior node.

Interior nodes each appear in two adjacent trapezoids, which is why each contributes a factor of 2 in the composite formula $C_{1,m}$.

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Accuracy and the Role of $m$

Note

As $m$ increases, $h$ decreases and each sub-interval becomes shorter. Over a short interval, any smooth function is nearly linear, so the single-trapezoid error on each piece is small. The total error for the Composite Trapezoidal Rule is $O(h^2)$ in terms of the sub-interval width $h$, meaning doubling $m$ (halving $h$) reduces the error by a factor of approximately 4. The Worked Examples section quantifies this convergence concretely using $f(x) = e^x$ on $[0, 2]$.