Error Analysis

Every Newton-Cotes formula replaces $f(x)$ with a polynomial $P_n(x)$, and the error in the integral is determined by how well $P_n(x)$ approximates $f(x)$ over $[a, b]$. From polynomial interpolation theory we already know that the pointwise error in $P_n(x)$ is controlled by $f^{(n+1)}(\xi)$ and the product of distances from $x$ to each node. Integrating that pointwise bound directly gives the error bound for $I_n(f)$.

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General Error Bound

The error in degree-$n$ Lagrange interpolation at any point $x \in [a, b]$ satisfies:

$f(x) - P_n(x) \leq \frac{f^{(n+1)}(\xi)}{(n+1)!}(x - x_0)(x - x_1)\cdots(x - x_n)$

for some unknown $\xi$ inside the interval spanned by the nodes. Integrating both sides over $[a, b]$:

$\int_a^b f(x)\,dx - \int_a^b P_n(x)\,dx \leq \frac{f^{(n+1)}(\xi)}{(n+1)!}\int_a^b (x - x_0)(x - x_1)\cdots(x - x_n)\,dx$

Since $I_n(f) = \int_a^b P_n(x)\,dx$, the left side is exactly $I(f) - I_n(f)$. This gives the general Newton-Cotes error bound:

$\boxed{I(f) - I_n(f) \leq \frac{f^{(n+1)}(\xi)}{(n+1)!}\int_a^b \prod_{k=0}^{n}(x - x_k)\,dx}$

Note

The unknown point $\xi$ lies somewhere inside the interval, but we cannot pinpoint it. To obtain a usable upper bound, we replace $|f^{(n+1)}(\xi)|$ with the maximum of $|f^{(n+1)}(x)|$ over $[a, b]$. The integral $\int_a^b \prod_{k=0}^n (x - x_k)\,dx$ depends only on the node positions and can be computed once for each rule.

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Error Bound for the Trapezoidal Rule ($n = 1$)

For $n = 1$ the two nodes are $x_0 = a$ and $x_1 = b$. The product term is $(x - a)(x - b)$, and the formula specializes to:

$I(f) - I_1(f) \leq \frac{f''(\xi)}{2!}\int_a^b (x - a)(x - b)\,dx$

Evaluating the integral (the product $(x-a)(x-b)$ is non-positive on $[a,b]$, so we use the absolute value for the bound):

$\int_a^b (x-a)(x-b)\,dx = -\frac{(b-a)^3}{6}$

Taking the absolute value:

$\left|I(f) - I_1(f)\right| \leq \frac{|f''(\xi)|}{2} \cdot \frac{(b-a)^3}{6} = \frac{|f''(\xi)|}{12}(b-a)^3$

To get a numerical bound, replace $|f''(\xi)|$ with $M_2 = \max_{x \in [a,b]} |f''(x)|$:

$\left|I(f) - I_1(f)\right| \leq \frac{M_2}{12}(b-a)^3$

For a convex function ($f''(x) > 0$ throughout $[a,b]$), the chord lies strictly above the curve, and the red region represents the entire overestimation $I_1 - I(f)$.

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Effect of Higher-Degree Rules

Rule	Order $n$	Leading Error Term	Error vs. $(b-a)$
Trapezoidal	1	$\dfrac{M_2}{12}(b-a)^3$	$O\!\left((b-a)^3\right)$
Simpson’s	2	$\dfrac{M_4}{90}(b-a)^5$	$O\!\left((b-a)^5\right)$

Two observations follow immediately:

1. Higher-degree rules are more accurate on a fixed interval. Simpson’s Rule introduces an error proportional to $(b-a)^5$ rather than $(b-a)^3$. For a moderately sized interval this is a dramatic reduction.

2. The Composite Trapezoidal Rule trades interval size for accuracy. Using $m$ sub-intervals each of width $h = (b-a)/m$, the error per sub-interval is $O(h^3)$ and the total error is $m \times O(h^3) = O(h^2)$. Doubling $m$ halves $h$ and reduces the total error by a factor of approximately 4.

Caution

The error bounds above assume $f$ is sufficiently differentiable. If $f$ or its derivatives are large over $[a,b]$, even a high-order rule may produce a poor approximation for a given number of function evaluations. In practice, the composite approach is often preferred over using a single high-order rule on a wide interval.