Editing Trapezoidal rule (section)

=== Proof ===
First suppose that <math>h=\frac{b-a}{N}</math> and <math>a_k=a+(k-1)h</math>. Let <math display="block"> g_k(t) = \frac{1}{2}  t[f(a_k)+f(a_k+t)] - \int_{a_k}^{a_k+t} f(x) \, dx</math> be the function such that <math> |g_k(h)| </math> is the error of the trapezoidal rule on one of the intervals, <math> [a_k, a_k+h] </math>. Then
<math display="block"> {dg_k \over dt}={1 \over 2}[f(a_k)+f(a_k+t)]+{1\over2}t\cdot f'(a_k+t)-f(a_k+t),</math>
and
<math display="block"> {d^2g_k \over dt^2}={1\over 2}t\cdot f''(a_k+t).</math>

Now suppose that <math> \left| f''(x) \right| \leq \left| f''(\xi) \right|, </math> which holds if <math> f </math> is sufficiently smooth. It then follows that
<math display="block"> \left| f''(a_k+t) \right| \leq f''(\xi)</math>
which is equivalent to
<math> -f''(\xi) \leq f''(a_k+t) \leq f''(\xi)</math>, or <math> -\frac{f''(\xi)t}{2} \leq g_k''(t) \leq \frac{f''(\xi)t}{2}.</math>

Since <math> g_k'(0)=0</math> and <math> g_k(0)=0</math>,
<math display="block"> \int_0^t g_k''(x)  dx = g_k'(t)</math> and <math display="block"> \int_0^t g_k'(x) dx = g_k(t).</math>

Using these results, we find
<math display="block"> -\frac{f''(\xi)t^2}{4} \leq g_k'(t) \leq \frac{f''(\xi)t^2}{4}</math>
and
<math display="block"> -\frac{f''(\xi)t^3}{12} \leq g_k(t) \leq \frac{f''(\xi)t^3}{12}</math>

Letting <math> t = h </math> we find
<math display="block"> -\frac{f''(\xi)h^3}{12} \leq g_k(h) \leq \frac{f''(\xi)h^3}{12}.</math>

Summing all of the local error terms we find
<math display="block"> \sum_{k=1}^{N} g_k(h) = \frac{b-a}{N} \left[ {f(a) + f(b) \over 2} + \sum_{k=1}^{N-1} f \left( a+k \frac{b-a}{N} \right) \right] - \int_a^b f(x)dx.</math>

But we also have
<math display="block"> - \sum_{k=1}^N \frac{f''(\xi)h^3}{12} \leq \sum_{k=1}^N g_k(h) \leq  \sum_{k=1}^N \frac{f''(\xi)h^3}{12}</math>
and
<math display="block"> \sum_{k=1}^N \frac{f''(\xi)h^3}{12}=\frac{f''(\xi)h^3N}{12},</math>

so that

<math display="block"> -\frac{f''(\xi)h^3N}{12} \leq \frac{b-a}{N} \left[ {f(a) + f(b) \over 2} + \sum_{k=1}^{N-1} f \left( a+k \frac{b-a}{N} \right) \right]-\int_a^bf(x)dx \leq \frac{f''(\xi)h^3N}{12}.</math>

Therefore the total error is bounded by

<math display="block"> \text{error} = \int_a^b f(x)\,dx - \frac{b-a}{N} \left[ {f(a) + f(b) \over 2} + \sum_{k=1}^{N-1} f \left( a+k \frac{b-a}{N} \right) \right] = \frac{f''(\xi)h^3N}{12}=\frac{f''(\xi)(b-a)^3}{12N^2}.</math>