# Quadrature Error for Trapezoidal Rule

Question: If the trapezoidal rule is used to approximate the integral of the function $f$ over the interval $[a,b]$ with 201 evenly spaced dissection points, estimate the quadrature error for

$$\int_0^1log(1+x^2)dx$$

For this style of question would I use the formula

$$error \le \sum_{i=1}^{N} \frac{-1}{12}\left(\frac{b-a}{N}\right)^3 f''(c_i)$$

to find the error?

Where N is the number of equal length intervals, [a,b] is the full inteval, $c_i$ is the midpoint of each interval and $f$ is $log(1+x^2)$.

If this is the case, how do I find $c_i$?

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Instead of adding up the estimates for the local errors, we can use the standard expression for the global error, which in general is $$-\frac{(b-a)^3}{12N^2}f''(c),$$ where $c$ is a number between $a$ and $b$. In our case, $a=0$ and $b=1$. The number $N$ of subintervals is $200$.
Now we need an estimate for $f''(c)$. The second derivative is, I think, $\dfrac{2(1-x^2)}{(1+x^2)^2}$. As $x$ increases, the numerator is decreasing, and the denominator is increasing. Thus $f''(x)$ is steadily decreasing. It is equal to $2$ at $x=0$, so $0\le f''(x)\lt 2$.
Remark: Or else use the local errors, and add them up. Since the second derivative is everywhere positive and $\lt 2$, the local error is always negative and $\lt \dfrac{1}{12}\cdot\dfrac{1}{(200)^3}\cdot 2$ in absolute value, so in absolute value the total error is $\le 200\cdot \dfrac{1}{12}\cdot\dfrac{1}{(200)^3}\cdot 2$.
Typically, the error estimates produced by this kind of calculation are unduly pessimistic. One can ordinarily get more useful information by comparing $T_{100}$ with $T_{200}$. The number $T_{100}$ is typically computed on the way to computing $T_{200}$, so it involves no additional cost.
You can use the "local" errors version. The $N$ individual errors are bounded as given by your formmula, so their sum is $\le N$ times the maximum local error. For that maximum, maximimize the second derivative as I did. – André Nicolas Oct 17 '12 at 19:18