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# 9.1: Mid-Point Rule

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The simplest numerical integration method is called the mid-point rule. Consider a definite 1D integral

$\mathcal{I} = \int_a^b f(x) \, dx.$

Let us divide the range $$a \le x \le b$$ into a set of $$N$$ segments of equal width, as shown in Fig. $$\PageIndex{1}$$ for the case of $$N=5$$. The mid-points of these segments are a set of $$N$$ discrete points $$\{x_0, \dots x_{N-1}\}$$, where

$x_n = a + \left(n + \frac{1}{2}\right)\,\Delta x ,\quad \Delta x \equiv \frac{b-a}{N}.$

We then estimate the integral as

$\mathcal{I}^{\mathrm{(mp)}} = \Delta x \; \sum_{n=0}^{N-1} f(x_n) \;\;\;\overset{N\rightarrow\infty}{\longrightarrow} \;\;\; \mathcal{I}.$ Figure $$\PageIndex{1}$$: Computing a definite integral using the mid-point rule.

The principle behind this formula is very simple to understand. As shown in Fig. $$\PageIndex{1}$$, $$I_{N}$$ represents the area enclosed by a sequence of rectangles, where the height of each rectangle is equal to the value of $$f(x)$$ at its mid-point. As $$N \rightarrow \infty$$, the spacing between rectangles goes to zero; hence, the total area enclosed by the rectangles becomes equal to the area under the curve of $$f(x)$$.

## 9.1.1 Numerical Error for the Mid-Point Rule

Let us estimate the numerical error resulting from this approximation. To do this, consider one of the individual segments, which is centered at $$x_n$$ with length $$\Delta x = (b-a)/N$$. Let us define the integral over this segment as

$\Delta \mathcal{I}_n \equiv \int_{x_n - \Delta x/2}^{x_n + \Delta x/2} f(x) dx.$

Now, consider the Taylor expansion of $$f(x)$$ in the vicinity of $$x_{n}$$:

$f(x) = f(x_n) + f'(x_n) (x-x_n) + \frac{f''(x_n)}{2} (x-x_n)^2 + \frac{f'''(x_n)}{6} (x-x_n)^3 + \cdots$

If we integrate both sides of this equation over the segment, the result is

\begin{align}\Delta\mathcal{I}_n \; = \; f(x_n) \Delta x \;\,&+\; f'(x_n) \int_{x_n - \Delta x/2}^{x_n + \Delta x/2} (x-x_n) dx\\ &+\; \frac{f''(x_n)}{2} \int_{x_n - \Delta x/2}^{x_n + \Delta x/2} (x-x_n)^2 dx\\ &+\; \cdots\end{align}

On the right hand side, every other term involves an integrand which is odd around $$x_{n}$$. Such terms integrate to zero. From the remaining terms, we find the following series for the integral of $$f(x)$$ over the segment:

$\Delta\mathcal{I}_n \; = \; f(x_n) \Delta x \;+\; \frac{f''(x_n) \Delta x^3}{24} + O(\Delta x^5).$

By comparison, the estimation provided by the mid-point rule is simply

$\Delta \mathcal{I}^{\mathrm{mp}}_n = f(x_n) \Delta x$

This is simply the first term in the exact series. The remaining terms constitute the numerical error in the mid-point rule integration, over this segment. We denote this error as

$\mathcal{E}_n = \left|\Delta \mathcal{I}_n - \Delta \mathcal{I}_n^{\mathrm{mp}}\right| \;\sim\; \frac{|f''(x_n)|}{24} \Delta x^3 \;\sim\; O\left(\frac{1}{N^3}\right).$

The last step comes about because, by our definition, $$\Delta x \sim O(1/N)$$.

Now, consider the integral over the entire integration range, which consists of $$N$$ such segments. In general, there is no guarantee that the numerical errors of each segment will cancel out, so the total error should be $$N$$ times the error from each segment. Hence, for the mid-point rule,

$\mathcal{E}_{\mathrm{total}} \sim O\left(\frac{1}{N^2}\right).$

9.1: Mid-Point Rule is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Y. D. Chong via source content that was edited to conform to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.