# Thoughts on Numerical Integration (Part 3): Derivation of left, right, and midpoint rules

Numerical integration is a standard topic in first-semester calculus. From time to time, I have received questions from students on various aspects of this topic, including:

• Why is numerical integration necessary in the first place?
• Where do these formulas come from (especially Simpson’s Rule)?
• How can I do all of these formulas quickly?
• Is there a reason why the Midpoint Rule is better than the Trapezoid Rule?
• Is there a reason why both the Midpoint Rule and the Trapezoid Rule converge quadratically?
• Is there a reason why Simpson’s Rule converges like the fourth power of the number of subintervals?

In this series, I hope to answer these questions. While these are standard questions in a introductory college course in numerical analysis, and full and rigorous proofs can be found on Wikipedia and Mathworld, I will approach these questions from the point of view of a bright student who is currently enrolled in calculus and hasn’t yet taken real analysis or numerical analysis.

For the sake of completeness, I discuss here the origins of the left-endpoint, right-endpoint, and midpoint rules of numerical integration. (These topics are often presented in calculus texts.) Consider the problem of finding $\displaystyle \int_a^b f(x) \, dx$, the area under a curve $f(x)$ between $x=a$ and $x=b$.

To start the process of numerical integration, the interval $[a,b]$ is divided into subintervals. Usually, for convenience, the intervals are chosen to be the same length, a convention that I’ll follow in this series. That said, if the function is known to vary wildly on some parts of the domain but not so wildly on other parts, then computational efficiency can be gained by varying the sizes of the subintervals, choosing smaller subintervals for the places where the function varies wildly.

In any event, we’ll choose equal-sized subintervals for the duration of this series.

One numerical approximation can be made by choosing left endpoints. In the picture below, the interval $[a,b]$ was divided into four equal subintervals. Let $h = (b-a)/4$, so that $x_0 = a$, $x_1 = x_0 +h$, $x_2 = x_0 + 2h$, $x_3 = x_0 + 3h$, and $x_4 = x_0 + 4h = b$. We then can draw rectangles using the left endpoints of each subinterval. The sum of the areas of these rectangles below is

$hf(x_0) + hf(x_1) + hf(x_2) +hf(x_3)$,

and so this serves as an approximation to the area under the curve. In general, if there are $n$ subintervals and $x_k = x_0 + kh$, then the integral may be approximated as

$\int_a^b f(x) \, dx \approx h \left[f(x_0) + f(x_1) + \dots + f(x_{n-1}) \right]$

That said, left endpoints were not necessary for making an approximation. We could have instead chosen the right endpoints of each subinterval. The sum of the areas of the rectangles below is

$hf(x_1) + hf(x_2) + hf(x_3) +hf(x_4)$,

and so this serves as an approximation to the area under the curve. In general, if there are $n$ subintervals, then the integral may be approximated as

$\int_a^b f(x) \, dx \approx h \left[f(x_1) + f(x_2) + \dots + f(x_n) \right]$

As a final approximation, any point in each subinterval could’ve been used for making an approximation. In the picture below, we use the midpoints of the subintervals, where $c_k = (x_k + x_{k-1})/2$. The sum of the areas of the rectangles below is

$hf(c_1) + hf(c_2) + hf(c_3) +hf(c_4)$,

and so this serves as an approximation to the area under the curve. In general, if there are $n$ subintervals, then the integral may be approximated as

$\int_a^b f(x) \, dx \approx h \left[f(c_1) + f(c_2) + \dots + f(c_n) \right]$

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