The Midpoint Rule
Textbook Reference
| Primary source | OpenStax Calculus Volume 2, Section 1.5: “Numerical Integration” |
| Direct link | https://openstax.org/books/calculus-volume-2/pages/1-5-numerical-integration |
| Textbook used in class | Stewart, Calculus, Section 4.2: “The Definite Integral” |
Opening Scenario
Left- and right-endpoint Riemann sums give approximations that err in opposite directions for monotone functions. Each rectangle’s height is the function value at one end of the subinterval, which is typically the worst possible choice: the function is at its extreme for that subinterval.
A simple improvement: evaluate the function at the midpoint of each subinterval. The midpoint is neither the maximum nor the minimum on that subinterval, and using it tends to cancel the errors from the two halves of the subinterval. The Midpoint Rule is often twice as accurate as the endpoint rules for the same number of rectangles.
Quick Reference
Midpoint Rule: Divide $[a, b]$ into $n$ equal subintervals of width $\Delta x = \dfrac{b-a}{n}$. Let $\bar{x}_i$ be the midpoint of the $i$-th subinterval: $$\bar{x}_i = \frac{x_{i-1} + x_i}{2} = a + \left(i - \tfrac{1}{2}\right)\Delta x.$$
$$\int_a^b f(x)\,dx \approx M_n = \sum_{i=1}^{n} f(\bar{x}_i)\,\Delta x = \Delta x\bigl[f(\bar{x}_1) + f(\bar{x}_2) + \cdots + f(\bar{x}_n)\bigr].$$
Key Concepts
1. Why Midpoints Are Better
On a small subinterval, a smooth function is nearly linear. For a linear function, the midpoint value equals the average of the left and right endpoint values, so the midpoint rectangle exactly captures the area of the trapezoid under the linear piece. Non-linear functions deviate from a straight line, but on a short subinterval the deviation is tiny. The result is that the midpoint rule error scales with the square of the subinterval width, just like the endpoint rules -- but with a smaller error constant in practice.
The technical statement (proved in more advanced courses) is:
- Error in $L_n$ and $R_n$ is $O(1/n)$ (halving $n$ halves the error).
- Error in $M_n$ is $O(1/n^2)$ (halving $n$ quarters the error).
2. Setting Up the Midpoints
The $i$-th subinterval runs from $x_{i-1} = a + (i-1)\Delta x$ to $x_i = a + i\,\Delta x$. Its midpoint is: $$\bar{x}_i = a + \left(i - \frac{1}{2}\right)\Delta x.$$
For $n = 4$ on $[0, 2]$: $\Delta x = \frac{1}{2}$, so the midpoints are $\bar{x}_1 = \frac{1}{4}$, $\bar{x}_2 = \frac{3}{4}$, $\bar{x}_3 = \frac{5}{4}$, $\bar{x}_4 = \frac{7}{4}$.
3. When to Use Numerical Rules
The Midpoint Rule (and its relatives, the Trapezoidal Rule and Simpson’s Rule) are useful when:
- An antiderivative of $f$ cannot be found in closed form.
- The function is known only from data (not a formula).
- A quick estimate is acceptable, with the number of subintervals chosen to meet a prescribed error tolerance.
The Fundamental Theorem of Calculus, when applicable, gives the exact answer in far fewer steps. The numerical rules are the fallback for integrands that resist antidifferentiation.
“The midpoint rule always overestimates or always underestimates.” Whether the midpoint rule over- or underestimates depends on the concavity of $f$, not monotonicity. For a concave-up function ($f'' > 0$), $M_n$ underestimates; for a concave-down function ($f'' < 0$), $M_n$ overestimates. The reasoning comes from the fact that the tangent line at the midpoint lies below a concave-up curve and above a concave-down curve.
Worked Example
Approximate $\displaystyle\int_1^3 \frac{1}{x}\,dx$ using $M_4$.
Step 1 -- Setup. $a = 1$, $b = 3$, $n = 4$, $\Delta x = \dfrac{2}{4} = \dfrac{1}{2}$.
Step 2 -- Midpoints. $\bar{x}_i = 1 + \left(i - \dfrac{1}{2}\right)\cdot\dfrac{1}{2}$:
$\bar{x}_1 = 1.25$, $\bar{x}_2 = 1.75$, $\bar{x}_3 = 2.25$, $\bar{x}_4 = 2.75$.
Step 3 -- Evaluate and sum.
$$M_4 = \frac{1}{2}\left[\frac{1}{1.25} + \frac{1}{1.75} + \frac{1}{2.25} + \frac{1}{2.75}\right]$$ $$= \frac{1}{2}[0.8000 + 0.5714 + 0.4444 + 0.3636]$$ $$= \frac{1}{2}(2.1794) \approx 1.0897.$$
Exact value: $\displaystyle\int_1^3 \frac{1}{x}\,dx = \ln 3 \approx 1.0986$.
Error: $|1.0986 - 1.0897| \approx 0.009$, less than 1%.
Boxed result: $M_4 \approx 1.090$ (exact: $\ln 3 \approx 1.099$).
Common Errors Summary
| Error | Example | Correction |
|---|---|---|
| Using left or right endpoints instead of midpoints | Setting $\bar{x}_i = a + i\,\Delta x$ | Midpoints are $a + (i - \frac{1}{2})\Delta x$; the $-\frac{1}{2}$ shifts each endpoint to the center |
| Forgetting to multiply the sum by $\Delta x$ | Reporting $f(\bar{x}_1)+\cdots+f(\bar{x}_n)$ as the approximation | The width $\Delta x$ must multiply each height; the full formula is $M_n = \Delta x\sum f(\bar{x}_i)$ |
| Assuming midpoint rule is always more accurate than trapezoidal | Stating $M_n$ is always better | Both have $O(1/n^2)$ error; which is closer depends on the specific function and interval |
Leveled Practice
Level 1 -- Setup and Compute
Problem 1. Approximate $\displaystyle\int_0^1 x^2\,dx$ using $M_4$.
Show answer
$\Delta x = \frac{1}{4}$. Midpoints: $\frac{1}{8}, \frac{3}{8}, \frac{5}{8}, \frac{7}{8}$.
$M_4 = \frac{1}{4}\left[\left(\frac{1}{8}\right)^2 + \left(\frac{3}{8}\right)^2 + \left(\frac{5}{8}\right)^2 + \left(\frac{7}{8}\right)^2\right]$ $= \frac{1}{4}\left[\frac{1}{64} + \frac{9}{64} + \frac{25}{64} + \frac{49}{64}\right]$ $= \frac{1}{4}\cdot\frac{84}{64} = \frac{84}{256} = \frac{21}{64} \approx 0.328$.
Exact: $\frac{1}{3} \approx 0.333$. Error $\approx 0.005$.
Level 2 -- Concavity and Error Direction
Problem 2. For $f(x) = e^x$ on $[0, 1]$, is $M_4$ an over- or underestimate? Justify using concavity.
Show answer
$f''(x) = e^x > 0$ on $[0, 1]$, so $f$ is concave up. For a concave-up function, the midpoint rectangle lies below the curve (the tangent line at the midpoint undershoots the curve on both sides). Therefore $M_4$ underestimates $\displaystyle\int_0^1 e^x\,dx$.
Mastery Checklist
Mental Model
Left and right endpoint rectangles are like pricing something at the store’s opening or closing price -- each is systematically high or low depending on the trend. The midpoint is like pricing at noon: it is the center of the day’s range, and for a nearly linear trend the noon price is the best single estimate of the day’s average. Over an entire day of small subintervals, the midpoint estimates accumulate less error than the endpoint estimates.
Connections
Looking back
- Riemann sums (Sections 4.1--4.2): The Midpoint Rule is one choice of sample points in a Riemann sum.
- Concavity (Section 3.3): Whether $f''$ is positive or negative determines the error direction.
Looking ahead
- Trapezoidal and Simpson’s rules: More advanced numerical methods with still higher accuracy.
- FTC (Section 4.3): When an antiderivative is available, use it for an exact answer instead.
Back to Integration Foundations | Next: Properties of the Definite Integral