Left and right sums are two choices among many. We may sample each strip anywhere: at the midpoint, at the highest point, at the lowest point. All such sums are called Riemann sums, and as the strips get thinner they all approach the same number.
Idea clave
A Riemann sum for $f$ on $[a, b]$ is $\displaystyle\sum_{i=1}^{n} f(x_i^*)\,\Delta x$, where $x_i^*$ is any sample point in the $i$th strip. The midpoint sum $M_n$ uses the middle of each strip. For a continuous $f$, every choice of sample points gives the same limit as $n \to \infty$.
Por qué funciona
For a continuous function on a thin strip, the highest and lowest values are close together, so every sample point gives nearly the same height. The upper sum (highest points) and lower sum (lowest points) trap every other Riemann sum between them, and their difference shrinks to 0 as the strips narrow. So all the sums share one limit. The midpoint often does best because, on a strip where the curve is nearly straight, its overshoot on one half cancels its undershoot on the other.
Ejemplo resuelto: A midpoint sum
Find $M_4$ for $f(x) = x^2$ on $[0, 2]$.
- $\Delta x = 0.5$; midpoints $0.25, 0.75, 1.25, 1.75$.
- Heights: $0.0625, 0.5625, 1.5625, 3.0625$, which add to $5.25$.
- $M_4 = 0.5 \cdot 5.25 = 2.625$.
Respuesta: $M_4 = 2.625$, much closer to $\frac{8}{3} \approx 2.667$ than $L_4$ or $R_4$.
Ejemplo resuelto: A sum in sigma notation
Evaluate $\displaystyle\sum_{i=1}^{4}\left(1 + \frac{i}{2}\right)^{2}\cdot\frac{1}{2}$, and name the integral it approximates.
- It is a right sum with $\Delta x = \frac{1}{2}$ for $f(x) = x^2$, starting at $a = 1$ (sample points $1.5, 2, 2.5, 3$), so $b = 3$.
- Terms: $2.25 + 4 + 6.25 + 9 = 21.5$.
- Times $\frac{1}{2}$: $10.75$.
Respuesta: $10.75$, an estimate of the area under $x^2$ from 1 to 3 (exactly $\frac{26}{3} \approx 8.67$; the right sum overshoots).
Ejemplo resuelto: Taking the limit
For $f(x) = x$ on $[0, 1]$, the right sum is $R_n = \frac{1}{n}\sum_{i=1}^{n}\frac{i}{n}$. Find $\lim_{n \to \infty} R_n$.
- Use $\sum_{i=1}^{n} i = \frac{n(n+1)}{2}$: $R_n = \frac{1}{n^2}\cdot\frac{n(n+1)}{2} = \frac{n + 1}{2n}$.
- As $n \to \infty$, $\frac{n+1}{2n} \to \frac{1}{2}$.
Respuesta: $\frac{1}{2}$, the area of the triangle under $y = x$, exactly as geometry says.
Error común
Using the left or right endpoints when the midpoint was asked, or using midpoints of the wrong strips. The midpoint of $[x_{i-1}, x_i]$ is $x_{i-1} + \frac{\Delta x}{2}$. Check: the first midpoint should be half a strip to the right of $a$.