How do you find the area of a region with curved boundaries? You can't use a simple formula like length times width. The brilliant idea behind calculus is to approximate the curved region with rectangles, then make the approximation better and better.
A Riemann sum is exactly this: the total area of rectangles that approximate the region under a curve. By increasing the number of rectangles, the approximation improves. In the limit, as we use infinitely many infinitely thin rectangles, we get the exact area.
The key insight: A Riemann sum isn't just a numerical approximation—it's a fundamental way of thinking about accumulation. Every integral you'll ever compute is, at its heart, a limit of Riemann sums.
| Property | Value |
|---|---|
| Concept | Integration |
| Course | MATH161 |
| Section | Stewart 4.2 |
| Difficulty | Intermediate |
| Time | ~20 minutes |
To approximate $\int_a^b f(x)\,dx$ using $n$ rectangles:
Step 1: Divide the interval $[a,b]$ into $n$ equal subintervals.
The width of each subinterval is: $$\Delta x = \frac{b - a}{n}$$
The endpoints are: $$x_0 = a, \quad x_1 = a + \Delta x, \quad x_2 = a + 2\Delta x, \quad \ldots, \quad x_n = b$$
In general: $x_i = a + i \cdot \Delta x$
Step 2: Choose sample points in each subinterval $[x_{i-1}, x_i]$.
Step 3: Form the sum of rectangle areas: $$\sum_{i=1}^{n} f(x_i^*) \cdot \Delta x$$
where $x_i^*$ is the sample point in the $i$th subinterval.
| Type | Sample Point | Formula for $x_i^*$ |
|---|---|---|
| Right Riemann Sum ($R_n$) | Right endpoint | $x_i^* = x_i = a + i\Delta x$ |
| Left Riemann Sum ($L_n$) | Left endpoint | $x_i^* = x_{i-1} = a + (i-1)\Delta x$ |
| Midpoint Sum ($M_n$) | Midpoint | $x_i^* = \bar{x}_i = \frac{x_{i-1} + x_i}{2}$ |
For $f(x) = x^2$ on $[0, 2]$ with $n = 4$ (right endpoints):
y
|
4 + ■
| ■■■■■
3 + ■
| ■■■■■
2 + ■
| ■■■■■
1 + ■■■■■
|■■■■■
0 +----+----+----+----+→ x
0 0.5 1 1.5 2
Each rectangle has width $\Delta x = 0.5$ and height $f(x_i)$ where $x_i$ is the right endpoint.
$$R_n = \sum_{i=1}^{n} f(x_i) \Delta x = \sum_{i=1}^{n} f\left(a + i \cdot \frac{b-a}{n}\right) \cdot \frac{b-a}{n}$$
Important observation: For an increasing function:
For a decreasing function, the opposite is true.
Riemann sums model accumulated quantities:
For the Riemann sum used to approximate $\int_1^5 (2x+3)\,dx$ with $n = 8$ subintervals:
Compute the right Riemann sum $R_4$ for $f(x) = x^2 - 1$ on $[0, 2]$ using $n = 4$ subintervals.
For $f(x) = \sqrt{x}$ on $[1, 9]$ with $n = 4$:
The function $g(x) = x - 2$ changes sign on the interval $[0, 4]$.
The Staircase Approximation: Imagine building a staircase that follows a curved hill. Each step (rectangle) has a fixed width but varies in height based on the hill at some point in that interval. A few wide steps give a rough approximation; many narrow steps follow the curve more closely. The Riemann sum is the total vertical material in your staircase. As steps get narrower and more numerous, your staircase becomes indistinguishable from the smooth hill.
Looking back:
Looking ahead:
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|---|---|---|
| Sigma Notation | Skills Index | Definite Integral Definition |
Last updated: 2026-01-22