From Rectangle Sums to a Definite Integral
A Riemann sum combines the heights of sample rectangles with their widths. With a coarse partition, the rectangles may give only a rough estimate of accumulated change. As the partition becomes finer, the widths shrink, and the sums can approach a fixed value.
The definite integral formalizes this limiting idea. Its value is not one particular rectangle sum: it is the number that the sums approach as the largest subinterval width tends to zero. The partition, its sample points, the widths, and the limit each have a distinct role.
Partitions, Tags, and Riemann Sums
Let \(f\) be a bounded function on a closed interval \([a,b]\), where \(a<b\). A partition of the interval is a finite, increasing list of endpoints
The partition divides \([a,b]\) into \(n\) subintervals. The \(i\)th subinterval is \([x_{i-1},x_i]\), and its width is \(\Delta x_i=x_i-x_{i-1}\). A tag, or sample point, \(\xi_i\) is chosen from that subinterval. The corresponding rectangle has width \(\Delta x_i\) and signed height \(f(\xi_i)\).
Each product \(f(\xi_i)\Delta x_i\) is one rectangle’s signed contribution. The sum adds those contributions across the interval. If \(f(\xi_i)\) is negative, that rectangle contributes a negative amount; the sum therefore represents signed accumulation rather than necessarily ordinary geometric area.
The Definite Integral as a Limit
The phrase “the sum approaches a limit as the partition gets finer” needs a precise meaning. It is not enough that the number of subintervals grows: a partition with many subintervals might still contain one wide subinterval. Instead, we require the mesh, the widest subinterval, to approach zero.
This definition says that all sufficiently fine tagged sums must be close to the same number \(I\). The tags may be chosen anywhere within their respective subintervals. The next tutorial examines why sample-point choice does not change the limit under the relevant conditions; here, the definition states what it means for a common limit to exist.
For the functions most often used in AP Calculus, a standard existence theorem applies:
Continuity on the whole closed interval is the condition in this theorem. Once integrability is established, the integral notation names the limiting value; it does not name an individual sum or require equal-width subintervals.
What Each Part of the Limit Means
| Part | Meaning |
|---|---|
| \([a,b]\) | The interval over which the accumulation is measured |
| \(x_0,\ldots,x_n\) | The partition points that divide the interval |
| \(\Delta x_i=x_i-x_{i-1}\) | The width of subinterval \(i\) |
| \(\xi_i\) | The sample point used to select the height on subinterval \(i\) |
| \(f(\xi_i)\Delta x_i\) | The signed contribution of one rectangle |
| \(\lVert P\rVert\to 0\) | The widest subinterval shrinks toward zero |
| \(\int_a^b f(x)\,dx\) | The common limiting value of sufficiently fine tagged sums, when it exists |
The variable \(x\) in the integral is a dummy variable: it indicates the input of \(f\) across the interval. The differential \(dx\) is part of the integral notation and is consistent with adding contributions over small widths. In an actual Riemann sum, those widths appear explicitly as \(\Delta x_i\).
Worked Example: A Tagged Sum on an Unequal Partition
Worked Example: A Tagged Sum on an Unequal Partition
Let \(f(x)=2x+1\) on \([0,3]\). Use the partition \(0,1,3\), with tags \(\xi_1=0.5\) and \(\xi_2=2\). Find the tagged Riemann sum and the mesh.
The subintervals are \([0,1]\) and \([1,3]\), so their widths are
Evaluate the function at each tag:
Multiply each sampled height by its matching width, then add:
The mesh is the larger of the two widths:
The sum is \(12\), but it is only the approximation for this particular partition and these tags. One finite sum by itself does not establish the limiting value of the definite integral. The calculation is checked by its two rectangle contributions, \(2\) and \(10\), whose total is \(12\).
Worked Example: A Right-Endpoint Limit
Worked Example: A Right-Endpoint Limit
Use \(n\) equal subintervals on \([0,1]\), and use each subinterval’s right endpoint as its tag for \(f(x)=x\). Find the limit of the resulting sums.
Each width is \(\Delta x=1/n\). The \(k\)th right endpoint is \(k/n\), for \(k=1,\ldots,n\). Thus the sum is
Use the finite-sum formula \(\sum_{k=1}^{n}k=n(n+1)/2\):
As \(n\) grows, the mesh \(1/n\) approaches zero, and \(1/(2n)\) approaches zero. Therefore, this sequence of sums approaches \(1/2\), giving
To check the limiting claim directly, let \(\varepsilon>0\). Choose a positive integer \(N\) with \(N>1/(2\varepsilon)\). If \(n\geq N\), then
So the sums do approach \(1/2\). This computation uses right endpoints for this particular sequence; it does not, on its own, prove that every possible choice of tags has the same limit.
Why the Definition Works
The sum is built to measure accumulated contributions. On a small subinterval, the function’s sampled value supplies a representative height, and the interval’s width supplies the horizontal amount. Their product approximates the contribution over that piece. Adding the products accounts for the entire interval.
The mesh condition controls the size of the pieces. Requiring the sums to approach one common value prevents the answer from depending on a particular coarse partition. The epsilon-delta statement makes this precise: however small a tolerance \(\varepsilon\) is requested, sufficiently small mesh guarantees that every eligible tagged sum is within that tolerance of the integral.
Choose partition points from \(a\) to \(b\), creating subintervals with widths \(\Delta x_i\).
Select a tag \(\xi_i\) in each subinterval and calculate \(f(\xi_i)\).
Multiply each sampled height by its width and add the signed contributions.
Make the widest subinterval approach zero and determine whether the sums approach a common value.
Worked Example: Reading a Limit Expression
Worked Example: Reading a Limit Expression
Interpret the expression below as a Riemann-sum limit. Identify the interval, subinterval width, tags, and function.
The repeated width factor is \(2/n\). With \(n\) equal subintervals, the total interval length is therefore \(n(2/n)=2\). Starting at \(0\), the interval is \([0,2]\), and its \(k\)th right endpoint is \(x_k=2k/n\).
The sampled height is \(3+2k/n=3+x_k\), so the function is \(f(x)=3+x\). The expression is the right-endpoint Riemann-sum limit for that function on \([0,2]\):
The interpretation is checked by matching every part: the \(n\) widths are \(2/n\), the right-endpoint tags are \(2k/n\), and evaluating \(3+x\) at those tags gives \(3+2k/n\). The limit notation expresses the integral; no single finite value of \(n\) is the integral itself.
Common Pitfall: Confusing a Sum with Its Limit
A common error is to treat one Riemann sum as if it were automatically the exact definite integral. For example, in the first worked example, the sum \(12\) came from a partition whose mesh was \(2\). That calculation gives one approximation, not a conclusion about the limiting value as the mesh approaches zero.
Another error is to omit the width. For the right-endpoint sum for \(f(x)=x\) on \([0,1]\), a mistaken calculation might use \(\sum_{k=1}^n k/n\) and claim the limit is the integral. But each rectangle’s width \(1/n\) must also be included. The correct sum is \(\sum_{k=1}^n (k/n)(1/n)\); without the second factor, the calculation does not represent rectangle areas and does not approach the stated integral.
Check Your Understanding
Use the definitions and interpretations in this tutorial to answer the following.
- For a partition with widths \(0.2\), \(0.5\), and \(0.3\), what is its mesh?
- What does the tag \(\xi_i\) determine in the \(i\)th contribution \(f(\xi_i)\Delta x_i\)?
- Why is one finite tagged Riemann sum not, by itself, the definite integral?
- In the expression \(\sum_{k=1}^{n} f(\xi_k)\Delta x_k\), what must happen to the mesh for the limit to define an integral?
- For a function continuous on a closed interval, what standard result guarantees that its Riemann integral exists?
