From an Equal Partition to a Right Sum
A right-endpoint sum estimates accumulation across an interval by dividing it into subintervals and using the function value at the right end of each one. When all subintervals have the same width, the sample locations follow a regular pattern. Deriving that pattern carefully makes the formula easier to use and helps prevent a common indexing error: including the left endpoint of the whole interval instead of its right endpoint.
Let the interval be \([a,b]\), where \(a<b\), and divide it into \(n\) equal subintervals. Here \(n\) is a positive integer and counts the subintervals, not the partition points. The right-endpoint sum uses one sample from each subinterval: specifically, its right endpoint.
The formula contains two related choices: the locations \(x_j\), which tell us where to evaluate \(f\), and the indices \(j=1,\ldots,n\), which tell us which subintervals to include. We will derive both from the partition.
Deriving the Partition Locations
Start at \(a\). The first partition point is one width to the right, the next is two widths to the right, and so on. After \(j\) widths, the location is \(a+j\Delta x\). Thus the partition points are
At \(j=0\), this gives \(x_0=a\). At \(j=n\), it gives \(x_n=a+n\Delta x=b\), because \(n\Delta x=b-a\). There are \(n+1\) partition points, which form exactly \(n\) adjacent subintervals.
The \(j\)th subinterval lies between \(x_{j-1}\) and \(x_j\). Its right endpoint is therefore \(x_j\), with \(j\) running from \(1\) through \(n\). The right sample locations are \(x_1,x_2,\ldots,x_n\): they begin one width after \(a\) and end at \(b\).
Worked Example: Four Right Samples on an Interval
Divide \([2,10]\) into \(n=4\) equal subintervals. The interval length is \(10-2=8\), so
The partition points are \(x_j=2+2j\), for \(j=0,1,2,3,4\). Substituting each index gives
The four subintervals are \([2,4]\), \([4,6]\), \([6,8]\), and \([8,10]\). Their right endpoints are \(4,6,8,10\). Therefore the right sum is
Check: there are four subintervals and four terms, each multiplied by the common width \(2\). The first sample is the right endpoint of \([2,4]\), and the last sample is the right endpoint of \([8,10]\), as required.
Why the Index Starts at One
The partition-point index starts at zero because \(x_0=a\). But the right-sum index does not start at zero: \(x_0\) is the left endpoint of the first subinterval, not its right endpoint. The first right endpoint is \(x_1\). At the other end, the final right endpoint is \(x_n=b\), so the sum includes index \(n\).
Each rectangle contributes its width times its sampled height. For subinterval \(j\), the width is \(x_j-x_{j-1}=\Delta x\), and the sampled height is \(f(x_j)\). Its contribution is \(f(x_j)\Delta x\). Adding the contributions for \(j=1\) through \(n\) gives the sum.
The total interval length \(b-a\) is divided into \(n\) equal pieces, so each has width \(\Delta x=(b-a)/n\).
Moving \(j\) widths from \(a\) gives \(x_j=a+j\Delta x\), for \(j=0,\ldots,n\).
Subinterval \(j\) runs from \(x_{j-1}\) to \(x_j\), so its right sample is \(x_j\), for \(j=1,\ldots,n\).
Each contribution is \(f(x_j)\Delta x\); summing them gives \(R_n=\sum_{j=1}^{n}f(x_j)\Delta x\).
This derivation relies on an equal partition of an interval with \(a<b\), and on \(f\) being defined at every selected sample location. The formula specifies a sum; it does not require the function to be positive. If a sampled value is negative, its contribution is negative, consistent with signed accumulation.
Right and Left Endpoint Indices
For comparison, the left endpoint of subinterval \(j\) is \(x_{j-1}\). As \(j\) runs from \(1\) to \(n\), these left samples are \(x_0,x_1,\ldots,x_{n-1}\). Replacing \(j-1\) with an index \(i\) gives the left-endpoint formula. The common width and partition points are unchanged; only the selected endpoint, and consequently the index range, changes.
| Sum | Sample locations | Index range |
|---|---|---|
| Left-endpoint sum | \(x_0,x_1,\ldots,x_{n-1}\) | \(i=0,\ldots,n-1\) |
| Right-endpoint sum | \(x_1,x_2,\ldots,x_n\) | \(j=1,\ldots,n\) |
Both formulas have exactly \(n\) terms. The left sum includes \(x_0=a\) and excludes \(x_n=b\); the right sum excludes \(x_0=a\) and includes \(x_n=b\). The symbols \(i\) and \(j\) are just index names. What matters is which partition point each term uses.
Worked Example: Right Samples on a Negative-to-Positive Interval
Divide \([-3,9]\) into \(n=3\) equal subintervals. The interval length is
Use \(x_j=-3+4j\). The partition points are
The subintervals are \([-3,1]\), \([1,5]\), and \([5,9]\). Their right endpoints are \(1,5,9\), so
Check: the final partition point is \(-3+3(4)=9\), the original right endpoint. The three right samples give one term for each of the three subintervals.
Writing the Formula Directly from the Interval
Substituting \(\Delta x=(b-a)/n\) and \(x_j=a+j(b-a)/n\) into the right-sum definition gives a formula using only the interval endpoints, the number of subintervals, and the function:
The factor outside the sum is the width of every rectangle. Inside, the first function input is \(a+(b-a)/n\), and the last is \(a+n(b-a)/n=b\). This form can be useful when the partition points are not written out separately, but it still has the same right-endpoint indexing: \(j=1,\ldots,n\).
Worked Example: Expressing the Sum with a General Index
Divide \([5,17]\) into \(n=4\) equal subintervals. First calculate
The partition points follow \(x_j=5+3j\). For \(j=0,1,2,3,4\), they are \(5,8,11,14,17\). The right samples use \(j=1,2,3,4\), giving \(8,11,14,17\). Thus
The same sum can be written with an index \(k\) starting at zero by setting \(k=j-1\), so that \(j=k+1\). Then \(5+3j=5+3(k+1)=8+3k\), and
Check: the inputs in this last expression are \(8,11,14,17\), exactly the four right endpoints found from the partition. Changing the index name or shifting its range is valid only when the function inputs shift consistently as well.
A Common Indexing Mistake
A tempting error is to use the partition-point range \(0,\ldots,n\) directly in a right sum. That range contains \(n+1\) points, including both ends of the whole interval, whereas a right-endpoint sum needs exactly one sample for each of the \(n\) subintervals.
For example, with four subintervals on \([2,10]\), the incorrect range \(j=0,\ldots,4\) would select \(2,4,6,8,10\): five sample locations for four rectangles. The corrected range \(j=1,\ldots,4\) selects \(4,6,8,10\), giving four samples. In general, counting terms is a quick check: the index range \(1,\ldots,n\) contains \(n\) integers, just as the partition has \(n\) subintervals.
Why the Distinction Matters
A sum is determined not only by the width and the function, but also by where each function value is sampled. The left and right sums share the same partition and the same width, yet they generally use different function values. In an accumulation estimate, those values represent the heights assigned to the rectangles. Using the wrong endpoint changes the estimate, even if the width is correct.
When deriving or checking a right sum, confirm three facts: the common width is \((b-a)/n\); the partition points are \(x_j=a+j\Delta x\); and the right sample for subinterval \(j\) is \(x_j\), with \(j\) from \(1\) through \(n\). These checks keep the interval endpoints, sample locations, and term count consistent.
Check Your Understanding
Use the partition-point notation \(x_j=a+j\Delta x\) to answer each question.
- For an equal partition of \([a,b]\) into \(n\) subintervals, what is the common width?
- What is the right endpoint of subinterval \(j\), and which values of \(j\) are used in the right sum?
- For \([1,13]\) divided into three equal subintervals, find the width and list the right sample locations.
- How do the index ranges for the left-endpoint sum and the right-endpoint sum differ?
- Why does a right-endpoint sum with \(n\) subintervals have \(n\) terms rather than \(n+1\)?
