From an Equal Partition to a Left Sum
When a quantity accumulates across an interval, a finite sum of rectangles can estimate the total. To construct that sum, we need to specify two things: the width of each rectangle and the point at which its height is sampled. In a left-endpoint sum, each height is taken from the left end of its subinterval.
The equal-width case makes the locations systematic. Once the interval and number of subintervals are given, the width and every left-endpoint sample can be written directly in terms of the partition count. This tutorial derives that formula and shows how to use it without shifting an index by mistake.
Partition Points and Rectangle Widths
Let \(a<b\) be the endpoints of an interval, and let \(n\in\mathbb{N}\) be the number of subintervals. The total interval length is \(b-a\). Dividing this length into \(n\) equal pieces gives the common width
Beginning at \(a\), move forward by one width to find the first partition point, by two widths to find the second, and so on. Number the partition points from \(0\) to \(n\), with \(x_0=a\) and \(x_n=b\). Then
There are \(n+1\) partition points because \(n\) subintervals have two endpoints each, with neighboring subintervals sharing an endpoint. The \(i\)th subinterval is \([x_{i-1},x_i]\), for \(i=1,\ldots,n\), and its width is \(x_i-x_{i-1}=\Delta x\).
The index \(i\) labels the subinterval, not its left endpoint number. Subinterval \(i\) starts at partition point \(x_{i-1}\), so the sample used for its rectangle is \(f(x_{i-1})\). Equivalently, if the left endpoints are indexed directly by \(j=0,\ldots,n-1\), the same sum is \(\Delta x\sum_{j=0}^{n-1} f(x_j)\).
Why the Formula Has These Indices
The sum formula follows directly from the construction of each rectangle. For subinterval \(i\), its left edge is \(x_{i-1}\), its height is \(f(x_{i-1})\), and its width is \(\Delta x\). Therefore its signed contribution is \(f(x_{i-1})\Delta x\). Adding the contributions from all \(n\) subintervals gives the stated formula.
Set \(\Delta x=(b-a)/n\), the common width of the \(n\) subintervals.
Starting from \(x_0=a\), use \(x_i=a+i\Delta x\) for \(i=0,\ldots,n\).
Subinterval \(i\) runs from \(x_{i-1}\) to \(x_i\), so its left sample is \(x_{i-1}=a+(i-1)\Delta x\).
Multiply each sampled value by \(\Delta x\), then sum for \(i=1,\ldots,n\).
This argument also explains why a sum of this form represents a finite accumulation estimate: each term is a sampled rate or height multiplied by the width over which it is used. If \(f\) is a rate, then rate times interval width has units of accumulated quantity. If some sampled values are negative, their rectangle contributions are negative as well; the sum is signed rather than a sum of unsigned geometric areas.
Worked Examples
Worked Example: Four Equal Subintervals on [2, 10]
Let \(f(x)=3x+2\). Construct and evaluate the left-endpoint sum using \(n=4\) equal subintervals.
First find the interval length and rectangle width:
The partition points are \(x_i=2+2i\), for \(i=0,\ldots,4\). Substitution gives
For subinterval \(i\), the left sample is \(x_{i-1}\). Thus the four sample locations, in order, are \(x_0=2\), \(x_1=4\), \(x_2=6\), and \(x_3=8\). Evaluate the function at each:
Multiply each height by the width \(2\) and add:
Check: the four rectangle contributions are \(2(8)=16\), \(2(14)=28\), \(2(20)=40\), and \(2(26)=52\). Their sum is \(16+28+40+52=136\), agreeing with the formula. If \(f\) is measured in units per unit of \(x\), then \(L_4\) is measured in units of the accumulated quantity.
Worked Example: Three Equal Subintervals on [0, 6]
Let \(f(x)=x^2+1\) and divide \([0,6]\) into \(n=3\) equal subintervals. The width is
The partition-point formula is \(x_i=0+2i\). Hence \(x_0=0\), \(x_1=2\), \(x_2=4\), and \(x_3=6\). The left sample locations are \(x_0,x_1,x_2\), namely \(0,2,4\). The corresponding heights are
Therefore the left sum is
Check by listing the contributions: \(1\cdot2=2\), \(5\cdot2=10\), and \(17\cdot2=34\). Their total is \(2+10+34=46\). Each of the three contributions uses width \(2\), and no sample is taken at the final point \(x_3=6\).
Worked Example: A Decreasing Function on [-1, 5]
Let \(f(x)=4-x\) and take \(n=3\) equal subintervals of \([-1,5]\). The interval length is \(5-(-1)=6\), so
The partition points are \(x_i=-1+2i\), giving
The left endpoints are \(-1,1,3\). Evaluate \(f\) at them:
Each rectangle has width \(2\), so
Check: the individual contributions are \(5\cdot2=10\), \(3\cdot2=6\), and \(1\cdot2=2\), and \(10+6+2=18\). The sample locations and widths are consistent with the interval: the first rectangle uses the value at \(-1\) over \([-1,1]\), the next uses the value at \(1\) over \([1,3]\), and the last uses the value at \(3\) over \([3,5]\).
A Common Indexing Mistake
A frequent error is to use \(x_i\) as the sample for subinterval \(i\). For example, on \([2,10]\) with \(n=4\), the width is \(2\) and the partition points are \(2,4,6,8,10\). Writing \(2[f(4)+f(6)+f(8)+f(10)]\) uses the right endpoint of each subinterval, not the left. With \(f(x)=3x+2\), that incorrect left-sum calculation would be
The correction is to match subinterval \(i\) with its left endpoint \(x_{i-1}\): the first subinterval starts at \(x_0\), not \(x_1\). Thus the correct calculation is \(2[f(2)+f(4)+f(6)+f(8)]=136\), as checked above. A reliable check is to confirm that the left-sum sample list starts at \(a=x_0\) and stops at \(x_{n-1}\); it does not include the final partition point \(x_n=b\).
What the Formula Keeps Track Of
The expression for \(L_n\) contains three linked choices: the interval endpoints \(a\) and \(b\), the positive integer \(n\), and the function values at the prescribed left samples. Changing \(n\) changes both the width and the sample locations. The sum always has \(n\) terms, each multiplied by the same width \((b-a)/n\).
Before evaluating a left sum, write out the width, partition points, and left samples. This makes the formula’s index visible and gives a direct way to catch errors: adjacent subintervals should share partition points, the samples should start at \(a\), and there should be exactly \(n\) rectangle contributions.
Check Your Understanding
For each question, identify the width and left sample locations before writing or evaluating a sum.
- Divide \([1,9]\) into \(4\) equal subintervals. What is \(\Delta x\), and what are all the partition points?
- For an equal partition of \([a,b]\) into \(n\) pieces, what is the left sample location for subinterval \(i\), where \(1\le i\le n\)?
- Write the equal-width left-endpoint sum for \(f(x)=2x+1\) on \([0,4]\) with \(n=2\), without evaluating it.
- For a left-endpoint sum with \(n\) subintervals, should the final sample be \(x_{n-1}\) or \(x_n\)? Explain briefly.
- If \(f(x)\) has units of quantity per unit of \(x\), what units does a term \(f(x_{i-1})\Delta x\) have?
