What Makes a Sum a Left-Endpoint Sum?
A Riemann sum estimates accumulated change over an interval by adding the areas of rectangles. To construct one, divide the interval into subintervals, choose a point in each subinterval, and use the function value at that point as the rectangle’s height. A left-endpoint sum uses the left endpoint of every subinterval as its sample point.
The endpoint choice matters because it determines the height of each rectangle. The width comes from the partition; the height comes from the function value at the selected point. Their product is the rectangle’s signed area.
Here \(L_n\) denotes the sum, \(f(x_{i-1})\) is the height of the rectangle on subinterval \(i\), and \(\Delta x\) is its width. If a function’s values are negative, its corresponding rectangle contributes a negative signed area. If the values are positive, those rectangles contribute positive signed area.
Finding the Left Endpoints
For an equal-width partition, the endpoints are \(a, a+\Delta x, a+2\Delta x,\ldots,b\). The first subinterval begins at \(a\), so its left endpoint is \(a\). The next begins at \(a+\Delta x\), and each following left endpoint advances by one width.
There are \(n\) rectangles but only \(n\) left endpoints used as sample points: the starting point of each subinterval. The rightmost partition point \(b\) is not used as a left endpoint. This is why the index \(i-1\) appears in the formula when the rectangles are numbered \(1\) through \(n\).
You can also number the subintervals starting at zero. If \(k=0,1,\ldots,n-1\), the left endpoints are \(a+k\Delta x\), so the same sum can be written as:
The two forms list the same sample points. The first starts the subinterval index at \(1\), while the second starts it at \(0\). In either form, check that exactly \(n\) function values are included.
Constructing a Sum from a Formula
When the function is given by a formula, first calculate the width and list the left endpoints. Then evaluate the function at those points, multiply each value by the width, and add the products. Keeping the endpoints and function values separate helps prevent indexing mistakes.
Worked Example: A Quadratic on Four Equal Subintervals
Find the left-endpoint sum for \(f(x)=x^2+2\) on \([1,5]\), using \(n=4\) equal subintervals.
Step 1: Find the width.
Step 2: List the left endpoints. They are \(1,2,3,4\). The endpoint \(5\) is the right endpoint of the final subinterval, not a left endpoint.
Step 3: Evaluate the function at each left endpoint.
Step 4: Multiply each height by the width and add.
Check: There are four rectangles, matching \(n=4\). Their widths are all \(1\), and their heights are \(3,6,11,18\), the function values at the four left endpoints. Thus the sum is \(38\).
Why the Formula Works
The formula follows directly from the definition of the rectangles. In an equal-width partition, the \(i\)th subinterval begins at \(x_{i-1}=a+(i-1)\Delta x\). A left-endpoint rectangle uses that point as its sample point, so its height is \(f(x_{i-1})\). Its width is \(\Delta x\). The signed area of that rectangle is therefore height times width, \(f(x_{i-1})\Delta x\).
Adding the signed areas of all \(n\) rectangles gives the sum. Substituting the expression for the left endpoint into each height gives the formula below. This derivation also explains why the left endpoint is indexed \(i-1\), even though the rectangle is numbered \(i\).
The \(i\)th subinterval starts at \(x_{i-1}=a+(i-1)\Delta x\).
Using the left endpoint as the sample point gives height \(f(x_{i-1})\).
Multiplying height by width gives \(f(x_{i-1})\Delta x\).
Summing for \(i=1\) through \(n\) produces \(L_n=\sum_{i=1}^{n} f(x_{i-1})\Delta x\).
This argument assumes that the function is defined at each left endpoint used. The result is a finite sum of rectangle areas; it is not necessarily the exact accumulated value of the function across the whole interval.
Using Function Values from a Table
A formula is not required if the needed function values are supplied. For an equal-width partition, identify the left endpoints from the interval and the number of subintervals. Then use the table entries at those endpoints as the rectangle heights. Do not include a value at \(b\) unless it is also the left endpoint of a subinterval, which it is not for this partition.
Worked Example: Selecting Heights from a Table
Suppose the interval is \([0,8]\), divided into \(n=4\) equal subintervals, and the function values are given below. Find the left-endpoint sum.
| \(x\) | 0 | 2 | 4 | 6 | 8 |
|---|---|---|---|---|---|
| \(f(x)\) | 5 | 4 | 3 | 2 | 1 |
Step 1: Find the width.
Step 2: Identify the left endpoints. The partition points are \(0,2,4,6,8\), so the four left endpoints are \(0,2,4,6\).
Step 3: Read the corresponding heights from the table. These are \(f(0)=5\), \(f(2)=4\), \(f(4)=3\), and \(f(6)=2\). The table’s value \(f(8)=1\) is not used.
Step 4: Multiply and add.
Check: The four selected heights match the four left endpoints, and every rectangle has width \(2\). The value \(28\) is the sum of those four signed rectangle areas.
A Sum Written Directly with Sigma Notation
The sigma form is compact, but you should be able to expand it into its individual terms. For an equal partition, writing out the left endpoints before substituting function values is a useful way to confirm that the indexing is correct.
Worked Example: Expanding a Left-Endpoint Sum
Let \(f(x)=2x+1\) on \([1,5]\), with \(n=4\). Write and evaluate the left-endpoint sum.
The width is:
Using \(x_{i-1}=1+(i-1)(1)\), the left endpoints for \(i=1,2,3,4\) are \(1,2,3,4\). The sum in sigma notation is:
Expanding and evaluating each term gives:
Check: The four heights are \(3,5,7,9\), and each width is \(1\), so the total rectangle area is \(24\), as calculated.
Common Mistake: Using the Wrong Endpoint
A frequent error is to use \(a+i\Delta x\) while numbering subintervals from \(i=1\) to \(n\). That expression gives the point one full subinterval to the right of the left endpoint. For the first subinterval, it gives \(a+\Delta x\) instead of \(a\). The left endpoint formula with this indexing is \(a+(i-1)\Delta x\).
Worked Example: Correcting an Indexing Error
For \(f(x)=x\) on \([0,4]\) with \(n=4\), the width is \(\Delta x=1\). A mistaken calculation uses \(f(1),f(2),f(3),f(4)\), producing:
This is not the left-endpoint sum because the first left endpoint is \(0\), not \(1\), and \(4\) is not a left endpoint in this partition. The correct left endpoints are \(0,1,2,3\), so:
Check: The corrected calculation uses four heights, one at the start of each of the four subintervals. Substituting those endpoints into \(f(x)=x\) gives \(0,1,2,3\), whose sum is \(6\).
Why Left-Endpoint Sums Are Useful
A left-endpoint sum turns information about a function into a finite calculation. It can be formed from an explicit formula or from listed function values, provided the needed values at the left endpoints are available. Each term has a clear interpretation: one rectangle’s signed area.
The sum is an approximation to accumulated change over the interval. In general, a rectangle’s height may differ from the function’s values elsewhere in its subinterval, so adding the rectangle areas does not guarantee the exact integral. The method’s precision depends on the function and the partition; the construction itself is determined by the chosen left endpoints and widths.
Check Your Understanding
For each question, identify the left endpoints before calculating the sum.
- What is the common width when \([2,10]\) is divided into \(4\) equal subintervals?
- Write the left endpoints for a partition of \([0,6]\) into \(3\) equal subintervals.
- For \(f(x)=x+2\) on \([0,3]\) with \(n=3\), what heights are used in the left-endpoint sum?
- A table gives function values at \(x=1,3,5,7\). If these are the partition points of \([1,7]\) with \(3\) equal subintervals, which table entries provide the left-endpoint heights?
- When subintervals are numbered \(i=1,\ldots,n\), why does the left endpoint have the form \(a+(i-1)\Delta x\) rather than \(a+i\Delta x\)?
