Why Sigma Notation Is Useful
A Riemann sum adds the areas of a collection of rectangles. Writing every rectangle separately works for a small number of terms, but becomes cumbersome when there are many subintervals. Sigma notation is a compact way to write a finite sum while keeping the pattern of its terms visible.
To interpret a Riemann sum correctly, read more than just the summation symbol. You need to identify the index and its range, the width of each subinterval, the sample point used in each rectangle, and the function value that gives its height.
Reading a Finite Sum
The notation \(\sum_{i=m}^{n} a_i\) means to add the terms \(a_i\) as the index \(i\) takes every integer value from \(m\) through \(n\), including both endpoints. The index is a counter: it labels the terms, and its values tell you which terms to include.
The number of terms is \(n-m+1\), not simply the upper index \(n\), unless the index starts at \(1\). For example, when \(i\) runs from \(2\) to \(5\), there are \(5-2+1=4\) terms.
For an equal-width partition of \([a,b]\) into \(n\) subintervals, the common width is \(\Delta x=(b-a)/n\). If \(x_i^*\) is the chosen sample point in subinterval \(i\), then the sum of the rectangle areas can be written as follows.
Here, \(i\) labels a subinterval and its rectangle; \(x_i^*\) is the sample point for that subinterval; \(f(x_i^*)\) is the rectangle's height, with its sign; and \(\Delta x\) is its width. The product \(f(x_i^*)\Delta x\) is that rectangle's signed area. The sigma symbol instructs us to add one such product for each index value.
Worked Example: Write Rectangle Areas in Sigma Notation
Worked Example: Write Rectangle Areas in Sigma Notation
On \([0,4]\), divide the interval into four equal subintervals. Use the sample points \(0.5,1.5,2.5,3.5\) for the function \(f(x)=2x+1\). Write the rectangle-area sum in sigma notation and evaluate it.
Step 1: Find the common width. The interval length is \(4-0=4\), and there are \(4\) subintervals, so \(\Delta x=4/4=1\).
Step 2: Match each sample point to its function value.
Step 3: Find the rectangle areas. Multiply each height by the common width \(1\). The areas are \(2(1)=2\), \(4(1)=4\), \(6(1)=6\), and \(8(1)=8\). Thus the finite sum is \(2+4+6+8=20\).
Step 4: Express the pattern compactly. The sample point associated with index \(i\) is \(x_i^*=i-\frac12\), for \(i=1,2,3,4\). Substituting into the function gives \(f(x_i^*)=2(i-\frac12)+1=2i\). Therefore, the sum is
Check: Expanding the sigma expression gives the same four rectangle areas, \(2,4,6,8\), and the total is \(20\).
How to Unpack a Riemann Sum
To translate from sigma notation to individual rectangle areas, substitute each permitted index value into the full term. Do not change the width or omit a factor while expanding. Then identify which part of each term is the width and which part is the function value.
List every integer from the lower index to the upper index, including both ends. This tells you how many rectangles are represented.
Find the factor \(\Delta x\), and identify the function value \(f(x_i^*)\) in the remaining part of each term.
Substitute the index values one at a time. Each term should correspond to one rectangle's signed area.
Confirm that each formula for \(x_i^*\) produces a point in the subinterval labeled by \(i\).
Worked Example: Decode a Sigma Expression
Worked Example: Decode a Sigma Expression
Interpret and evaluate \(\sum_{i=1}^{3}2\big((2i-2)+3\big)\) as a Riemann sum. Identify the width, sample points, and function values.
Step 1: Identify the width and height. The constant factor \(2\) is the common width, so \(\Delta x=2\). The expression \((2i-2)+3\) can be read as \(f(x_i^*)\), with \(x_i^*=2i-2\) and \(f(x)=x+3\).
Step 2: Check the sample points. The index values are \(1,2,3\). Substitution gives \(x_1^*=2(1)-2=0\), \(x_2^*=2(2)-2=2\), and \(x_3^*=2(3)-2=4\). These are sample points in the three width-\(2\) subintervals of \([-1,5]\): \([-1,1]\), \([1,3]\), and \([3,5]\).
Step 3: Find the heights and areas.
Multiplying each height by the width \(2\) gives the areas \(3(2)=6\), \(5(2)=10\), and \(7(2)=14\). The expanded sum is \(6+10+14=30\).
Step 4: Verify by expansion.
Check: There are three index values, three rectangles, and three area terms. Each term uses width \(2\) and the function value at its listed sample point.
Why the Notation Matches the Rectangle Areas
The meaning of sigma notation follows directly from the definition of a finite sum. For each index \(i\), the rectangle has width \(\Delta x\) and signed height \(f(x_i^*)\). Its signed area is their product. Adding those products over all the rectangle labels adds exactly the areas of the rectangles.
For example, if the index runs from \(1\) to \(n\), the definition of summation gives
Every term on the right is the area contribution for one subinterval. There are no missing terms because the index takes every value from \(1\) through \(n\), and no extra terms because it takes no other values. This is why the sigma expression and the finite list of rectangle areas represent the same sum.
Worked Example: Build Sigma Notation from a Finite Sum
Worked Example: Build Sigma Notation from a Finite Sum
On \([0,3]\), use three equal subintervals of width \(1\), sample points \(x_i^*=i-\frac14\), and \(f(x)=2x\). Write and evaluate the Riemann sum in sigma notation.
Step 1: Confirm the width and sample points. The width is \(\Delta x=(3-0)/3=1\). The sample-point formula gives \(x_1^*=\frac34\), \(x_2^*=\frac74\), and \(x_3^*=\frac{11}{4}\). These belong, respectively, to \([0,1]\), \([1,2]\), and \([2,3]\).
Step 2: Find the heights.
Step 3: Write and evaluate the finite sum. Multiplying each height by the width \(1\) gives \(\frac32\), \(\frac72\), and \(\frac{11}{2}\). Their sum is \(\frac{3+7+11}{2}=\frac{21}{2}\).
Check: Substituting \(i=1,2,3\) into \(2(i-\frac14)\) gives \(\frac32,\frac72,\frac{11}{2}\), precisely the three rectangle-area terms because the width is \(1\).
A Common Indexing Mistake
A frequent error is to treat the upper index as the number of terms, regardless of where the index begins. For instance, \(\sum_{i=2}^{5}(i+1)\) does not have five terms. Its index values are \(2,3,4,5\), so it has \(5-2+1=4\) terms.
The correct expansion is
A mistaken expansion that starts at \(i=1\) adds an extra term: \(2+3+4+5+6=20\). That is not the sum written, because the index value \(1\) is outside the stated range. In a Riemann sum, this error would incorrectly include an extra rectangle or attach the wrong function value to a rectangle.
Reading the Sum as a Model of Accumulation
A Riemann sum is useful because it keeps the pieces of an approximation organized. The index tells you which piece you are considering; the sample point specifies where to evaluate the function; the function value supplies the signed height; and the width converts that height into a rectangle's signed area.
When the function value is positive, the rectangle contributes positive signed area. When it is negative, its contribution is negative. The sigma notation adds these contributions, so positive and negative terms can cancel. The result is an accumulated signed-area approximation, not necessarily the sum of the absolute geometric areas.
When translating either direction, use the same checklist: number of terms, width, sample-point rule, function value, and product giving each area. If each expanded term matches one rectangle and the index includes each rectangle exactly once, the notation is consistent.
Check Your Understanding
For each question, identify the terms carefully before evaluating or interpreting the sum.
- How many terms are in \(\sum_{i=3}^{7}a_i\)?
- Expand and evaluate \(\sum_{i=1}^{3}(2i-1)\).
- A Riemann sum is \(\sum_{i=1}^{4} f(x_i^*)\left(\frac12\right)\). What is the width of each rectangle, and how many rectangles are represented?
- For \(f(x)=x+1\), width \(\Delta x=2\), and sample points \(x_i^*=i\) for \(i=1,2,3\), write the sum in sigma notation and expand it.
- Explain why \(\sum_{i=2}^{5} f(x_i^*)\Delta x\) has four terms rather than five.
