Write a Riemann Sum in Sigma Notation

To write a Riemann sum in sigma notation, you must pack three numbers into one compact expression: the subinterval width Δx, the location of each sample point xᵢ*, and the number of terms n. The formula for a uniform partition on [a,b] is Σ_{i=1}^{n} f(xᵢ*) Δx, where Δx = (b-a)/n and xᵢ* depends on whether you are using a left, right, or midpoint rule. The single most useful fact: the index i runs from 1 to n, not from 0 to n-1, and the sample point formula shifts by one index between left and right sums.

Take f(x) = x² on [0,2] with n = 4 uniform subintervals. Compute Δx = (2-0)/4 = 0.5. For a left Riemann sum, the sample point in the i-th subinterval is xᵢ* = 0 + (i-1)×0.5. So the summand is f(0 + (i-1)×0.5)×0.5, and the full sigma expression is Σ_{i=1}^{4} ( (i-1)×0.5 )² × 0.5. Evaluate: i=1 gives 0²×0.5 = 0; i=2 gives 0.5²×0.5 = 0.125; i=3 gives 1²×0.5 = 0.5; i=4 gives 1.5²×0.5 = 1.125. Sum = 1.75.

For a right sum, xᵢ* = 0 + i×0.5, so the sigma notation becomes Σ_{i=1}^{4} ( i×0.5 )² × 0.5. That yields 0.5²×0.5 + 1²×0.5 + 1.5²×0.5 + 2²×0.5 = 0.125 + 0.5 + 1.125 + 2 = 3.875. The same sigma structure, a one-index shift in the sample point, changes the approximation from 1.75 to 3.875. The true area under x² from 0 to 2 is 8/3 ≈ 2.667, so the left sum underestimates and the right sum overestimates because f is increasing.

The three components, width, sample points, and term count, become a single sigma expression through substitution. First, Δx is always (b-a)/n and becomes a constant factor outside f if the partition is uniform. Second, the sample point formula plugs into f. Third, the term count n becomes the upper limit of the summation. For a midpoint sum on the same f(x)=x², [0,2], n=4: xᵢ* = 0 + (i-0.5)×0.5, so the sum is Σ_{i=1}^{4} ( (i-0.5)×0.5 )² × 0.5. Evaluate: i=1 gives 0.25²×0.5 = 0.03125; i=2 gives 0.75²×0.5 = 0.28125; i=3 gives 1.25²×0.5 = 0.78125; i=4 gives 1.75²×0.5 = 1.53125. Sum = 2.625, closer to 2.667 than either endpoint rule.

How the Index Controls the Sample Point

The index i counts subintervals, not endpoints. For left sums, xᵢ* = a + (i-1)Δx; for right sums, xᵢ* = a + iΔx. This is the most common indexing error: using i instead of i-1 for a left sum shifts every rectangle one subinterval to the right, producing a right sum by accident.

What Sigma Notation Does Not Show

The sigma notation Σ_{i=1}^{n} f(xᵢ*)Δx does not encode whether xᵢ* is left, right, or midpoint. The same notation represents all three rules; only the definition of xᵢ* changes. That is why a Riemann sum problem must always state which rule to use. A student who writes Σ_{i=1}^{4} f(xᵢ)Δx without specifying xᵢ has not fully defined the sum.

Non-Uniform Partitions Break the Pattern

If the subinterval widths vary, Δxᵢ cannot be factored out. The sigma notation becomes Σ_{i=1}^{n} f(xᵢ*) Δxᵢ, with Δxᵢ inside the summand. This is rarer in AP problems but appears in college-level definitions of the integral. The width is no longer a single constant; each term has its own width. For a table of values with non-uniform intervals, you must list each Δxᵢ explicitly.

To build the compact summation, follow three steps in order. Step one: compute Δx = (b-a)/n. Step two: write the sample point formula for the chosen rule. Step three: assemble the summand as f(sample point) × Δx. The index i runs 1 to n. For a left sum on [0,2] with n=4, this gives Σ_{i=1}^{4} f(a + (i-1)Δx)Δx. That is the entire compact expression. The width Δx, the sample point pattern, and the count n are all present.

When the calculation becomes tedious for large n, a summation notation calculator can handle the arithmetic, but the setup, the sigma expression itself, is what carries the meaning. The calculator evaluates; the sigma notation communicates.

An AP free-response question gives a velocity table with times 0, 3, 7, 12 minutes and velocities 30, 45, 50, 40 mph. The subinterval widths are 3, 4, and 5 minutes, not equal. A student who writes Σ v(tᵢ) × (12/3) has used the wrong width for every term. The correct sigma notation is Σ_{i=1}^{3} v(tᵢ*) Δtᵢ, with Δt₁=3, Δt₂=4, Δt₃=5. For a left sum, tᵢ* = tᵢ₋₁, so the sum is v(0)×3 + v(3)×4 + v(7)×5 = 30×3 + 45×4 + 50×5 = 90 + 180 + 250 = 520 miles. A right sum uses tᵢ* = tᵢ, giving v(3)×3 + v(7)×4 + v(12)×5 = 45×3 + 50×4 + 40×5 = 135 + 200 + 200 = 535 miles. The constant-Δx shortcut fails completely.

AP Calculus AB and BC students need to write Riemann sums in sigma notation for free-response questions where the setup earns more points than the arithmetic. First-year college calculus students need the same skill for limit definitions of the integral. Physics and engineering students use it to translate velocity tables into distance approximations. Skip this if you cannot yet evaluate a function at a point, or if you need an exact antiderivative, a Riemann sum is always an approximation, not a closed form. The single thing that most often goes wrong: treating the finite sum as the exact area instead of as the finite step before the limit.