Guides / Integrals

Riemann sums: left, right, midpoint, and trapezoid, and which way each one misses

A Riemann sum estimates the area under a curve with thin rectangles. It is how a definite integral is defined, and it is how you estimate one when all you have is a table of measurements. The arithmetic is short. The mistakes are in counting: how many heights go in, where the points are, and how much weight the ends get.

Why it works

Each rectangle has width Δx and a height taken from the curve somewhere in its strip, so its area is close to the area under the curve over that strip. Adding them up gives an estimate of the whole area, and as the strips get thinner the gap at the top of each one shrinks. The definite integral is defined as the limit of these sums as n → ∞, and the fundamental theorem of calculus is what lets you skip the limit when you know an antiderivative.

Which way the sum misses depends on the shape of the curve. If f is increasing, the left edge of every strip is its lowest point, so left rectangles sit under the curve. If f is concave up, the chord across a strip lies above the curve, so the trapezoid rule overestimates, and the midpoint rectangle underestimates.

Example 1: left and right sums

Estimate ∫₀³ (x² + 1) dx with 3 strips, using left and right sums.

1. Δx = 3 − 03 = 1; points 0, 1, 2, 3; heights 1, 2, 5, 10Three strips, four points.
2. L₃ = 1 · (1 + 2 + 5) = 8The first three heights, one for the left edge of each strip.
3. R₃ = 1 · (2 + 5 + 10) = 17The last three heights.
4. 8 < ∫₀³ (x² + 1) dx = 12 < 17x² + 1 is increasing on [0, 3], so the left sum is too small and the right sum too big. The exact value is [x³/3 + x] from 0 to 3, which is 9 + 3.

Example 2: midpoint and trapezoid

The same integral with 3 strips, using the midpoint sum and the trapezoid rule.

1. Midpoints 0.5, 1.5, 2.5; heights 1.25, 3.25, 7.25Halfway along each strip.
2. M₃ = 1 · (1.25 + 3.25 + 7.25) = 11.750.25 too small: the curve is concave up.
3. T₃ = 12(1 + 2 · 2 + 2 · 5 + 10) = 12.5Ends once, inside points twice. It is also (L₃ + R₃)/2 = (8 + 17)/2.
4. M₃ = 11.75 (error 0.25), T₃ = 12.5 (error −0.5)Both are far closer than L₃ or R₃, and they miss in opposite directions, with the trapezoid error exactly twice the midpoint error, as it always is for a quadratic.

Example 3: the limit of right sums

Evaluate ∫₁³ 2x dx as a limit of right sums.

1. Δx = 2n, x_i = 1 + 2inx_i = a + iΔx, and here a = 1, not 0.
2. R_n = Σ_i=1ⁿ 2(1 + 2in) · 2n = 4n · n + 8n² · n(n + 1)2Split the sum, and use 1 + 2 + … + n = n(n + 1)/2.
3. = 4 + 4(n + 1)n = 8 + 4nSimplify before taking the limit.
4. lim_n→∞ R_n = 8Check: [x²] from 1 to 3 is 9 − 1 = 8. Every right sum is 4/n too big, because 2x is increasing.

Example 4: a table of data

Water flows into a tank at 10, 14, 16 and 15 liters per minute, measured at t = 0, 2, 4, 6 minutes. Estimate how much flowed in over the 6 minutes.

1. Δt = 2; L₃ = 2(10 + 14 + 16) = 80, R₃ = 2(14 + 16 + 15) = 90The amount is the integral of the rate, and there are three 2-minute strips.
2. T₃ = 22(10 + 2 · 14 + 2 · 16 + 15) = 85The trapezoid rule is usually the best of the three for data like this.
3. about 85 litersThe rate goes up and then down, so you can't tell from the shape whether the left or the right sum is the low one. Over or under is decided by the function, not by the name of the sum.

Common mistakes, and the exact line they happen on

1. Using every point in a left sum

The left sum from Example 1.

1. Δx = 1, heights 1, 2, 5, 10Right so far.
2. L₃ = 1 · (1 + 2 + 5 + 10) = 18This is the mistake. That is four rectangles on three strips: the last one sits over [3, 4], outside the interval. It's even bigger than the right sum, 17, which is a sign something is off. A left sum uses the first n heights and gives 8.

2. Writing x_i without the starting point

The limit from Example 3.

1. Δx = 2nRight so far.
2. x_i = 2in, so R_n = Σ 2 · 2in · 2n = 4 + 4n → 4This is the mistake. These points run from 0 to 2, so this is the integral over [0, 2], not [1, 3]. The points start at a: x_i = 1 + 2i/n, and the limit is 8.

3. Giving the end heights full weight in the trapezoid rule

The tank from Example 4.

1. Rates 10, 14, 16, 15 every 2 minutesRight so far.
2. T₃ = 2(10 + 14 + 16 + 15) = 110This is the mistake. That is four full rectangles over three strips. Each end reading belongs to one trapezoid and each inside reading to two, and each trapezoid takes half of each of its two heights: (2/2)(10 + 28 + 32 + 15) = 85.

Practice

1. Find L₄ and R₄ for f(x) = x on [0, 4].

   Show answer

   Δx = 1: L₄ = 0 + 1 + 2 + 3 = 6 and R₄ = 1 + 2 + 3 + 4 = 10. The exact area is 8, and here the trapezoid rule, (6 + 10)/2, gets it exactly.

2. Find M₂ for x³ on [0, 2].

   Show answer

   Midpoints 0.5 and 1.5, Δx = 1: 0.125 + 3.375 = 3.5. The exact value is 4.

3. Estimate ∫₁³ dxx with the trapezoid rule and 2 strips.

   Show answer

   12(1 + 2 · 12 + 13) = 76 ≈ 1.167, a little above ln 3 ≈ 1.099, because 1/x is concave up.

4. Evaluate lim_n→∞ Σ_i=1ⁿ 1n(in)³.

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   It is a right sum for x³ on [0, 1], so the limit is ∫₀¹ x³ dx = 14.

5. Write ∫₂⁵ x² dx as a limit of right sums, and evaluate it.

   Show answer

   Δx = 3/n, x_i = 2 + 3i/n, and R_n = Σ (2 + 3i/n)² · 3/n = 39 + 632n + 92n² → 39, which matches (125 − 8)/3.