Lesson 13: Riemann Sums & the Definite Integral

What You'll Learn

• Build left, right, and midpoint Riemann sums and the trapezoidal approximation from formulas, graphs, and tables — including unequal-width subintervals.
• Decide whether an approximation is an overestimate or underestimate using increasing/decreasing (for left/right sums) and concavity (for the trapezoid).
• Read the definite integral as the limit of Riemann sums and as signed (net) area, evaluate integrals geometrically from a graph, and apply the integral properties.

Content

Rectangles approximate area

To estimate the area under f on [a, b], slice the interval into n subintervals and cap each with a rectangle. A Riemann sum totals Σ f(xᵢ*)·Δx, where xᵢ* is the sample point that sets each rectangle's height. For a uniform partition, Δx = (b − a)/n. Take f(x) = x² on [0, 4] with n = 4 (Δx = 1, grid 0,1,2,3,4):

• Left sum: heights at 0,1,2,3 → 1·(0+1+4+9) = 14.
• Right sum: heights at 1,2,3,4 → 1·(1+4+9+16) = 30.
• Midpoint sum: heights at 0.5,1.5,2.5,3.5 → 1·(0.25+2.25+6.25+12.25) = 21.

The trapezoidal sum connects consecutive points with straight segments; each trapezoid is (width)·(average of the two heights). On a uniform grid it equals the average of the left and right sums: (14 + 30)/2 = 22. The exact value is ∫₀⁴ x² dx = 64/3 ≈ 21.33, so the midpoint and trapezoid bracket it tightly.

Over- or underestimate

Two separate rules — do not cross them:

A chord on a concave-up curve lies above it, so trapezoids overcount. Left/right depend on slope direction; the trapezoid depends on concavity.

Unequal-width tables

Data often arrives in a table with unequal spacing — then there is no single Δx. Multiply each height (or each trapezoid's average height) by its own subinterval width and add. For R(t) in gal/min:

Widths are 2, 3, 2, 5. Left sum = 2(10)+3(18)+2(24)+5(28) = 262. Trapezoid = 2(14)+3(21)+2(26)+5(31.5) = 300.5. Pairing a height with the wrong width is the most common table error.

The integral as a limit — and as signed area

The definite integral is the exact area, the limit of Riemann sums as n → ∞:

∫ₐᵇ f(x) dx = lim_{n→∞} Σ f(xᵢ*)·Δx

Geometrically it is signed (net) area: regions above the axis count positive, regions below count negative.

[GRAPH: piecewise f on [0, 8]
- [0, 2]: triangle above the axis, area 2  → +2
- [2, 6]: lower semicircle of radius 2, area ½π(2)² = 2π, below the axis  → −2π
- [6, 8]: triangle above the axis, area 2  → +2]

So ∫₀⁸ f(x) dx = 2 − 2π + 2 = 4 − 2π ≈ −2.28. Evaluate integrals from a graph by finding each region's area (triangle, rectangle, semicircle), attaching a sign, and adding. Net area (the integral) can be negative; total area = ∫|f| counts every region positive. For f(x) = x − 2 on [0, 4], the net area is −2 + 2 = 0 but the total area is 2 + 2 = 4.

Integral properties

∫ₐᵃ f dx = 0                      ∫ᵦᵃ f dx = −∫ₐᵇ f dx
∫ₐᵇ k·f dx = k∫ₐᵇ f dx           ∫ₐᵇ [f ± g] dx = ∫ₐᵇ f dx ± ∫ₐᵇ g dx
∫ₐᶜ f dx = ∫ₐᵇ f dx + ∫ᵦᶜ f dx

The additivity rule holds even when the split point is not between the endpoints: if ∫₂⁵ f = 10 and ∫₂⁸ f = 4, then ∫₅⁸ f = ∫₂⁸ f − ∫₂⁵ f = −6.

Key Takeaways

• Riemann sum = Σ (height)·(width); left drops the last point, right drops the first, midpoint uses interval centers.
• With unequal widths there is no common Δx — use each subinterval's own width.
• Increasing/decreasing governs left/right direction; concavity governs the trapezoid.
• The definite integral is signed area: below the axis is negative. Net area ≠ total area = ∫|f|.
• Properties: reversing limits negates; constants factor out; integrals split and recombine additively.

Practice Questions