Short on time? This page is the recap. For explanations, the diagram and worked examples, use the full study guide. This topic is shared by Calculus AB and Calculus BC.
Notation: Σ (i = 1 to n) aᵢ means a₁ + a₂ + … + aₙ, and ∫ (a to b) f(x) dx means the definite integral of f from a to b.
Recap
- Sigma notation has an index, a first value, a last value and a term. Σ (i = 1 to 4) (2i + 1) = 3 + 5 + 7 + 9 = 24.
- A Riemann sum needs a partition a = x₀ < x₁ < … < xₙ = b. In each subinterval pick a sample point xᵢ*, multiply f(xᵢ*) by the width Δxᵢ, then add.
- For f continuous on [a, b], the definite integral is the limit of Riemann sums as the largest width shrinks to 0. Any choice of sample points gives the same limit.
- With equal widths, “largest width → 0” means n → ∞.
- The integral is a signed area: parts below the axis count as negative.
Key relationships
| Idea | Formula or rule |
|---|---|
| Width (equal pieces) | Δx = (b − a)/n |
| Partition points | xᵢ = a + iΔx, so x₀ = a and xₙ = b |
| Right sum | Σ (i = 1 to n) f(a + iΔx) Δx |
| Left sum | Σ (i = 1 to n) f(a + (i − 1)Δx) Δx |
| Definition | ∫ (a to b) f(x) dx = lim (n → ∞) Σ (i = 1 to n) f(a + iΔx) Δx |
| Sum → integral | Δx becomes dx; a + iΔx becomes x; b = a + (b − a) |
| Constant factor | Σ c·aᵢ = c · Σ aᵢ |
Assumptions behind the method
- f is continuous on [a, b], so the limit exists and does not depend on the sample points.
- In a sum with equal widths, the factor (number)/n is the width, and the number is b − a.
- A limit of sums can match more than one integral (for example ∫ (2 to 5) √x dx and ∫ (0 to 3) √(2 + x) dx). Each must give the same value.
Mistakes to avoid
- Treating the number in (number)·i/n as b. It is b − a.
- Shifting twice, e.g. writing ∫ (2 to 5) √(2 + x) dx.
- Dropping or changing the width factor. A 1/n in place of 3/n changes the value.
- Letting n → ∞ with unequal widths without checking that the widest piece shrinks.
- Miscounting terms when the index starts at 0.
- Leaving out dx.
Quick self-check
- Find Σ (i = 1 to 3) i². (1 + 4 + 9 = 14)
- Write lim (n → ∞) Σ (i = 1 to n) (1 + 2i/n)² · (2/n) as an integral. (∫ (1 to 3) x² dx, or ∫ (0 to 2) (1 + x)² dx)
- Write ∫ (0 to 2) 3x dx as the limit of a right Riemann sum, then find its value with geometry. (lim (n → ∞) Σ (i = 1 to n) 3(2i/n) · (2/n); a triangle with base 2 and height 6, so the value is 6)
Next: practice questions.