Riemann Sums and the Definite Integral

You’ll be able to

Compute left, right, midpoint and trapezoidal approximations from a formula or a table
Decide whether an approximation over- or underestimates from monotonicity and concavity
Express a definite integral as the limit of a Riemann sum

Four approximations, one idea

Cut [a, b] into subintervals and approximate the area over each by a rectangle or a trapezoid. A left sum uses the function value at each subinterval's left edge as the height, a right sum the right edge, a midpoint sum the middle. A trapezoidal sum averages the two endpoint heights instead — which makes it the average of the left and right sums when the widths are equal. Widths need not be equal, and on table problems they usually are not, so each subinterval contributes height × its own width.

Over or under, decided by shape

For an increasing function, a left sum underestimates and a right sum overestimates; for a decreasing function, the reverse. Concavity decides the trapezoidal and midpoint cases: on a concave-up function the trapezoids sit above the curve, so the trapezoidal sum overestimates while the midpoint sum underestimates; on a concave-down function both flip. Two different properties are in play — monotonicity for left/right, concavity for trapezoid/midpoint — and mixing them up is the standard error.

The definite integral as a limit

Let the number of subintervals grow without bound and the approximation converges to the exact area: *∫ₐᵇ f(x) dx = lim(n→∞) Σ f(xᵢ)Δx**, where Δx = (b − a)/n. This is the definition, and the exam tests the translation in both directions — recognizing lim(n→∞) Σ (from i = 1 to n) of (2 + 3i/n)·(3/n) as ∫₂⁵ x dx by reading Δx = 3/n as the width and 2 + 3i/n as the sample point.

Trapezoidal sum, unequal widths

Σ [(f(xᵢ) + f(xᵢ₊₁))/2] · (xᵢ₊₁ − xᵢ) · equal widths: (Δx/2)[f(x₀) + 2f(x₁) + … + 2f(xₙ₋₁) + f(xₙ)]

The h/2 shortcut with doubled interior terms is valid only when every subinterval has the same width.

Worked example

A table gives f(0) = 4, f(2) = 7, f(5) = 9, f(6) = 10. Use a trapezoidal sum over the three subintervals to estimate ∫₀⁶ f(x) dx, and say whether it over- or underestimates if f is concave down.

1.The widths are unequal: 2, 3 and 1. Handle each subinterval separately.
2.[0, 2]: average height (4 + 7)/2 = 5.5, times width 2 → 11.
3.[2, 5]: average height (7 + 9)/2 = 8, times width 3 → 24.
4.[5, 6]: average height (9 + 10)/2 = 9.5, times width 1 → 9.5.
5.Total: 11 + 24 + 9.5 = 44.5. Since f is concave down, each trapezoid's top chord lies below the curve, so 44.5 is an underestimate.

Answer: ∫₀⁶ f(x) dx ≈ 44.5, an underestimate because f is concave down and the chords lie below the graph.

Watch out

Applying the equal-width formula (Δx/2)[f₀ + 2f₁ + 2f₂ + f₃] to this table gives a number, and it is wrong. Check the spacing before reaching for the shortcut — AP tables are deliberately unequal.

Checkpoint

f is increasing and concave up on [1, 5]. Which is true of a left Riemann sum for ∫₁⁵ f(x) dx?

Checkpoint

lim(n→∞) Σ (i = 1 to n) of (1 + 2i/n)³ · (2/n) equals which definite integral?

On the exam

Riemann-sum questions on the calculator-free section reward writing the sum before evaluating it: the setup line carries its own point, and an arithmetic slip afterward costs less than a missing setup.

Answer the 2 checkpoints as you read.