Difficulty: Intermediate | Prerequisites: limits, sigma notation, definition of the derivative.
Unit 7 sits at the bridge between differential and integral calculus. You have spent several units learning what derivatives are and how to compute them; now you are learning how to work backwards, estimating accumulated quantities from rate data. This is the conceptual core of integration: given a rate of change, how much total change occurred? If you are comfortable with limits, function notation, and the idea that a derivative gives an instantaneous rate of change, you are ready for this material.
Riemann sums (left, right, midpoint, trapezoidal) let you approximate a definite integral when you only have a table of values or a graph rather than a neat formula. Concavity tells you whether your approximation overshoots or undershoots. Converting between a definite integral and its equivalent limit of a Riemann sum is a standard exam skill.
Definite integral
The limit of a Riemann sum as the number of subintervals approaches infinity. Written as the integral from a to b of f(x) dx. In simple terms, it gives you the net signed area between the curve and the x-axis over an interval.
Riemann sum
An approximation of a definite integral formed by summing the areas of rectangles (or trapezoids) over subintervals. Think of it as slicing the area under a curve into strips and adding them up.
Left Riemann sum
A Riemann sum where the height of each rectangle is the function value at the left endpoint of each subinterval. If the function is increasing, the left sum underestimates; if decreasing, it overestimates.
Right Riemann sum
A Riemann sum where the height of each rectangle is the function value at the right endpoint of each subinterval. The opposite bias to the left sum: overestimates for increasing functions, underestimates for decreasing.
Trapezoidal sum
An approximation that uses trapezoids rather than rectangles. Equivalent to the average of the left and right Riemann sums. For a concave-down function, the trapezoidal sum is an underestimate; for concave-up, it is an overestimate.
Concavity
Describes whether a curve bends upward (concave up, second derivative positive) or downward (concave down, second derivative negative). Concavity determines whether trapezoidal and midpoint approximations over- or underestimate the true integral.
Intermediate Value Theorem (IVT)
If f is continuous on [a, b] and N is any value between f(a) and f(b), then there exists at least one c in (a, b) such that f(c) = N. In plain terms, a continuous function cannot skip over values; it must hit everything in between.
Subinterval
One of the smaller intervals you divide [a, b] into when building a Riemann sum. The width of each subinterval is often called delta-x.
When you have a table of values for a function but no formula, you estimate the derivative at a point by computing the slope between the two nearest data points that bracket it.
For example, given values of r(t) at t = 0, 3, 5, 6, 9, to estimate r'(4):
Use the interval [3, 5] (the closest points that bracket t = 4)
r'(4) is approximately (r(5) - r(3)) / (5 - 3) = (112 - 95) / 2 = 8.5 rotations per minute squared
The units on a derivative are always (units of output) / (units of input). Here, rotations per minute divided by minutes gives rotations per minute squared.
When a table has unequal spacing, the subinterval widths (delta-t values) differ. For a left Riemann sum, you use the function value at the left endpoint of each subinterval, multiplied by that subinterval's width.
Worked example: approximate the integral from 0 to 9 of r(t) dt using four subintervals from the table (t = 0, 3, 5, 6, 9):
Subintervals: [0,3], [3,5], [5,6], [6,9]
Widths: 3, 2, 1, 3
Left values: r(0) = 72, r(3) = 95, r(5) = 112, r(6) = 77
Sum: 3(72) + 2(95) + 1(112) + 3(77) = 216 + 190 + 112 + 231 = 749
Interpretation with units: the integral from 0 to 9 of r(t) dt represents the total number of rotations the wheel made between t = 0 and t = 9 minutes. The units are (rotations/minute)(minutes) = rotations.
A trapezoidal sum averages the left and right function values on each subinterval, then multiplies by the width.
For each subinterval [a, b]: (1/2)(f(a) + f(b))(b - a).
Worked example with the same data:
(1/2)(3)(72 + 95) + (1/2)(2)(95 + 112) + (1/2)(1)(112 + 77) + (1/2)(3)(77 + 50)
= 250.5 + 207 + 94.5 + 190.5 = 742.5 rotations
Whether an approximation overshoots or undershoots depends on the function's concavity:
Concave down (r'' < 0): the trapezoidal sum underestimates the integral, and the midpoint sum overestimates
Concave up (r'' > 0): the trapezoidal sum overestimates, and the midpoint sum underestimates
In the worked example, r is concave down, so the trapezoidal sum of 742.5 is an underestimate.
The IVT applies whenever a function is continuous on a closed interval and you need to show that it hits a particular value.
Worked example: is there a time t in [0, 5] where r(t) = 106?
r is continuous on [0, 5] (given that it is differentiable)
r(3) = 95 and r(5) = 112
Since 95 < 106 < 112, by the IVT there must be some t in (3, 5) with r(t) = 106
Always state: (1) the function is continuous on the interval, (2) the target value lies between the function values at the endpoints, (3) therefore by the IVT such a point exists.
The definite integral from a to b of f(x) dx equals the limit as n approaches infinity of the sum from i = 1 to n of f(x_i*) times delta-x, where delta-x = (b - a)/n and x_i* is a sample point in the i-th subinterval.
For right-endpoint Riemann sums on [a, b]:
delta-x = (b - a) / n
x_i = a + i * delta-x
The sum becomes: limit as n approaches infinity of the sum from i = 1 to n of f(a + i * delta-x) * delta-x
Convert the integral from 0 to 5 of (x + 1) dx into a Riemann sum limit.
Here a = 0, b = 5, so delta-x = 5/n
Right endpoints: x_i = 0 + i(5/n) = 5i/n
f(x_i) = (5i/n) + 1
Answer: limit as n approaches infinity of the sum from i = 1 to n of (5i/n + 1)(5/n)
This matches option (C) from the practice set.
A single limit-of-sum expression can represent multiple integrals through substitution.
Consider the limit as n approaches infinity of the sum from k = 1 to n of (-2 + 8k/n)^3 * (8/n).
delta-x = 8/n, so b - a = 8
x_k = -2 + 8k/n, and the starting value (k = 1, n large) approaches -2, so a = -2, b = 6
This equals the integral from -2 to 6 of x^3 dx
With the substitution u = -2 + x on the integral from 0 to 8 of (-2 + x)^3 dx, you get the same thing. Similarly, with u = 8x on the integral from 0 to 1 of 8(-2 + 8x)^3 dx, you also get the same thing. All three integral forms are equivalent.
Left Riemann sum (unequal subintervals)
Sum of f(t_i) * (t_{i+1} - t_i) for each subinterval, using the left endpoint.
Right Riemann sum (unequal subintervals)
Sum of f(t_{i+1}) * (t_{i+1} - t_i) for each subinterval, using the right endpoint.
Trapezoidal sum (unequal subintervals)
Sum of (1/2)(f(t_i) + f(t_{i+1}))(t_{i+1} - t_i) for each subinterval.
Equal subintervals on [a, b]
delta-x = (b - a) / n
Right-endpoint Riemann sum as a limit
Integral from a to b of f(x) dx = limit as n approaches infinity of the sum from i = 1 to n of f(a + i * delta-x) * delta-x
Derivative estimate from a table
f'(c) is approximately (f(b) - f(a)) / (b - a), using the two data points closest to c.
Riemann sums are the backbone of any situation where you measure a rate over time and need to know the total accumulated quantity. A speedometer reading gives your speed at each moment; a Riemann sum of those readings estimates the total distance you have driven. Engineers use trapezoidal approximations to estimate total fluid flow from sensor readings taken at discrete intervals. In medicine, the area under a drug-concentration curve (AUC) is estimated from blood samples taken at specific times, using exactly the trapezoidal method covered here.
Students often forget to use unequal subinterval widths when the table has unequal spacing. Every delta-t must match the actual gap between consecutive t-values. Do not assume every subinterval has the same width.
Students confuse whether a left sum overestimates or underestimates. It depends on whether the function is increasing or decreasing, not on whether the function is positive or negative.
Students mix up the trapezoidal rule's bias. The trapezoidal sum underestimates for concave-down functions and overestimates for concave-up functions. A common error is to get this backwards.
When estimating a derivative from a table, students sometimes use points that do not bracket the target value. Always pick the two data points on either side of the point where you want the derivative, not just the two closest in the table.
Table-based Riemann sum problems appear on nearly every AP Calculus and university Calculus I exam. You will almost certainly see one.
Always include units in your interpretation of a definite integral. Examiners award separate marks for correct units and a correct contextual sentence.
The IVT justification question is a staple. You must state that the function is continuous, identify the interval, and show the target value falls between the endpoint values. Missing any of these three pieces loses marks.
Know how to identify over- and underestimates from concavity. Expect a follow-up question asking you to explain your reasoning.
True or false: a left Riemann sum always underestimates the integral.
False. It underestimates only when the function is increasing. When the function is decreasing, the left sum overestimates.
Fill in the blank: for a concave-down function, the trapezoidal sum is an ______ of the integral.
Underestimate.
True or false: to apply the Intermediate Value Theorem, the function must be differentiable on the interval.
False. The function only needs to be continuous. Differentiability is a stronger condition and is not required.
Fill in the blank: if [a, b] is divided into n equal subintervals, delta-x = ______.
(b - a) / n.
True or false: the units of the integral from 0 to 9 of r(t) dt, where r is in rotations per minute and t is in minutes, are rotations per minute.
False. The units are (rotations/minute)(minutes) = rotations.
Q: A function g is continuous and concave up on [1, 5]. You approximate the integral from 1 to 5 of g(x) dx using a trapezoidal sum. Is your result an overestimate or an underestimate?
A: Overestimate. For a concave-up function, the trapezoid tops sit above the curve.
Q: Given the table below, estimate f'(7) using a difference quotient.
x | 5 | 8 |
|---|---|---|
f(x) | 20 | 35 |
A: f'(7) is approximately (35 - 20) / (8 - 5) = 15/3 = 5.
Q: Write the definite integral from 2 to 6 of x^2 dx as a limit of a right Riemann sum.
A: delta-x = 4/n, x_i = 2 + 4i/n. The limit is: lim (n to infinity) of the sum from i = 1 to n of (2 + 4i/n)^2 * (4/n).
Q: A car's velocity v(t) in km/h is recorded at t = 0, 2, 5, 8 minutes as 60, 75, 80, 70. Use a left Riemann sum with three subintervals to approximate the integral from 0 to 8 of v(t) dt. What does the result represent?
A: 2(60) + 3(75) + 3(80) = 120 + 225 + 240 = 585. The units are (km/h)(min). To get km, you would convert minutes to hours. The result approximates the total displacement of the car over the 8-minute interval.
Q: The function h is continuous on [0, 10], h(3) = 4, and h(7) = 12. Must there be a value c in (3, 7) where h(c) = 9? Justify.
A: Yes. h is continuous on [3, 7], and 4 < 9 < 12, so by the Intermediate Value Theorem there exists some c in (3, 7) with h(c) = 9.
This material connects directly to the Fundamental Theorem of Calculus (Unit 8), which replaces the limit-of-sums approach with antiderivatives for exact evaluation. The IVT work here also ties back to continuity and the formal definition of limits from earlier units. Understanding over- and underestimates from concavity connects to the second derivative tests you studied in the applications-of-derivatives unit.
Riemann sum, left Riemann sum, right Riemann sum, trapezoidal rule, trapezoidal sum, trapezoidal approximation, midpoint sum, definite integral, integral estimation, sigma notation, limit of a sum, delta-x, subinterval, partition, concave up, concave down, overestimate, underestimate, IVT, Intermediate Value Theorem, derivative from table, rate of change, accumulation, area under a curve, Calculus I, AP Calculus AB, Unit 7, MTH 132, Michigan State University