The Distance Problem
Textbook Reference
| Primary source | OpenStax Calculus Volume 2, Section 1.1: “Approximating Areas” |
| Direct link | https://openstax.org/books/calculus-volume-2/pages/1-1-approximating-areas |
| Textbook used in class | Stewart, Calculus, Section 4.1: “The Area and Distance Problems” |
Opening Scenario
When a car travels at a constant speed of 60 mph for 2 hours, the distance is simply $60 \times 2 = 120$ miles. But what if the speedometer needle is always moving? The formula distance = rate × time fails because there is no single rate.
The insight that launches integral calculus: slice the time interval into many short pieces, treat the velocity as approximately constant on each short piece, add up all the small distances, and take the limit as the pieces shrink to zero. That limit equals the exact distance traveled -- and it also equals the area under the velocity curve.
Quick Reference
Constant velocity: $d = v \cdot t$.
Variable velocity -- approximation with $n$ intervals: Divide $[a, b]$ into $n$ subintervals of width $\Delta t = \dfrac{b-a}{n}$, pick any sample time $t_i^*$ in the $i$-th subinterval, approximate velocity as constant at $v(t_i^*)$: $$\text{Distance} \approx \sum_{i=1}^{n} v(t_i^*)\,\Delta t.$$
Exact distance (limit): $$d = \lim_{n \to \infty} \sum_{i=1}^{n} v(t_i^*)\,\Delta t.$$
Geometric meaning: This limit equals the area under the velocity curve $y = v(t)$ on $[a, b]$.
Key Concepts
1. Why the Constant-Velocity Formula Breaks Down
When velocity changes continuously, the product $v \cdot t$ gives the right answer only if $v$ is the same throughout. Over a short enough time interval, however, velocity barely changes. “Short enough” is the key phrase: as the interval length approaches zero, the error in treating velocity as constant over that interval also approaches zero.
This is the same logic used in the area problem: approximate with simple shapes, then improve the approximation by making the shapes smaller and more numerous.
2. The Sum of Small Distances
Divide the time interval $[a, b]$ into $n$ equal pieces of width $\Delta t$. During the $i$-th piece, the velocity is approximately $v(t_i^*)$ for some time $t_i^*$ in that piece. The distance traveled during that piece is approximately $v(t_i^*)\,\Delta t$.
Adding all $n$ pieces: $$\text{Total distance} \approx v(t_1^*)\,\Delta t + v(t_2^*)\,\Delta t + \cdots + v(t_n^*)\,\Delta t = \sum_{i=1}^{n} v(t_i^*)\,\Delta t.$$
As $n \to \infty$ (and $\Delta t \to 0$), the approximation becomes exact.
3. Distance Equals Area Under the Velocity Curve
The limit $\displaystyle\lim_{n \to \infty} \sum_{i=1}^{n} v(t_i^*)\,\Delta t$ has exactly the same form as the limit that defines the area under a curve. That is not a coincidence. If you draw the graph of $v(t)$, the area of the region between the $t$-axis and the curve from $t = a$ to $t = b$ is precisely the distance traveled during $[a, b]$ (assuming $v \geq 0$).
This is the first hint of why calculus is so unified: the area problem and the distance problem are the same problem in different clothing.
“Distance traveled equals the area under the position curve.” Distance is the area under the velocity curve, not the position curve. Position tells you where the object is; velocity tells you how fast it is moving. It is velocity that multiplies time to give distance.
Worked Example
A car’s velocity in ft/s is recorded every second:
| $t$ (s) | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| $v(t)$ (ft/s) | 0 | 10 | 18 | 24 | 28 |
Estimate the distance traveled over $[0, 4]$ using left-endpoint and right-endpoint sums.
Left-endpoint sum ($L_4$): Use velocities at $t = 0, 1, 2, 3$ with $\Delta t = 1$.
$$L_4 = (0 + 10 + 18 + 24)(1) = 52 \text{ ft}.$$
Right-endpoint sum ($R_4$): Use velocities at $t = 1, 2, 3, 4$ with $\Delta t = 1$.
$$R_4 = (10 + 18 + 24 + 28)(1) = 80 \text{ ft}.$$
Since velocity is increasing on this interval, $L_4$ underestimates and $R_4$ overestimates the true distance. The distance is somewhere between 52 ft and 80 ft.
Boxed result: $52 \text{ ft} < d < 80 \text{ ft}$.
Common Errors Summary
| Error | Example | Correction |
|---|---|---|
| Using the position formula $d = v \cdot t$ with a non-constant $v$ | Plugging in $v(2)$ as if velocity were constant | The formula applies only to constant velocity; variable velocity requires summing small products $v(t_i^*)\,\Delta t$ |
| Confusing position with velocity | Computing area under a position graph | Area under $v(t)$ gives distance; the position graph gives location, not change in location |
| Treating the estimate as exact | Reporting $L_4 = 52$ ft as the distance | $L_4$ is an approximation; the exact distance requires $\lim_{n\to\infty}$ |
Leveled Practice
Level 1 -- Reading Table Data
Problem 1. A jogger’s speed in mph is measured every half hour:
| $t$ (hr) | 0 | 0.5 | 1.0 | 1.5 |
|---|---|---|---|---|
| $v(t)$ | 4 | 5 | 6 | 5 |
Estimate the distance run over $[0, 1.5]$ using right-endpoint sums.
Show answer
$\Delta t = 0.5$, right endpoints give velocities $5, 6, 5$.
$R_3 = (5 + 6 + 5)(0.5) = 16 \times 0.5 = 8$ miles.
Level 2 -- Formula Setup
Problem 2. A ball is dropped from rest and falls with velocity $v(t) = 9.8t$ m/s. Set up the right-endpoint sum $R_n$ for the distance fallen in the first 3 seconds, simplify using the formula $\sum_{i=1}^{n} i = \dfrac{n(n+1)}{2}$, and evaluate $\lim_{n \to \infty} R_n$.
Show answer
$\Delta t = \frac{3}{n}$, $t_i = \frac{3i}{n}$, $v(t_i) = \frac{9.8 \cdot 3i}{n} = \frac{29.4i}{n}$.
$R_n = \sum_{i=1}^{n} \frac{29.4i}{n} \cdot \frac{3}{n} = \frac{88.2}{n^2} \cdot \frac{n(n+1)}{2} = \frac{44.1(n+1)}{n}$.
$\lim_{n\to\infty}\frac{44.1(n+1)}{n} = 44.1$ m.
Boxed answer: Distance = $44.1$ m. (Compare with the kinematics formula $d = \frac{1}{2}(9.8)(3^2) = 44.1$ m.)
Mastery Checklist
Mental Model
Imagine the speedometer of a car frozen in a photograph taken once per second. In each 1-second window, treat the car as traveling at that frozen speed. Add up all the distances. The sum is a rough estimate. Take a photograph once per millisecond instead, and the estimate improves dramatically. In the limit -- an infinite number of infinitely brief snapshots -- the sum equals the exact distance, which is the area under the continuous speedometer graph.
Connections
Looking back
- Constant velocity: The formula $d = vt$ is the base case; the Riemann sum is its generalization.
- Limits (Chapter 2): The exact distance is defined as the limit of approximating sums.
Looking ahead
- The definite integral (Section 4.2): Formalizes the limit $\lim_{n\to\infty}\sum v(t_i^*)\,\Delta t$ as $\displaystyle\int_a^b v(t)\,dt$.
- Net change theorem (Section 4.4): States that $\displaystyle\int_a^b v(t)\,dt$ equals the net displacement, not just distance when $v$ can be negative.