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The Area and Distance Problems

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Reference: Stewart §4.1

Two Problems, One Big Idea

How do you find the area of a region with curved boundaries? And how do you find the distance traveled by a car if its velocity keeps changing?

These seem like completely different questions, but they lead to the same mathematical structure: adding up infinitely many infinitely small pieces. This is the fundamental idea behind integration.

We know how to find areas of rectangles (length × width) and distances with constant velocity (velocity × time). The insight of calculus is that we can handle curved boundaries and changing velocities by approximating with rectangles, then taking a limit as the rectangles get smaller and more numerous.

Prerequisite Map

PrerequisitesEvaluating FunctionsLimits at Infinity
This skillThe Area and Distance Problems
Leads toRiemann SumsSigma Notation for Sums

Quick Reference

Property Value
Section Stewart §4.1
Course MATH161
Difficulty Beginner
Time ~15 minutes

Key Concepts

The Area Problem

Goal: Find the area $A$ of the region $S$ under the curve $y = f(x)$ from $x = a$ to $x = b$.

     y
     |
     |    ___________
     |   /           \
     |  /             \    y = f(x)
     | /               \
     |/     AREA = ?    \
     +------------------+--→ x
     a                  b

The challenge: We know how to find areas of rectangles, but the region $S$ has a curved boundary.

The solution: Approximate with rectangles!

Approximating with Rectangles

Step 1: Divide $[a, b]$ into $n$ equal subintervals of width $\Delta x = \frac{b-a}{n}$

Step 2: Build a rectangle on each subinterval

Step 3: Add up the areas of all rectangles

     y
     |
     |   +--+--+--+--+
     |   |  |  |  |  |   ← Rectangles approximate
     |  /|  |  |  |  |\      the curved region
     | / |  |  |  |  | \
     |/  |  |  |  |  |  \
     +---+--+--+--+--+---→ x
     a   x₁ x₂ x₃ x₄  b
           Δx

Left vs Right Endpoints

Right-endpoint sum $R_n$: Height of each rectangle is $f$ evaluated at the right endpoint $$R_n = f(x_1)\Delta x + f(x_2)\Delta x + \cdots + f(x_n)\Delta x$$

Left-endpoint sum $L_n$: Height of each rectangle is $f$ evaluated at the left endpoint $$L_n = f(x_0)\Delta x + f(x_1)\Delta x + \cdots + f(x_{n-1})\Delta x$$

Key observation: If $f$ is increasing:

If $f$ is decreasing, it’s the opposite!

The Big Idea: Area as a Limit

As $n \to \infty$, the rectangles get thinner and more numerous. The approximation gets better and better.

$$\boxed{A = \lim_{n \to \infty} R_n = \lim_{n \to \infty} L_n}$$

This limit defines the area! It can be proven that for continuous functions, both limits exist and are equal.

Example: Area Under $y = x^2$

For $f(x) = x^2$ on $[0, 1]$:

$n$ $L_n$ $R_n$ True area is between
4 0.21875 0.46875 $0.219 < A < 0.469$
8 0.2734 0.3984 $0.273 < A < 0.398$
100 0.32835 0.33835 $0.328 < A < 0.338$
1000 0.333167 0.333835 Very close to $\frac{1}{3}$!

As $n \to \infty$: Both $L_n$ and $R_n$ approach $\frac{1}{3}$.

Conclusion: The area under $y = x^2$ from 0 to 1 is exactly $\frac{1}{3}$.

The Distance Problem

Goal: Find the distance traveled during time interval $[a, b]$ if velocity $v(t)$ varies.

The connection: If velocity were constant: $\text{distance} = \text{velocity} \times \text{time}$

But velocity varies! So we approximate:

  1. Divide $[a, b]$ into small time intervals
  2. Assume velocity is approximately constant on each small interval
  3. Add up the distances: $v(t_i) \cdot \Delta t$
  4. Take limit as intervals shrink

$$d = \lim_{n \to \infty} \sum_{i=1}^{n} v(t_i) \Delta t$$

This has the same form as the area formula!

Why They’re the Same

The distance traveled equals the area under the velocity curve.

  v(t)
   |
   |      ___
   |     /   \
   |    /     \___     Distance = Area
   |___/          \      under v(t)
   +----------------→ t
   a               b

This isn’t a coincidence: it’s a fundamental principle. Whenever you’re accumulating something (area, distance, volume, work, ...), you end up with the same mathematical structure.

Practice Problems

Level 1 Computing Rectangle Areas

Estimate the area under $f(x) = x$ from $x = 0$ to $x = 2$ using:

(a) Two rectangles with right endpoints (b) Two rectangles with left endpoints (c) What is the true area (hint: it’s a triangle)?

Thought Process

With $n = 2$ rectangles on $[0, 2]$:

  • $\Delta x = \frac{2-0}{2} = 1$
  • Subintervals: $[0, 1]$ and $[1, 2]$
  • Right endpoints: $x = 1, 2$
  • Left endpoints: $x = 0, 1$
Show Answer

$\Delta x = 1$, subintervals are $[0, 1]$ and $[1, 2]$

(a) Right endpoints at $x = 1$ and $x = 2$: $$R_2 = f(1) \cdot 1 + f(2) \cdot 1 = 1 + 2 = 3$$

(b) Left endpoints at $x = 0$ and $x = 1$: $$L_2 = f(0) \cdot 1 + f(1) \cdot 1 = 0 + 1 = 1$$

(c) The region is a right triangle with base 2 and height 2: $$A = \frac{1}{2}(2)(2) = 2$$

Notice: $L_2 < A < R_2$ (since $f(x) = x$ is increasing)

Level 2 More Rectangles, Better Estimate

For the same function $f(x) = x$ on $[0, 2]$, now use four rectangles:

(a) Compute $R_4$ (b) Compute $L_4$ (c) Compare to your answers with two rectangles. Which is closer to the true area of 2?

Thought Process

With $n = 4$:

  • $\Delta x = \frac{2}{4} = 0.5$
  • Subintervals: $[0, 0.5], [0.5, 1], [1, 1.5], [1.5, 2]$
  • Right endpoints: $0.5, 1, 1.5, 2$
  • Left endpoints: $0, 0.5, 1, 1.5$
Show Answer

$\Delta x = 0.5$

(a) Right endpoints: $$R_4 = f(0.5)(0.5) + f(1)(0.5) + f(1.5)(0.5) + f(2)(0.5)$$ $$= (0.5 + 1 + 1.5 + 2)(0.5) = 5(0.5) = 2.5$$

(b) Left endpoints: $$L_4 = f(0)(0.5) + f(0.5)(0.5) + f(1)(0.5) + f(1.5)(0.5)$$ $$= (0 + 0.5 + 1 + 1.5)(0.5) = 3(0.5) = 1.5$$

(c) Comparison:

  • With $n = 2$: $L_2 = 1$, $R_2 = 3$ (true area trapped between)
  • With $n = 4$: $L_4 = 1.5$, $R_4 = 2.5$ (closer to true area of 2)

More rectangles give better approximations!

Level 3 Distance from Velocity Data

A runner’s velocity (in m/s) is recorded every 2 seconds:

Time (s) 0 2 4 6 8 10
Velocity (m/s) 0 3 5 6 5 4

(a) Estimate the distance traveled using left endpoints (b) Estimate using right endpoints (c) Which estimate is an overestimate, and which is an underestimate? (Consider the velocity behavior.)

Thought Process

$\Delta t = 2$ seconds for each interval.

For left endpoints: use velocity at start of each interval. For right endpoints: use velocity at end of each interval.

The velocity increases then decreases, so neither will be consistently over or under.

Show Answer

$\Delta t = 2$ seconds

(a) Left endpoints (velocities at $t = 0, 2, 4, 6, 8$): $$L_5 = (0 + 3 + 5 + 6 + 5)(2) = 19(2) = 38 \text{ meters}$$

(b) Right endpoints (velocities at $t = 2, 4, 6, 8, 10$): $$R_5 = (3 + 5 + 6 + 5 + 4)(2) = 23(2) = 46 \text{ meters}$$

(c) The velocity is not monotonic (it increases then decreases), so neither estimate is guaranteed to be an over/underestimate overall.

However, the true distance is likely between these values: approximately $38 < d < 46$ meters, or we can estimate $d \approx 42$ meters (averaging the two estimates).

Level 4 Over and Under Estimates

Let $f$ be a continuous, increasing function on $[a, b]$.

(a) Explain why $L_n < A < R_n$ for any $n$.

(b) Show that $R_n - L_n = \frac{b-a}{n}[f(b) - f(a)]$

(c) Use part (b) to explain why both $L_n$ and $R_n$ approach the same limit as $n \to \infty$.

Thought Process

(a) Since $f$ is increasing, on each subinterval the left endpoint gives the minimum value and the right endpoint gives the maximum value.

(b) Write out both sums and subtract: most terms cancel!

(c) As $n \to \infty$, the difference $R_n - L_n \to 0$, so they must have the same limit.

Show Answer

(a) On each subinterval $[x_{i-1}, x_i]$:

  • $f(x_{i-1}) \leq f(x) \leq f(x_i)$ for all $x$ in the subinterval (since $f$ is increasing)
  • The rectangle with height $f(x_{i-1})$ lies entirely under the curve
  • The rectangle with height $f(x_i)$ entirely contains the region under the curve

Summing over all subintervals: $L_n < A < R_n$

(b) $$R_n = [f(x_1) + f(x_2) + \cdots + f(x_n)]\Delta x$$ $$L_n = [f(x_0) + f(x_1) + \cdots + f(x_{n-1})]\Delta x$$

Subtracting: $$R_n - L_n = [f(x_n) - f(x_0)]\Delta x = [f(b) - f(a)] \cdot \frac{b-a}{n}$$

(c) Since $f(b) - f(a)$ is a fixed constant: $$\lim_{n \to \infty} (R_n - L_n) = \lim_{n \to \infty} \frac{(b-a)[f(b)-f(a)]}{n} = 0$$

Since the difference goes to 0, and $L_n \leq A \leq R_n$, both $L_n$ and $R_n$ must converge to the same value $A$. $\square$

Level 5 Connecting Area and Distance

A car accelerates from rest, and its velocity at time $t$ seconds is $v(t) = 3t$ m/s.

(a) Find a formula for the distance traveled in the first $T$ seconds by evaluating $\lim_{n \to \infty} R_n$ where $R_n$ is the right Riemann sum for $v(t)$ on $[0, T]$.

(b) Verify your answer by computing the area of the region under $v(t) = 3t$ from $t = 0$ to $t = T$ geometrically.

(c) If the car’s velocity were instead $v(t) = 3t^2$, what would the distance formula be? (You may use the formula $\sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}$.)

Thought Process

(a) Set up $R_n$ with $\Delta t = T/n$, $t_i = iT/n$. Then use summation formulas.

(b) The region is a triangle with base $T$ and height $3T$.

(c) Same process but with $v(t_i) = 3t_i^2$: will need the sum of squares formula.

Show Answer

(a) On $[0, T]$ with $n$ subintervals:

  • $\Delta t = \frac{T}{n}$
  • Right endpoints: $t_i = \frac{iT}{n}$
  • $v(t_i) = 3 \cdot \frac{iT}{n} = \frac{3iT}{n}$

$$R_n = \sum_{i=1}^{n} v(t_i) \Delta t = \sum_{i=1}^{n} \frac{3iT}{n} \cdot \frac{T}{n} = \frac{3T^2}{n^2} \sum_{i=1}^{n} i$$

Using $\sum_{i=1}^n i = \frac{n(n+1)}{2}$:

$$R_n = \frac{3T^2}{n^2} \cdot \frac{n(n+1)}{2} = \frac{3T^2(n+1)}{2n}$$

$$d = \lim_{n \to \infty} \frac{3T^2(n+1)}{2n} = \lim_{n \to \infty} \frac{3T^2}{2}\left(1 + \frac{1}{n}\right) = \frac{3T^2}{2}$$

Distance: $d = \frac{3T^2}{2}$ meters

(b) The region under $v(t) = 3t$ from 0 to $T$ is a triangle:

  • Base = $T$
  • Height = $v(T) = 3T$
  • Area = $\frac{1}{2}(T)(3T) = \frac{3T^2}{2}$ ✓

(c) With $v(t) = 3t^2$:

$$R_n = \sum_{i=1}^{n} 3\left(\frac{iT}{n}\right)^2 \cdot \frac{T}{n} = \frac{3T^3}{n^3} \sum_{i=1}^{n} i^2$$

$$= \frac{3T^3}{n^3} \cdot \frac{n(n+1)(2n+1)}{6} = \frac{T^3(n+1)(2n+1)}{2n^2}$$

$$d = \lim_{n \to \infty} \frac{T^3(n+1)(2n+1)}{2n^2}$$

Expanding: $(n+1)(2n+1) = 2n^2 + 3n + 1$

$$= T^3 \lim_{n \to \infty} \frac{2n^2 + 3n + 1}{2n^2} = T^3 \lim_{n \to \infty} \left(1 + \frac{3}{2n} + \frac{1}{2n^2}\right) = T^3 \cdot 1 = T^3$$

Distance: $d = T^3$ meters

Common Misconceptions

Common misconception

the Riemann-sum approximation IS the area under the curve.

This is the concept-image-conflicts-definition error. The finite rectangle sum $R_n$ or $L_n$ is never the exact area; it is only an approximation. The area is defined as the limit of these sums as $n \to \infty$. For $f(x) = x^2$ on $[0,1]$, the right sum $R_4 = 0.46875$ exceeds the true area $\tfrac{1}{3}$ because finitely many rectangles always leave gaps or overlaps. Only the limiting process removes the error.

Common misconception

distance equals the product of velocity and elapsed time.

This is the rate-as-fixed-number error. The formula distance = velocity $\times$ time applies only when velocity is constant. When velocity varies, each small time slice $\Delta t$ contributes approximately $v(t_i)\,\Delta t$ to the distance, and the total is the accumulated sum of all such slices. Taking the limit of these sums as $\Delta t \to 0$ gives the exact distance; no single velocity value captures the whole motion.

Mastery Checklist

Mental Model

The Infinite Subdivision:

Imagine cutting a loaf of bread into slices:

Area under a curve works the same way. Each rectangle is a “slice” of the area. With infinitely many infinitely thin slices, we capture the curved boundary exactly.

The remarkable fact is that this same slicing idea works for distance, volume, work, and countless other quantities.

Connections

Looking back:

Looking ahead:

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Last updated: 2026-01-22