The Area and Distance Problems

MATH161

Reference: Stewart §4.1 • Chapter: 4 • Section: 1

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The Area and Distance Problems

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

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:

$L_n$ underestimates the true area (rectangles fit inside)
$R_n$ overestimates the true area (rectangles stick out)

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:

Divide $[a, b]$ into small time intervals
Assume velocity is approximately constant on each small interval
Add up the distances: $v(t_i) \cdot \Delta t$
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 $R_n - L_n = \frac{b-a}{n}[f(b) - f(a)]$

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

Mastery Checklist

[ ] I understand why we approximate curved regions with rectangles
[ ] I can compute left and right endpoint sums for simple functions
[ ] I understand that area is defined as a limit of these sums
[ ] I can estimate distance from velocity data using rectangles
[ ] I recognize that area under velocity = distance traveled
[ ] I understand why more rectangles give better approximations

Mental Model

The Infinite Subdivision:

Imagine cutting a loaf of bread into slices:

With a few thick slices, the surface is rough
With many thin slices, the surface is smooth

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:

Limits give us the tool to make "infinitely many" precise
Continuous functions on closed intervals are bounded, ensuring rectangle heights are finite

Looking ahead:

Sigma Notation gives us compact notation for these sums
Riemann Sums formalize the rectangle approximation
The Definite Integral is the limit we've been computing

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Closed Interval Method	Section Index	Sigma Notation

Last updated: 2026-01-22