The Area 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
Here is a question with no obvious answer: what is the area of the region under the curve $y = x^2$ from $x = 0$ to $x = 1$? The region has a straight bottom, straight sides, but a curved top. Every formula from geometry assumes straight edges. So we need a new strategy.
The strategy is to approximate the curved region with shapes whose areas we already know -- rectangles -- and then improve the approximation by using more, thinner rectangles. The limit of this process defines the area exactly.
Quick Reference
Setup. Given a continuous function $f \geq 0$ on $[a, b]$, divide $[a, b]$ into $n$ equal subintervals of width $\Delta x = \dfrac{b - a}{n}$.
Right-endpoint sum: $\displaystyle R_n = \sum_{i=1}^{n} f(x_i)\,\Delta x$ where $x_i = a + i\,\Delta x$.
Left-endpoint sum: $\displaystyle L_n = \sum_{i=1}^{n} f(x_{i-1})\,\Delta x$ where $x_{i-1} = a + (i-1)\,\Delta x$.
Area as a limit: $\displaystyle A = \lim_{n \to \infty} R_n = \lim_{n \to \infty} L_n$.
Key Concepts
1. Why Rectangles Work
A rectangle with base $\Delta x$ and height $f(x^*)$ -- for any $x^*$ in a subinterval -- has area $f(x^*)\,\Delta x$. This is an approximation to the area of the curved strip above that subinterval.
As the number of subintervals $n$ grows, each subinterval gets narrower. Over a narrower base, $f$ barely changes, so a rectangle closely hugs the curve. The sum of all rectangle areas approaches the true area beneath the curve.
2. Left vs. Right Endpoints
The choice of where in each subinterval to evaluate $f$ produces different approximations.
- Right-endpoint sum $R_n$: each rectangle’s height is the function value at the right end of the subinterval.
- Left-endpoint sum $L_n$: each rectangle’s height is the function value at the left end.
- Midpoint sum $M_n$: each rectangle’s height is the function value at the center. (Developed in Stewart 4.2.)
For an increasing function, the left-endpoint rectangles all lie below the curve, so $L_n$ underestimates the area. The right-endpoint rectangles all extend above the curve, so $R_n$ overestimates. For a decreasing function, the inequalities reverse. For a function that is neither always increasing nor always decreasing, neither sum is guaranteed to be an over- or underestimate.
3. Area Defined as a Limit
The definition of area is the key step. We do not assume the area exists and then compute it; we define it to be the common limit of the approximating sums (when that limit exists).
For a continuous, non-negative function on a closed interval, it can be proven that the right-sum limit, the left-sum limit, and the limit using any other sample points within each subinterval all agree. That shared limit is declared to be the area.
$$A = \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i^*)\,\Delta x$$
where $x_i^*$ is any point in the $i$-th subinterval. This expression is called a Riemann sum (studied further in Section 4.2).
“More rectangles always gives a bigger sum.” Not true. Adding rectangles reduces the width of each one, changing both the number and the shape of the rectangles. Whether the sum goes up or down depends on the function. For an increasing function $R_n$ decreases toward the true area as $n$ grows, while $L_n$ increases toward it. The sums from both sides squeeze toward the same limit.
Worked Example
Find the area under $f(x) = x^2$ on $[0, 1]$ using right-endpoint sums, then take the limit.
Step 1 -- Setup.
$\Delta x = \dfrac{1}{n}$, right endpoints $x_i = \dfrac{i}{n}$ for $i = 1, 2, \ldots, n$.
Step 2 -- Build the sum.
$$R_n = \sum_{i=1}^{n} f(x_i)\,\Delta x = \sum_{i=1}^{n} \left(\frac{i}{n}\right)^2 \cdot \frac{1}{n} = \frac{1}{n^3}\sum_{i=1}^{n} i^2.$$
Step 3 -- Apply the summation formula.
Using $\displaystyle\sum_{i=1}^{n} i^2 = \dfrac{n(n+1)(2n+1)}{6}$:
$$R_n = \frac{1}{n^3} \cdot \frac{n(n+1)(2n+1)}{6} = \frac{(n+1)(2n+1)}{6n^2}.$$
Step 4 -- Take the limit.
$$A = \lim_{n \to \infty} \frac{(n+1)(2n+1)}{6n^2} = \lim_{n \to \infty} \frac{2n^2 + 3n + 1}{6n^2} = \frac{2}{6} = \frac{1}{3}.$$
Boxed answer: The area under $y = x^2$ from 0 to 1 is $\dfrac{1}{3}$.
Common Errors Summary
| Error | Example | Correction |
|---|---|---|
| Confusing left and right endpoints | Labeling right-endpoint heights as $f(x_{i-1})$ | Right endpoints are $x_i = a + i\,\Delta x$; left endpoints are $x_{i-1} = a + (i-1)\,\Delta x$ |
| Forgetting to take the limit | Reporting $R_{10}$ as the area | The area is defined as $\lim_{n\to\infty} R_n$, not any finite approximation |
| Assuming $L_n < A < R_n$ always | Applying this for a non-monotone function | The inequality $L_n < A < R_n$ holds only when $f$ is increasing on $[a,b]$ |
Leveled Practice
Level 1 -- Reading a Sum
Problem 1. For $f(x) = 3x$ on $[0, 2]$ with $n = 4$ rectangles, compute $R_4$.
Show answer
$\Delta x = \frac{1}{2}$, right endpoints: $\frac{1}{2}, 1, \frac{3}{2}, 2$.
$R_4 = f\!\left(\tfrac12\right)\cdot\tfrac12 + f(1)\cdot\tfrac12 + f\!\left(\tfrac32\right)\cdot\tfrac12 + f(2)\cdot\tfrac12$ $= \left(\tfrac32 + 3 + \tfrac92 + 6\right)\cdot\tfrac12 = 15\cdot\tfrac12 = \dfrac{15}{2}$.
Level 2 -- Formula and Limit
Problem 2. Set up $R_n$ for $f(x) = 2x$ on $[0, 3]$ and evaluate $\lim_{n\to\infty} R_n$ using the formula $\sum_{i=1}^{n} i = \dfrac{n(n+1)}{2}$.
Show answer
$\Delta x = \frac{3}{n}$, $x_i = \frac{3i}{n}$.
$R_n = \sum_{i=1}^{n} \frac{6i}{n}\cdot\frac{3}{n} = \frac{18}{n^2}\sum_{i=1}^{n} i = \frac{18}{n^2}\cdot\frac{n(n+1)}{2} = \frac{9(n+1)}{n}$.
$\lim_{n\to\infty}\frac{9(n+1)}{n} = 9$.
Boxed answer: $A = 9$. (This matches the area of the triangle with base 3 and height 6: $\frac{1}{2}(3)(6) = 9$.)
Mastery Checklist
Mental Model
Think of the rectangles as floor tiles laid under a curved roof. With wide tiles, the gaps between the tile tops and the curved ceiling are large -- the approximation is rough. With narrower tiles, the gaps shrink. In the limit, the tiles become infinitely thin and their combined area equals the area beneath the curve exactly.
Connections
Looking back
- Limits (Chapter 2): The area is defined as a limit; the tools for evaluating limits of sequences apply here.
- Function evaluation: Computing $f(x_i)$ at finitely many sample points is the core arithmetic step.
Looking ahead
- Sigma notation (Section 4.1): Compact notation for the sums $R_n$ and $L_n$.
- Riemann sums (Section 4.2): The general framework, allowing any sample points.
- Definite integral (Section 4.2): The formal limit of Riemann sums, extended to all continuous functions (not just non-negative ones).
Back to Integration Foundations | Next: The Distance Problem