The Timpson Lyceum · Calculus 1 (2024)

Areas, Finite Sums, and Limits of Finite Sums - Lecture 15 (Calculus 1)

Overview

This lecture begins the study of integration by tackling the question: how do we find the area under a curve between two points? The strategy is to approximate that area with a collection of rectangles, then let the number of rectangles grow without bound so the approximation becomes exact. Along the way we develop sigma notation for writing large sums compactly, learn its algebraic properties, and memorize three closed-form formulas for summing $k$ , $k^2$ , and $k^3$ . The payoff is the definition of the definite integral as a limit of finite (Riemann) sums — the fourth and final "big idea" of calculus after limits, continuity, and derivatives. Understanding this limit is what everything that follows builds on.

Key Concepts

Approximating area with rectangles

To estimate the area under $f(x)$ on $[a,b]$ , slice the interval into rectangles, compute each rectangle's area (base $\times$ height), and add them up.
The base of each rectangle comes from how the interval is split; the height comes from the value of $f$ at some chosen point in that subinterval.
Common choices of sample point:
- Upper sum — use the highest value of $f$ on each subinterval (overestimate).
- Lower sum — use the lowest value of $f$ on each subinterval (underestimate).
- Left-endpoint sum — height taken at each subinterval's left edge.
- Right-endpoint sum — height taken at each subinterval's right edge.
- Midpoint sum — height taken at the middle of each subinterval.
As the rectangles shrink (more of them), the choice of sample point stops mattering: every choice converges to the same area.

Sigma notation

$\displaystyle\sum_{k=1}^{n} a_k$ means: start the index $k$ at the bottom value, increase by 1, plug each $k$ into the term $a_k$ , and add the results, stopping at the top value $n$ .

\sum_{k=1}^{n} a_k = a_1 + a_2 + a_3 + \cdots + a_n

The starting index is arbitrary; the same sum can be written many ways (e.g. starting at $0$ or even a negative index) by adjusting the term. A common convention is to start at $k=1$ .

Algebraic properties of sums

Sum/difference rule: a sum of two terms splits into two sums.

\sum_{k=1}^{n} (a_k \pm b_k) = \sum_{k=1}^{n} a_k \;\pm\; \sum_{k=1}^{n} b_k

Constant-multiple rule: a constant factor pulls outside the sum.

\sum_{k=1}^{n} c\,a_k = c \sum_{k=1}^{n} a_k

Sum of a constant: adding a constant $c$ to itself $n$ times gives $\sum_{k=1}^{n} c = nc$ . (Watch the "gotcha": if the term itself is $n$ , then $\sum_{k=1}^{n} n = n \cdot n = n^2$ , since $n$ does not depend on $k$ .)

Closed-form summation formulas (used to collapse big sums quickly)

\begin{aligned} \sum_{k=1}^{n} k &= \frac{n(n+1)}{2} \\ \sum_{k=1}^{n} k^2 &= \frac{n(n+1)(2n+1)}{6} \\ \sum_{k=1}^{n} k^3 &= \left(\frac{n(n+1)}{2}\right)^2 \end{aligned}

The first formula follows from pairing the sum with its reverse: $1+2+\cdots+n$ added to $n+\cdots+2+1$ gives $n$ copies of $(n+1)$ , so twice the sum is $n(n+1)$ .

The definite integral as a limit of finite sums

Divide $[a,b]$ into $n$ equal subintervals. With equal widths and right endpoints, each base is $\dfrac{b-a}{n}$ and the $k$ -th sample point is $a + k\cdot\dfrac{b-a}{n}$ . The area is

\int_a^b f(x)\,dx = \lim_{n \to \infty} \sum_{k=1}^{n} f\!\left(a + k\,\frac{b-a}{n}\right)\frac{b-a}{n}

General (Riemann) definition: the subintervals need not be equal and the sample point $c_k$ may be anywhere in the $k$ -th subinterval of width $\Delta x_k$ . A partition is a sequence of points $a = x_0 < x_1 < \cdots < x_n = b$ , and the norm $\lVert P \rVert$ is the width of the largest subinterval. Then

\int_a^b f(x)\,dx = \lim_{\lVert P \rVert \to 0} \sum_{k=1}^{n} f(c_k)\,\Delta x_k

The equal-width, right-endpoint version is a convenient special case.

Related ideas

Average value (height): the average height of $f$ on $[a,b]$ is $\dfrac{\text{area}}{\text{base}}$ — the height of a rectangle with the same base and the same area as the region.
Signed area: regions below the $x$ -axis contribute negative area (negative height $\times$ positive base), so an integral can cancel positive and negative parts.
Connection to antiderivatives: integrating a velocity over time gives displacement. The antiderivative evaluated at the endpoints, $F(b) - F(a)$ , gives the same number as the area computed by the limit of sums — a preview of the Fundamental Theorem.

Worked Example

Compute $\displaystyle\int_0^3 (x+1)\,dx$ from the definition as a limit of right-endpoint sums.

With $a=0$ , $b=3$ , the base is $\Delta x = \dfrac{3}{n}$ and the $k$ -th sample point is $x_k = \dfrac{3k}{n}$ , so $f(x_k) = \dfrac{3k}{n} + 1$ .

\begin{aligned} \int_0^3 (x+1)\,dx &= \lim_{n\to\infty} \sum_{k=1}^{n} \left(\frac{3k}{n}+1\right)\frac{3}{n} \\ &= \lim_{n\to\infty} \frac{3}{n}\left[\frac{3}{n}\sum_{k=1}^{n} k \;+\; \sum_{k=1}^{n} 1\right] \\ &= \lim_{n\to\infty} \frac{3}{n}\left[\frac{3}{n}\cdot\frac{n(n+1)}{2} \;+\; n\right] \\ &= \lim_{n\to\infty} \left[\frac{9(n+1)}{2n} + 3\right] \\ &= \lim_{n\to\infty} \left[\frac{9}{2} + \frac{9}{2n} + 3\right] \\ &= \frac{9}{2} + 0 + 3 = \frac{15}{2} \end{aligned}

This matches the geometric area: a $3\times 1$ rectangle (area $3$ ) plus a triangle of base $3$ and height $3$ (area $\tfrac{9}{2}$ ), totaling $\tfrac{15}{2}$ .

Summary

The area under a curve is approximated by rectangles; summing base $\times$ height over all rectangles gives an estimate, and letting the number of rectangles tend to infinity makes it exact.
Upper, lower, left, right, and midpoint sums all converge to the same area as the rectangles shrink — the sample-point choice stops mattering in the limit.
Sigma notation $\sum_{k=1}^{n} a_k$ compactly writes long sums; it obeys the sum/difference rule, the constant-multiple rule, and $\sum_{k=1}^{n} c = nc$ .
Three closed-form formulas — for $\sum k$ , $\sum k^2$ , $\sum k^3$ — let you evaluate finite sums without adding term by term, which is essential for computing integrals by hand.
The definite integral $\int_a^b f(x)\,dx$ is defined as the limit of Riemann sums $\sum f(c_k)\,\Delta x_k$ as the partition's norm goes to zero; equal widths with right endpoints is the standard working case.
Average value equals area over base, regions below the axis count as negative area, and evaluating an antiderivative as $F(b)-F(a)$ yields the same result as the limit of sums.