Answers · The mathematics underneath

How can you add up infinitely many infinitely thin slices and get a finite number?

In shortYou add small positive pieces, then take the finite total they approach as the pieces shrink; the pieces never all become zero before the adding.

You do not add actual zero-width pieces. You add pieces of small positive width, then ask what the finite totals approach as the width shrinks; the answer is the number they close in on. For a height rule the pieces are thin rectangles, and for a speed rule they are short distances, but the arithmetic is the same in both cases.

T \approx \sum_{i=1}^{n} f(x_i)\,\Delta x \qquad\longrightarrow\qquad T=\lim_{n\to\infty}\sum_{i=1}^{n}f(x_i)\,\Delta x

Rectangles, not zeroes

Read each term as a rectangle. The value $f(x_i)$ is a height and $\Delta x$ is a width, so the product is the area of one thin rectangle. On the interval from zero to one with the squaring rule, four rectangles taken at the left edges have total area $7/32$ , and four taken at the right edges have $15/32$ . The curve's area has to lie between those two numbers. The rectangles are not zero wide; they are small, and there are many. As they get thinner their tops follow the curve more closely, and the error is the area of the slivers between the rectangles and the curve. The upper-minus-lower gap measures exactly that error.

L_4=\frac14\left(0^2+\left(\frac14\right)^2+\left(\frac12\right)^2+\left(\frac34\right)^2\right)=\frac14\left(0+\frac1{16}+\frac14+\frac9{16}\right)=\frac{14}{64}=\frac7{32}\\[4pt] U_4=\frac14\left(\left(\frac14\right)^2+\left(\frac12\right)^2+\left(\frac34\right)^2+1^2\right)=\frac14\left(\frac1{16}+\frac14+\frac9{16}+1\right)=\frac{30}{64}=\frac{15}{32}

The true total for the squaring rule on zero to one is trapped between $7/32$ and $15/32$ . The gap is $15/32-7/32=8/32=1/4$ . Doubling the number of slices halves the gap. At sixteen slices the lower sum is $155/512$ and the upper sum is $187/512$ , so the gap is $32/512=1/16=0.0625$ . The number $1/3=0.333\ldots$ lies inside both intervals, and the intervals are closing on it.

FIG 1

number of slices	slice width	lowest-value sum	highest-value sum	gap
2	1/2	1/8	5/8	1/2
4	1/4	7/32	15/32	1/4
8	1/8	35/128	51/128	1/8
16	1/16	155/512	187/512	1/16

The natural guess is that infinitely many pieces of zero width must add to zero, or that infinitely many positive pieces must add to infinity. Neither follows, because the width is not set to zero before the adding. If you set $\Delta x=0$ first, every term is zero and there is no finite number of terms left to add; the expression has no value. The construction keeps $n$ finite, computes a real sum, and only then lets $n$ grow. The width and the count move together: at $n$ slices the width is $1/n$ , the piece near $x$ is about $x^2/n$ , and there are $n$ pieces. At $x=1$ that piece is $1/n$ ; at $n=16$ it is $1/16=0.0625$ , not zero. The final total is not the sum of the limiting pieces, because there are no final pieces. It is the value the finite sums are squeezed towards.

The squeeze and the notation

The squeeze is what makes the limit exist. Take the lowest value of the rule on each slice and the sum is certainly too small. Take the highest and the sum is certainly too big. For a rising rule the two sums differ by the total rise times the slice width. Here the rise is $1$ , so the gap is $1/n$ , and $1/n$ tends to zero. At $n=16$ the gap is $1/16=0.0625$ ; at $n=1000$ it is $1/1000=0.001$ . Exactly one number can remain inside all the traps, and that number is written with an integral sign: $\int_0^1 x^2\,dx$ . The formula for it is $b^3/3$ with $b=1$ , which gives $1^3/3=1/3$ .

\int_0^b x^2\,dx=\frac{b^3}{3}\qquad\text{so at }b=1,\quad \int_0^1x^2\,dx=\frac{1}{3}

The integral sign is an elongated S because it stands for a sum, but the sum it stands for is the limit of finite sums, not the sum of zeroes. The width $dx$ is a reminder of what was multiplied by the height before the limit was taken. The answer $1/3$ is finite because the interval has finite width, the heights stay bounded, and the error bound shrinks like $1/n$ .

Infinitely many positive pieces

Infinitely many positive pieces can also add to a finite total when they thin fast enough. Start at zero and at each step add $0.3$ of what is still missing. The pieces are $0.3$ , $0.21$ , $0.147$ , $0.1029$ , and so on, every one positive. After four steps the total is $0.3+0.21+0.147+0.1029=0.7599$ . The formula $S_k=1-(1-0.3)^k$ gives $1-0.7^4=1-0.2401=0.7599$ , and as $k$ grows $0.7^k$ tends to zero, so the infinite sum is $1$ . The pieces shrink fast enough. If they did not, the total would run away; infinitely many positive terms do not always have a finite sum.

Where it stops

The construction needs the trap to close. If the rule is unbounded on the interval, or oscillates so wildly that the upper and lower sums keep a fixed gap, there is no single trapped number and the total is not defined in this sense. Over an infinite interval even a simple positive rule can fail to have a finite total, because the slices may not thin in total width. So a finite answer is not a property of infinite sums as such. It is a property of a particular limit, where the width tends to zero and the number of pieces grows in the coordinated way that makes the error bound shrink.

Worth keeping

An integral is a limit of finite sums of height times width, not a sum of literal zero-width pieces.
For a rising rule, the gap between the lowest-value sum and the highest-value sum is bounded by the total rise times the slice width, so it closes as slices thin.
Infinitely many positive pieces can add to a finite total when they shrink fast enough, as a geometric series does.

Where this goes next

This question is one step of Change and Accumulation, a course that works through its subject from the beginning. The page that teaches this part of it is Adding Up Thin Slices, and it reads the whole way through whether or not you have an account.

What an account changes is the practice: problems standing under every idea, marked with a reason rather than a tick, in a copy of the course that is yours.

Read that page The whole course