Area as a limit
“…the smaller the bits the more of them there will be.”
From bounds to area
In the previous lesson, lower and upper rectangles bounded the area, with every sampled sum between those bounds. Narrower strips suggested a way to bring the bounds together. We now make that limiting step precise.
Continue with a continuous, nonnegative function \(f\) on \([a,b]\), where \(a<b\).
Adding division points while keeping the old ones is called refining a partition. A refinement cannot decrease the lower sum or increase the upper sum. On each smaller strip, the minimum can only rise and the maximum can only fall; the smaller widths still add up to the old width. Thus subdivision tightens the bounds, or leaves them unchanged.
To make the bounds approach one another, we need control over the width of every strip.
With unequal widths, simply increasing the number of strips is not enough. For example, keep \([0,1/2]\) as one strip while repeatedly subdividing \([1/2,1]\). The partition has more and more strips, but its mesh stays \(1/2\): half the interval never receives a finer approximation.
Theorem
The area limit
If \(f\) is continuous and nonnegative on \([a,b]\), its lower and upper sums approach a common limit as \(\lVert P\rVert\to0\), independent of the partitions. We define the area under the graph to be that common limit:
\[\begin{aligned} A&=\lim_{\lVert P\rVert\to0}\underline{S}(f,P)\\ &=\lim_{\lVert P\rVert\to0}\overline{S}(f,P). \end{aligned}\]In particular, equal-width lower and upper sums both approach \(A\) as \(n\to\infty\). Their gap tends to zero.
Every Riemann sum lies between these bounds, so the squeeze theorem also gives
\[\lim_{\lVert P\rVert\to0}\sum_{i=1}^n f(x_i^*)\Delta x_i=A.\]The sample points may change freely within their strips as the partition changes. The limit is independent of those choices.
As every strip becomes narrow, the lower and upper bounds close in on the same area.
The general theorem guarantees existence under the stated assumptions. We can also find a common limit directly in a particular case, without using an antiderivative.
Example 1
An exact area from the bounds
For \(f(x)=x^2\) on \([0,2]\), let \(n\) be any positive integer. Then \(\Delta x=2/n\) and \(x_i=2i/n\). Because the function increases, the maximum on strip \(i\) is \(f(x_i)\). Hence
\[\begin{aligned} \overline{S}_n&=\sum_{i=1}^n\left(\frac{2i}{n}\right)^2\frac2n\\ &=\frac8{n^3}\sum_{i=1}^n i^2. \end{aligned}\]The finite-sum identity
\[\sum_{i=1}^n i^2=\frac{n(n+1)(2n+1)}6\]lets us rewrite this as
\[\begin{aligned} \overline{S}_n&=\frac{4(n+1)(2n+1)}{3n^2}\\ &=\frac83+\frac4n+\frac4{3n^2}. \end{aligned}\]Derivation of the sum identity (optional)
First, pairing \(1,2,\ldots,n\) with the same list in reverse gives \(n\) pairs, each adding to \(n+1\). Therefore
\[\sum_{i=1}^n i=\frac{n(n+1)}2.\]Next expand the difference of consecutive cubes:
\[i^3-(i-1)^3=3i^2-3i+1.\]Add this equality for \(i=1,\ldots,n\). On the left, consecutive cubes cancel, leaving \(n^3\). On the right, add like terms:
\[\begin{aligned} n^3&=3\sum_{i=1}^n i^2\\ &\quad-\frac{3n(n+1)}2+n. \end{aligned}\]Solving for the sum of squares gives the stated identity. This is a finite algebraic calculation; no integration formula is being assumed.
The minimum on strip \(i\) is \(f(x_{i-1})\). Subtracting the lower sum from the upper sum cancels all interior division-point heights, leaving
\[\begin{aligned} \overline{S}_n-\underline{S}_n&=\Delta x\bigl(f(2)-f(0)\bigr)\\ &=\frac2n(4-0)=\frac8n. \end{aligned}\]Thus
\[\underline{S}_n=\frac83-\frac4n+\frac4{3n^2}.\]Both bounds have the same limit:
\[\begin{aligned} \lim_{n\to\infty}\underline{S}_n&=\frac83,\\ \lim_{n\to\infty}\overline{S}_n&=\frac83. \end{aligned}\]Since \(\underline{S}_n\le A\le\overline{S}_n\) for every \(n\), squeezing the area between these bounds gives
\[A=\frac83.\]Here \(n\) tends to infinity through positive integers. No finite number of these lower or upper rectangles gives the exact area; their limits do.
What the products measure
The axes need not both measure length. On a speed–time graph, a rectangle’s height is a speed and its width is a duration. Their product has units \((\mathrm{m/s})\,\mathrm s=\mathrm m\) and represents a distance.
Example 2
Distance from speed readings
A cyclist’s speed increases continuously over six seconds. A sensor records:
| Time (s) | Speed (m/s) |
|---|---|
| \(0\) | \(0\) |
| \(2\) | \(3\) |
| \(4\) | \(5\) |
| \(6\) | \(6\) |
How far did the cyclist travel? Since the speed never decreases, each interval’s first and last readings are its minimum and maximum. Multiplying each by the two-second duration and adding gives
\[\begin{aligned} \underline{S}&=2(0+3+5)=16\,\mathrm m,\\ \overline{S}&=2(3+5+6)=28\,\mathrm m. \end{aligned}\]Thus the distance \(D\) satisfies \(16\,\mathrm m\le D\le28\,\mathrm m\).
Below the axis: signs matter
Riemann sums and the lower and upper sum formulas still make sense for a continuous function with negative values. A negative height times a positive width contributes a negative number, so the sums now track signed contributions. Minimum heights still give the lower sum and maximum heights the upper sum, enclosing every sampled sum on the same partition.
For the line \(f(x)=x-1\) on \([0,3]\), the triangular region below the axis has area \(1/2\); the one above has area \(2\). The signed total is \(2-1/2=3/2\), while the total geometric area is \(2+1/2=5/2\). Using \(\lvert f(x)\rvert\) instead of \(f(x)\) makes every rectangle contribution nonnegative.
The next lesson gives the common limit of lower and upper sums its standard name, the definite integral, and develops its properties. The connection with antiderivatives will come with the fundamental theorem of calculus.