Constant Functions from Riemann Sums

Why Constant Functions Are a Useful Starting Point

A Riemann sum combines function values with the widths of subintervals. For a general function, the function values may vary from one tag to another. For a constant function, every sampled value is the same. This makes it possible to see exactly how the subinterval widths combine, even when the partition is uneven.

The resulting calculation is simple, but it captures the central mechanism in the definition of the definite integral: tagged sums over partitions with increasingly small mesh approach a single value. In this case, the sums already have that value for every partition and every permitted choice of tags.

Partitions, Tags, and Tagged Sums

Let \(a<b\). A partition \(P\) of \([a,b]\) is a finite increasing list of points

The \(i\)th subinterval is \([x_{i-1},x_i]\), and its width is \(\Delta x_i=x_i-x_{i-1}\). A tag \(\xi_i\) for that subinterval is any point in \([x_{i-1},x_i]\). The tagged Riemann sum of a function \(f\) for this partition and these tags is

The mesh of the partition, written \(\|P\|\), is the largest subinterval width:

In the tagged-sum definition of the Riemann integral, \(f\) is integrable on \([a,b]\) with integral \(I\) if, for every \(\varepsilon>0\), there is a \(\delta>0\) such that every partition \(P\) with \(\|P\|<\delta\), and every allowed choice of tags, satisfies \(|S(f,P,\xi)-I|<\varepsilon\).

The Tagged Sum for a Constant Function

If \(f(x)=c\), then every tag satisfies \(f(\xi_i)=c\). Therefore the tagged sum becomes

The sum of the widths is \(b-a\). To see why, write out its terms: each interior partition point appears once with a positive sign and once with a negative sign, so those terms cancel. Only the right endpoint and left endpoint remain.

Consequently, the tagged sum has the same value for every partition and all choices of tags. This is an exact equality, not an approximation that becomes accurate only when the mesh is small.

Worked Example: Unequal Widths and Different Tags

Worked Example: A Negative Constant

Deriving the Integral from Its Definition

The tagged-sum identity directly proves the integral formula. To verify the definition, let \(\varepsilon>0\) be arbitrary and set \(I=c(b-a)\). Choose any \(\delta>0\), for example \(\delta=1\). Whenever \(P\) has mesh \(\|P\|<\delta\), the identity above gives \(S(f,P,\xi)=I\) for every choice of tags. Hence

This verifies the required condition for every positive \(\varepsilon\). In fact, the mesh restriction is not needed for a constant function: the equality holds even for partitions whose subintervals are large. We may therefore conclude directly from the definition that the function is Riemann integrable and its integral is \(c(b-a)\).

Worked Example: Applying the Formula on Any Partition

A Common Mistake: Counting Subintervals Instead of Width

A frequent error is to multiply the constant by the number of subintervals while forgetting that each function value must also be multiplied by its subinterval width. For instance, on the partition \(0<1<3\), the function \(f(x)=4\) has two subintervals. The incorrect calculation \(4+4=8\) treats both widths as if they were \(1\).

The actual widths are \(1-0=1\) and \(3-1=2\). The correct tagged sum, for any tags, is

Checking the total width gives \(1+2=3\), and \(4\cdot3=12\), confirming the result. The number of subintervals is not generally the length of the interval; the widths, which may differ, are what must be summed.

A related point is that tags are used to select function values, not to determine widths. Here each tag gives the value \(4\), but the contribution of a subinterval is still that value multiplied by the width of that subinterval.

Why the Exact Identity Matters

The constant function provides a clear example of how a tagged-sum limit works. For a varying function, changing tags can change individual sampled values, and the mesh condition is used to control the resulting sums. For a constant function, there is no variation in sampled values, and the width calculation settles the entire question at once.

The result also gives a basic reference point for later computations: on an interval of length \(b-a\), a constant value \(c\) contributes exactly \(c(b-a)\). The formula applies whether \(c\) is positive, negative, or zero, because the proof uses only the equality of all function values and the telescoping sum of the widths.