Integrals Of Constant Functions

The Integral of a Constant

A constant function has the same value at every point of its domain. Its integral over an interval is therefore a particularly direct application of the definition of the Riemann integral: each subinterval contributes the constant value multiplied by that subinterval’s width.

The order of the endpoints matters. The interval length used in the formula is the oriented length, \(b-a\), not the nonnegative distance between the endpoints. This distinction handles reversed endpoints and negative constants without changing the formula.

To use the integral formula, the function must be defined throughout the interval between the endpoints. Thus, if \(f(x)=c\) on \(I\), we require \(a,b\in I\). Because \(I\) is an interval, this ensures that every point between \(a\) and \(b\) lies in its domain. Without this condition, the integral may not be defined.

The Constant-Integrand Formula

For a constant function defined on an interval containing \(a\) and \(b\), the integral is the constant value multiplied by the oriented interval length:

The expression \(b-a\) records both the size and direction of the interval. When \(a<b\), it is positive; when \(b<a\), it is negative; and when \(a=b\), it is zero. The same formula applies in all three cases.

Why the Formula Works

First suppose \(a<b\). A tagged partition of \([a,b]\) consists of points \(a=x_0<x_1<\cdots<x_n=b\), together with a sample point \(\xi_i\in[x_{i-1},x_i]\) for each subinterval. The width of the \(i\)-th subinterval is \(\Delta x_i=x_i-x_{i-1}\). The corresponding Riemann sum is

Since \(f\) has the constant value \(c\), the choice of each sample point makes no difference: \(f(\xi_i)=c\). Therefore every tagged Riemann sum, for every partition, has the same value:

The sum of the subinterval widths telescopes: each interior partition point occurs once with a positive sign and once with a negative sign. Hence

It follows that every tagged Riemann sum equals \(c(b-a)\). In particular, as the partition mesh shrinks, the Riemann sums converge to \(c(b-a)\), so the Riemann integral exists and has that value. This argument works for any real \(c\), including \(c=0\) and negative \(c\).

For reversed endpoints, the signed-orientation convention gives \(\int_a^b f(x)\,dx=-\int_b^a f(x)\,dx\) when \(b<a\). Applying the result just proved to the increasing interval from \(b\) to \(a\) yields

If \(a=b\), the zero-length-interval rule gives \(\int_a^a c\,dx=0=c(a-a)\). Together, these cases establish the formula for every permitted pair of endpoints.

Worked Examples

Keeping Track of Endpoint Order

The most reliable way to avoid sign errors is to write the endpoint difference as \(b-a\) exactly as it appears in the formula. Do not replace it by the distance between the endpoints. The distance is nonnegative, whereas the oriented length records whether the bounds are in increasing or decreasing order.

The constant \(c\) and the oriented length \(b-a\) are separate factors. Either factor can be negative. If both are negative, their product is positive; if exactly one is negative, their product is negative. This is simply the sign arithmetic in \(c(b-a)\), not a reason to change the formula.

For example, the incorrect calculation \(\int_5^2 3\,dx=3(5-2)=9\) uses the distance \(5-2\) instead of the oriented difference. The upper endpoint is \(b=2\), and the lower endpoint is \(a=5\), so the correct calculation is

Check: By signed orientation, \(\int_5^2 3\,dx=-\int_2^5 3\,dx\). The integral with increasing endpoints is \(3(5-2)=9\), so reversing its bounds gives \(-9\), as required.

Using the Formula Reliably

For a constant-integrand integral, identify the constant and preserve the endpoint order. Then multiply the constant by the upper endpoint minus the lower endpoint. Before accepting the result, check that the constant function is defined throughout the interval between the endpoints and recompute the endpoint difference with its sign.

This method is useful because it turns a whole class of integral calculations into one exact formula. The derivation also explains why no choice of sample points or finer partition can change the answer: every subinterval uses the same function value, and the subinterval widths add to the total oriented length.