Average Values of Polynomial Functions

From Polynomial Values to an Interval Average

A polynomial can rise and fall across an interval, so its values at the endpoints alone do not generally describe its average value throughout that interval. The definite integral accounts for the function across the whole interval. For a polynomial, the integral is especially manageable: integrate each power term by term, evaluate at the bounds, and divide by the interval length.

This method produces an exact average whenever the polynomial coefficients and interval endpoints are given exactly. It also makes clear which part of the calculation represents the total accumulated value and which part converts that total into an average.

Every polynomial is continuous, so this definition applies to every polynomial on every closed interval with distinct endpoints. The average is a single number, not a new function. In particular, it is not usually the average of the endpoint values.

Integrating a Polynomial Term by Term

Write a polynomial in the form \(p(x)=\sum_{k=0}^{n}c_kx^k\), where \(n\) is a nonnegative integer and the \(c_k\) are constants. The power rule for integration gives an antiderivative term by term: each term \(c_kx^k\) contributes \(c_kx^{k+1}/(k+1)\). By the Fundamental Theorem of Calculus and the linearity of the definite integral, the average can therefore be found by evaluating that antiderivative at the interval endpoints and dividing by \(b-a\).

This expression is valid for all real \(a<b\) and all polynomial coefficients. It follows from the power rule and the Fundamental Theorem of Calculus; no assumption that the polynomial is increasing, decreasing, or symmetric is needed. When the coefficients and endpoints are rational numbers, the formula yields a rational number, which can be kept in exact form.

A Formula for the Average of a Power

For an individual monomial \(x^k\), with \(k\) a nonnegative integer, the formula becomes particularly direct. Integrating \(x^k\) gives \(x^{k+1}/(k+1)\). Evaluating at \(a\) and \(b\), then dividing by the interval length, gives:

To justify the formula, the function \(x^{k+1}/(k+1)\) is an antiderivative of \(x^k\), since its derivative is \(x^k\). The Fundamental Theorem of Calculus therefore gives \(\int_a^b x^k\,dx=(b^{k+1}-a^{k+1})/(k+1)\). Dividing by \(b-a\), which is positive because \(a<b\), proves the stated expression. Applying linearity to a finite sum then gives the polynomial formula.

There is another useful way to read the monomial formula. Since \(b^{k+1}-a^{k+1}=(b-a)(b^k+b^{k-1}a+\cdots+a^k)\), the interval-length factor cancels:

The sum in the numerator has \(k+1\) terms. This form can be convenient when the endpoints are simple numbers, but direct integration is just as reliable. The two forms are equivalent; use whichever keeps the arithmetic clearer.

Worked Examples

A Reliable Calculation Sequence

For an unfamiliar polynomial average, a consistent sequence prevents common errors. Keep the bounds and the interval length visible, and do not round intermediate values when an exact result is available.

Common Pitfall: Forgetting to Divide by the Length

The definite integral gives the accumulated value over the interval, not the average value by itself. For example, in the first worked example the integral is \(32/3\), but the interval has length \(2\). Reporting \(32/3\) as the average would confuse the total with the average.

The incorrect calculation is $$ p_{\mathrm{avg}}=\int_0^2 p(x)\,dx=\frac{32}{3}. $$ The correction is to include the interval-length divisor: $$ p_{\mathrm{avg}} = \frac{1}{2-0}\int_0^2p(x)\,dx = \frac{1}{2}\cdot\frac{32}{3} = \frac{16}{3}. $$ A quick check exposes the mistake: multiplying the claimed average \(32/3\) by the interval length \(2\) gives \(64/3\), not the actual integral \(32/3\). The correct average \(16/3\), multiplied by \(2\), gives \(32/3\).

Why the Method Is Useful

Polynomials are common models for quantities whose rates of change vary smoothly, and their averages summarize behavior across a whole interval with one number. The term-by-term rule also makes the calculation systematic: each coefficient contributes separately, and the bounds determine the contribution of each power. This is useful even when the polynomial has several terms or the interval does not begin at zero.

The integral formula does not require a graph or a table of sampled values. A few point evaluations can miss what happens between the points; the definite integral instead accounts for the function over the entire interval. Exact arithmetic also helps distinguish a true result from a decimal approximation. If a decimal is eventually needed for an application, simplify the exact value first and round only at the final step.

For every example, the final check has the same meaning: the computed average, multiplied by the interval length, recovers the definite integral. This relationship is a practical way to catch a missing divisor, an incorrect subtraction at the bounds, or an arithmetic error.