Definite Polynomial-Exponential Integrals by Tables

Why Endpoint Accounting Matters

Tabular integration by parts is especially efficient when one factor is a polynomial: repeated differentiation eventually reduces the polynomial to zero. For a product involving an exponential, repeated integration of the exponential also has a simple pattern. In a definite integral, however, finding an antiderivative is only part of the calculation. The final expression must be evaluated at both endpoints, with every term included and the subtraction in the correct order.

This tutorial develops a systematic table and endpoint formula for integrals of a polynomial times \(e^{ax}\). The coefficient \(a\) is assumed to be nonzero unless stated otherwise. As with any application of the Fundamental Theorem of Calculus, the polynomial and exponential are continuous on the finite interval of integration, so an antiderivative can be evaluated at the endpoints.

The Tabular Pattern

Start a table with successive derivatives of \(P(x)\) in one column and successive antiderivatives of \(e^{ax}\) in the other. Each integration of the exponential contributes another factor of \(1/a\). The signs on the diagonal products alternate, beginning with a positive sign.

If \(P\) has degree \(n\), then \(P^{(n+1)}(x)=0\), so the derivative column terminates. Multiplying across each row and applying its sign gives the antiderivative

For a definite integral, this entire antiderivative is evaluated at \(v\) and then at \(u\). Equivalently, each diagonal product contributes its own upper-endpoint value minus its lower-endpoint value. Keeping those endpoint evaluations visible helps prevent sign errors.

The expression in parentheses is always the value at the upper endpoint minus the value at the lower endpoint. The alternating sign and the power \(a^{k+1}\) belong to the \(k\)th derivative term.

Why the Formula Works

The table is a compact form of repeated integration by parts. Its signs alternate because each new integration-by-parts step subtracts an integral involving one more derivative of the polynomial. The following verification also proves that the displayed expression is an antiderivative.

This check is useful in practice: after forming the polynomial factor \(Q(x)\), verify that \(aQ(x)+Q'(x)=P(x)\). A correct table expression must pass this test before its endpoint values are trusted.

Worked Examples

Common Pitfall: Losing the Lower-Endpoint Sign

A frequent error is to evaluate the antiderivative at the upper endpoint and stop, or to add rather than subtract the lower-endpoint value. For example, in the first worked example, the upper value is \(\frac34e^2\), but the lower value is \(\frac34\), not zero. Writing the integral as \(\frac34e^2\) omits a boundary contribution. The correct calculation is

The subtraction applies to the whole lower-endpoint value. If that value is negative, subtracting it produces a plus sign, as in the second example: \(-5-(-4e)=4e-5\). Writing out both endpoint substitutions before simplifying makes this sign visible.

Conditions and a Practical Checklist

The table formula uses powers of \(1/a\), so it requires \(a\ne0\). If \(a=0\), then \(e^{ax}=1\) and the problem is simply \(\int_u^v P(x)\,dx\), evaluated with a polynomial antiderivative; the exponential table is unnecessary. For \(a\ne0\), the polynomial must have finite degree so repeated differentiation eventually gives zero.

This method matters because a definite integral can be computed without repeating integration by parts separately at each stage. The table organizes the repeated derivatives, the exponential integrations, and the alternating signs in one place. The endpoint formula then makes clear exactly what must be evaluated. These habits remain useful whenever tabular integration is applied to a finite interval.