Taylor Approximation of Definite Integrals

Why Approximate an Integral with a Taylor Polynomial?

A definite integral can be difficult to evaluate exactly even when its integrand is easy to approximate near a chosen point. If a Taylor polynomial is accurate throughout the interval of integration, integrating that polynomial gives a simple approximation to the integral. The central tasks are to compute the polynomial integral correctly and to justify that the approximation is valid across the whole interval.

There are two distinct questions to keep in view. First, can the Taylor polynomial approximate the function on the interval with a controlled remainder? Second, if we start from an infinite Taylor series, does that series converge to the function throughout the interval? A polynomial can be integrated whether or not it approximates the function well; an error argument is what connects its integral to the desired one.

Integrating a Taylor Polynomial

Let \(f\) have derivatives through order \(N\) at a center \(c\). Its Taylor polynomial of degree \(N\) about \(c\) is

Integrating each of its finitely many terms gives an exact formula for the integral of the polynomial:

This formula does not by itself say how close the result is to \(\int_a^b f(x)\,dx\). To make that comparison, define the pointwise remainder \(R_N(x)=f(x)-T_N(x)\). Linearity of the integral gives

Thus the integral error is the integral of the pointwise Taylor remainder. In particular, cancellations in the remainder may make the signed error smaller than a bound obtained by integrating its absolute value.

When Does a Taylor Expansion Apply Throughout the Interval?

For a function represented by a power series about \(c\), the radius of convergence \(R\) describes where that series converges: it converges when \(|x-c|<R\), diverges when \(|x-c|>R\), and requires separate examination at points where \(|x-c|=R\). This statement concerns convergence of the series. It does not by itself establish that the sum equals a particular function \(f\).

For instance, a smooth function can have a Taylor series that converges everywhere but fails to equal the function except at its center. So an argument using an infinite Taylor series must establish both that the series converges on the interval and that its sum there is \(f\). One common sufficient setting is that \(f\) is analytic on a region containing the interval and its Taylor series about \(c\) represents \(f\) there.

If the interval \([a,b]\) lies strictly inside the radius of convergence, so that \(\max_{x\in[a,b]}|x-c|<R\), the power series converges uniformly on \([a,b]\). If its terms are continuous there, its partial sums are continuous, and their uniform limit \(f\) is continuous as well. Under these conditions, term-by-term integration is justified: the integrals of the partial sums converge to the integral of \(f\). If an endpoint lies on the radius boundary, convergence and the conditions for integration must be checked separately.

There is also a useful finite-polynomial approach that does not require an infinite series to represent \(f\). If \(c\in[a,b]\) and \(f^{(N+1)}\) is continuous on \([a,b]\), Taylor's theorem gives a pointwise remainder bound. In particular, if \(|f^{(N+1)}(x)|\leq M\) for every \(x\in[a,b]\), then

Integrating this inequality gives a bound on the definite-integral error:

This bound requires the stated derivative bound on the entire interval, not merely near \(c\). That requirement is important: a Taylor approximation is constructed locally, while the integral depends on values throughout \([a,b]\).

Worked Example: Approximating an Exponential Integral

Why the Integral Error Bound Works

The error estimate follows from the basic inequality for integrals: if \(R_N\) is integrable, then the absolute value of its integral cannot exceed the integral of its absolute value. A pointwise Taylor estimate can therefore be converted into an integral estimate without assuming that the remainder has a constant sign.

The method works because it preserves the distinction between local approximation and global integration. The Taylor theorem controls the error at each point; integration then accumulates those pointwise errors over the interval.

Worked Example: Integrating a Logarithm Near Its Expansion Center

A Series Must Represent the Function on the Whole Interval

A tempting shortcut is to integrate the terms of a Taylor series even when the interval extends beyond its radius of convergence. The next example shows why that is unsafe, and how changing the center can restore a valid expansion across the interval.

Common Pitfall: Treating Convergence as Equality

A power series can converge on an interval without being the Taylor series representation of the function there. Convergence only describes the behavior of the series; the equality between its sum and \(f(x)\) is an additional mathematical claim. Therefore, integrating a convergent Taylor series term by term is not justified until that equality and the conditions for integration have been established.

For example, it is not enough to argue, “the Taylor series has radius \(R\), and the interval lies inside that radius, so its integral is the desired integral.” The radius guarantees convergence inside the disk or interval, not equality with an arbitrary function having those Taylor coefficients. A safe alternative for a finite-degree approximation is to use Taylor's theorem with a derivative bound on the entire interval, as in the first two examples.

Another mistake is to bound the remainder only at the center. A bound at \(x=c\) says nothing about the rest of the interval. The integral estimate requires a bound valid for every \(x\in[a,b]\), or a carefully derived pointwise bound that can itself be integrated.