Why Integrate a Taylor Approximation?
Some functions have definite integrals that are difficult to evaluate exactly, even when their Taylor series are familiar. A Taylor polynomial can replace the function by a finite sum of powers, which is straightforward to integrate. To make that estimate useful, we also need to control how much the polynomial differs from the function across the entire interval.
The key idea is to track the remainder before integrating. If the remainder is small everywhere on an interval, then its integral must also be small. A uniform bound gives a direct guarantee; a bound that varies with the input can often give a sharper guarantee.
From a Taylor Polynomial to an Integral Estimate
Suppose \(f\) has a Taylor polynomial \(T_n(x)\) of degree \(n\), centered at \(a\). Write the approximation with its remainder as
Here \(R_n(x)=f(x)-T_n(x)\). For an interval \([u,v]\), where \(u\le v\), define the exact integral and its Taylor-polynomial estimate by
By linearity of integration,
Thus the error in the integral estimate is precisely the integral of the Taylor remainder. If \(|R_n(x)|\le M\) everywhere on \([u,v]\), then the absolute integral error is at most the interval length times \(M\).
A Taylor remainder bound often comes from Taylor’s theorem with the Lagrange remainder. If \(f\) has \(n+1\) continuous derivatives on the interval between the center \(a\) and each \(x\) in \([u,v]\), then
for some \(\xi_x\) between \(a\) and \(x\). If \(|f^{(n+1)}(t)|\le K\) throughout the relevant interval, it follows that
This pointwise bound can be integrated directly. Alternatively, one can replace \(|x-a|^{n+1}\) by its maximum on \([u,v]\), giving a uniform bound \(M\), and then use the theorem above.
Why the Error Bound Works
The estimate follows from a basic property of the definite integral: the absolute value of an integral is no greater than the integral of the absolute value. Since \(R_n=f-T_n\), we have
This proves the uniform-bound result. If instead we know \(|R_n(x)|\le q(x)\), where \(q(x)\ge0\) on the interval, the same reasoning gives the potentially sharper estimate
The estimate does not require the remainder to keep the same sign. It bounds the error even if positive and negative contributions might cancel.
Worked Example: Estimating an Exponential Integral
Worked Example: Estimating an Exponential Integral
Use the degree-three Taylor polynomial for \(e^x\) centered at \(0\) to estimate \(\int_0^1 e^x\,dx\). Give a guaranteed error bound.
The degree-three Taylor polynomial is
Integrate each term on \([0,1]\):
To check the integration, an antiderivative of \(T_3\) is \(x+x^2/2+x^3/6+x^4/24\). Its derivative is \(T_3(x)\), and its change from \(0\) to \(1\) is \(41/24\).
For a degree-three Taylor polynomial, the Lagrange remainder is
for some \(\xi_x\in[0,x]\). Since \(e^{\xi_x}\le e\) and \(x^4\le1\) on \([0,1]\),
The interval has length \(1\), so the integral error is at most \(1\cdot e/24=e/24\). Therefore
In particular, the exact integral is guaranteed to lie between \(41/24-e/24\) and \(41/24+e/24\). The estimate is justified by a bound valid throughout the full interval, not just at one selected input.
Worked Example: A Pointwise Bound for a Composite Function
Worked Example: Estimating an Integral Involving \(\sin(x^2)\)
Estimate \(\int_0^1\sin(x^2)\,dx\) using the degree-four Taylor polynomial for \(\sin t\) centered at \(0\), and bound the error.
That Taylor polynomial is \(t-t^3/6\); the even-degree term is zero. Substituting \(t=x^2\) gives the approximation
Integrating the polynomial gives
An antiderivative of the polynomial is \(x^3/3-x^7/42\). Differentiating gives \(x^2-x^6/6\), and evaluating the antiderivative from \(0\) to \(1\) confirms the estimate \(13/42\).
The fifth derivative of \(\sin t\) is \(\cos t\), whose absolute value is at most \(1\). Taylor’s theorem therefore gives
for \(t\in[0,1]\). Since \(t=x^2\), the remainder in the integrand is bounded by \(x^{10}/120\). Integrating this pointwise bound gives
Thus \(13/42\) is an estimate with error at most \(1/1320\). Integrating the pointwise bound preserves the factor \(x^{10}\); replacing it by its maximum first would give a valid but less precise estimate.
Worked Example: Bounding the Error for a Logarithm
Worked Example: Estimating an Integral of \(\ln(1+x)\)
Estimate \(\int_0^{1/2}\ln(1+x)\,dx\) using its degree-three Taylor polynomial at \(0\), and find an error bound by integrating a pointwise remainder bound.
The polynomial is
Integrating term by term gives
An antiderivative of \(T_3\) is \(x^2/2-x^3/6+x^4/12\). Its derivative is \(x-x^2/2+x^3/3\), and its change from \(0\) to \(1/2\) is \(7/64\).
The fourth derivative of \(f(x)=\ln(1+x)\) is
On \([0,1/2]\), its absolute value is at most \(6\). The degree-three Lagrange remainder therefore satisfies
Integrating this bound over the interval,
Hence the exact integral lies within \(1/640\) of \(7/64\). For comparison, replacing \(x^4\) by its maximum \(1/16\) before integrating would give the larger bound \((1/2)(1/64)=1/128\).
When Term-by-Term Integration Applies
A Taylor polynomial is finite, so it can always be integrated term by term. If instead we use an infinite power series to represent a function, term-by-term integration requires an appropriate convergence condition. In particular, a power series converges uniformly on every closed interval lying strictly inside its interval of convergence, and there it may be integrated term by term.
For an approximation, the practical procedure is to choose a Taylor polynomial, integrate its terms, and then use Taylor’s theorem to bound the remainder on the whole interval of integration. If the interval reaches a boundary of the power series’ interval of convergence, the endpoint behavior must be checked; convergence inside the interval alone does not settle what happens at the endpoint.
Use a center and degree for which the polynomial and a valid remainder estimate are available.
Compute its definite integral term by term and keep exact values where possible.
Use the maximum derivative required by Taylor’s theorem, or a pointwise bound that depends on \(x\).
Integrate the pointwise bound, or multiply a uniform bound by the interval length.
A Common Error: Using a Derivative Bound at Only the Center
Taylor’s remainder uses a derivative value at an intermediate point, not necessarily at the center. The derivative bound must hold throughout the relevant interval. For example, for the degree-three approximation to \(e^x\) on \([0,1]\), it would be incorrect to use \(e^0=1\) as a bound for the fourth derivative throughout the interval and claim \(|R_3(x)|\le x^4/24\). The fourth derivative is \(e^x\), which increases from \(1\) to \(e\), so it is not bounded above by \(1\) on \([0,1]\).
The correct bound uses \(e^x\le e\) throughout that interval and gives \(|R_3(x)|\le ex^4/24\). A related pitfall is to report only an integral approximation without explaining its error bound. The polynomial integral supplies an estimate; the remainder analysis supplies the guarantee about its accuracy.
Using the Estimate
If the polynomial integral is \(I_n\) and the error bound is \(E\), then the exact integral is guaranteed to satisfy
The estimate need not equal the exact integral. Its value is that the error is controlled under stated conditions. In applications, a sharper bound may be useful when a required tolerance is small. The pointwise remainder method can sharpen a uniform estimate because it accounts for how the error changes across the interval instead of treating it as equally large everywhere.
Check Your Understanding
For each question, identify both the polynomial integral and the reasoning that controls its error.
- If \(|f(x)-T_n(x)|\le 0.02\) on an interval of length \(3\), what bound follows for the integral error?
- Why must a Lagrange remainder estimate use a bound on the relevant derivative throughout the interval rather than only at the Taylor center?
- For a pointwise remainder bound \(|R(x)|\le x^2\) on \([0,1/2]\), what integral error bound follows by integrating the pointwise estimate?
- What additional issue must be checked before integrating a power series term by term on an interval that includes an endpoint of its interval of convergence?
- In the logarithm example, why does integrating the pointwise remainder bound give a sharper error estimate than using its maximum over the whole interval?
