Problem: Approximate the integral ∫_{0}^{1} e^{x^{2}} dx to 2 significant figures.

Method: We will approximate the integrand by a Taylor polynomial — so that we can integrate it! — and use Taylor’s theorem with remainder to justify the answer.

Substituting x^{2} for z in the Taylor expansion of e^{z}, we get

e^{x^{2}} = 1 + x^{2} + \frac{x^{4}}{2} + \frac{x^{6}}{6} + R_{3}(x^{2}).

We can write out as many terms as we need; if this turns out to be not enough, we’ll come back for more later. Therefore,

∫_{0}^{1} e^{x^{2}} dx = ∫_{0}^{1}[1 + x^{2} + ⋯] dx

= [x + \frac{x^{3}}{3} + \frac{x^{5}}{10} + \frac{x^{7}}{42} ]_{0}^{1} + ∫_{0}^{1} R_{3}(x^{2}) dx.

= 1 + \frac{1}{3} + \frac{1}{10} + \frac{1}{42} + ∫_{0}^{1} e^{c} \frac{x^{8}}{24} dx.

At the last step we have used the standard formula for the Taylor remainder (“Version 2” in the terminology of this Web page), and the fact that f^{(4)}(z) = e^{z} if f(z) = e^{z}. The number c may depend on x, but in this problem it is always guaranteed to be between 0 and 1; therefore, e^{c} is less than e, and the remainder term in the integral is less than

e ∫_{0}^{1} \frac{x^{8}}{24} dx = \frac{e}{216} .

So, doing the arithmetic with the fractions, we conclude that

∫_{0}^{1} e^{x^{2}} dx ≈ 1.45714 ≈ 1.46

with a maximum error of

\frac{e}{216} = 0.01258.

So the accuracy is right on the edge of what we wanted. To be safe, one should go back and include one more term in the appoximation:

∫_{0}^{1} e^{x^{2}} dx ≈ ⋯ + \frac{1^{9}}{216} ≈ 1.46

with a maximum error of

∫_{0}^{1} R_{4}(x^{2}) dx ≤ \frac{e}{11 × 5!} ≈ 0.002.

You might compare this result with the value for the integral calculated numerically by Maple.