In this explanation, ∑ stands for the sum from n = 0 to infinity

The Maclaurin series for e^{x} is ∑ x^{n}/n!

Then, the Maclaurin series for e^{-x^2} is ∑ (-1)^{n}x^{2n}/n!

Using I[a,b] for the integral from a to b and E[a,b] for the evaluation from a to b,

I[0,1] x^{2n} dx is x^{2n+1} /(2n+1) E[0,1] = 1/(2n+1)

Thus, the integral from 0 to 1 of ∑(-1)^{n}x^{2n}/n! is ∑ (-1)^{n}/((2n+1)n!)

Then, as this is an alternating series, we only need to consider the next term (disregarding (-1)^{n}) . When we find the n such that the term is less than 10^{-6}, we need one less term; however, since we start with n = 0, that end up being the number of terms in the summation.

1/((2n+1)(n!)) < 10^{-6}

(2n+1)(n!) > 10^{6}

17*8! = 685440, and 19*9! = 6894720

We need to add, from n = 0 to 8, (-1)^{n}/((2n+1)n!) =

1 - 1/(3*1!) + 1/(5*2!) - 1/(7*3!) + 1/(9*4!) - 1/(11*5!) + 1/(13*6!) - 1/(15*7!) + 1/(17*8!) =

1 - 1/3 + 1/10 - 1/42 + 1/216 - 1/1320 + 1/9360 - 1/75600 + 1/685440 =

.746824