Maclaurin Series and Term-By-Term Integration

Question

Approximate the integral from 0 to 1 of e^(-x^2)dx with an accuracy of 10^-6 using the maclaurin series of e^x , replacement of x by -x^2, and term-by-term integration. You will have to figure out how many terms of the series must be used.

Answer 1

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

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

Then, the Maclaurin series for e^{-x^2} is ∑ (-1)ⁿx²ⁿ/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] x²ⁿ dx is x²ⁿ⁺¹ /(2n+1) E[0,1] = 1/(2n+1)

Thus, the integral from 0 to 1 of ∑(-1)ⁿx²ⁿ/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!) > 10⁶

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