6.4.5

Instructor: Example 6-- express the antiderivative of e to the negative x squared dx as an infinite series. Well to begin with, we want to use the tactic we had in some previous questions to determine what e to the negative x squared is. E to the negative x squared would be the e to the x series, except we're going to replace x to the n over n factorial with negative x squared to the n or n factorial. 

So that should be-- well, it's alternating for one, negative 1 to the n, x to 2n or n factorial. I'm going to go and write this out term by term. That would be 1 minus x squared over 1 factorial plus x to the fourth over 2 factorial minus x to the sixth over three factorial. So notice my denominators are going 1, 2, 3, 1, 2, 3 because of this. 

But the exponents are even. And of course, we're alternating. So all of the even terms are going to be positive. And the others are going to be negative. All terms will be negative. 

All right, so that continues on-- actually, let's do one more. X to the 8 over n factorial, that continues in that pattern. So that the antiderivative would be we first have our constant integration plus that would be an x term, minus x cubed. Adding 1, dividing by 2 over 2 terms 1 factorial, dividing by 3 plus x to the fifth over 5 times 2 factorial minus to the 7 over 7 times 3 factorial plus x to the 9 over 9 times 4 factorial minus. So there it is as an infinite series. 

Now let's say if we can use that to evaluate the integral from 0 to 1 of e to the negative x squared dx to within an error of 0.01. If we evaluate 0 first, we would get 0. This entire thing-- we ignore the constant immigration when we're doing this. So this would just be 0. So we'll have minus 0. 

Now evaluating that same series at 1, we would have 1 minus 1/3 plus 1/10 minus 1/42 plus 1/216 minus some other terms after that. Now I stop there because theorem 5.14 tells me that because 1/216 is less than 0.01, the remainder of that fourth term is less than 0.1. So that means that that's where the error comes in. 

1/216-- so the error of this is less than or equal to this, which is less than 0.01, which tells me that I can stop at that 1/216 term. So this integral it's approximately 1 minus 1/3 plus 1/10 minus 1/42 plus 1/216, which is 0.74749.