The power series for #e^x# is #sum_(n=0)^oox^n/(n!)#. Substituting in #x^2# for #x#, we get
#e^(x^2) = sum_(n=0)^oo(x^2)^n/(n!) = sum_(n=0)^oox^(2n)/(n!)#
Substituting this in, we have
#int_0^1e^(x^2)dx = int_0^1sum_(n=0)^oox^(2n)/(n!)dx = sum_(n=0)^ooint_0^1x^(2n)/(n!)dx#
where the second equality is valid due to #x^(2n)/(n!) >= 0# for all #x, n#.
Evaluating the new integral, we have
#sum_(n=0)^ooint_0^1x^(2n)/(n!)dx = sum_(n=0)^oo1/(n!)int_0^1x^(2n)dx#
#=sum_(n=0)^oo1/(n!)[x^(2n+1)/(2n+1)]_0^1#
#=sum_(n=0)^oo1/(n!(2n+1))#
At this point we only need to calculate enough terms of the above sum to get an answer accurate to #4# decimal places. We could either keep calculating them until it becomes obvious, or do a little more work to get a bound on how many terms we need. In either case, calculating enough terms shows us that
#sum_(n=0)^oo1/(n!(2n+1)) = 1.46265... ~~1.4627#
