We know that the Taylor series expansion of #e^t# around #t=0# is:
#e^t = sum_(n=0)^oo t^n/(n!)#
Substituting #t=-x^3#:
#e^(-x^3) = sum_(n=0)^oo (-1)^n x^(3n)/(n!)#
and since this series has radius of convergence #R=oo# we can integrate term by term:
#int_0^1 e^(-x^3) dx = sum_(n=0)^oo (-1)^n int_0^1x^(3n)/(n!)dx#
#int_0^1 e^(-x^3) dx = sum_(n=0)^oo (-1)^n/(n!) [x^(3n+1)/(3n+1)]_0^1#
#int_0^1 e^(-x^3) dx = sum_(n=0)^oo (-1)^n/(n!) 1/(3n+1)#
If we stop the sum at the term #n=N#, Lagrange's theorem states that:
#int_0^1 e^(-x^3) dx = sum_(n=0)^N (-1)^n/(n!) 1/(3n+1)+ (-1)^(N+1)/((N+1)!) xi^(3N+1)/(3N+1)# with #0<=xi<=1#
As #xi<=1 => xi^(3N+1) <= 1# we have that:
#int_0^1 e^(-x^3) dx = sum_(n=0)^N (-1)^n/(n!) 1/(3n+1)+R_N#
where:
#abs(R_N) <= 1/((N+1)!)1/(3N+1)#
So:
#abs(R_1) <= 1/2*1/4 = 1/8#
#abs(R_2) <= 1/6*1/7 = 1/42#
#abs(R_3) <= 1/24*1/10 = 1/240 < 0.01#
So to have an error smaller than #0.01# we need four terms:
#int_0^1 e^(-x^3) dx ~= 1-1/4+1/14-1/60 = (420-105+30-7)/420=338/420=169/210#
