Question #4cacf

1 Answer
Apr 10, 2017

We have:

#int_0^1 e^(-x^3) dx = sum_(n=0)^oo (-1)^n/(n!) 1/(3n+1)#

and we can stop the sum at #n=3# to have an error less than #0.01#:

#int_0^1 e^(-x^3) dx ~= 169/210#

Explanation:

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#