We know that the Taylor series expansion of e^tet around t=0t=0 is:
e^t = sum_(n=0)^oo t^n/(n!)et=∞∑n=0tnn!
Substituting t=-x^3t=−x3:
e^(-x^3) = sum_(n=0)^oo (-1)^n x^(3n)/(n!)e−x3=∞∑n=0(−1)nx3nn!
and since this series has radius of convergence R=ooR=∞ 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∫10e−x3dx=∞∑n=0(−1)n∫10x3nn!dx
int_0^1 e^(-x^3) dx = sum_(n=0)^oo (-1)^n/(n!) [x^(3n+1)/(3n+1)]_0^1∫10e−x3dx=∞∑n=0(−1)nn![x3n+13n+1]10
int_0^1 e^(-x^3) dx = sum_(n=0)^oo (-1)^n/(n!) 1/(3n+1)∫10e−x3dx=∞∑n=0(−1)nn!13n+1
If we stop the sum at the term n=Nn=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)∫10e−x3dx=N∑n=0(−1)nn!13n+1+(−1)N+1(N+1)!ξ3N+13N+1 with 0<=xi<=10≤ξ≤1
As xi<=1 => xi^(3N+1) <= 1ξ≤1⇒ξ3N+1≤1 we have that:
int_0^1 e^(-x^3) dx = sum_(n=0)^N (-1)^n/(n!) 1/(3n+1)+R_N∫10e−x3dx=N∑n=0(−1)nn!13n+1+RN
where:
abs(R_N) <= 1/((N+1)!)1/(3N+1)|RN|≤1(N+1)!13N+1
So:
abs(R_1) <= 1/2*1/4 = 1/8|R1|≤12⋅14=18
abs(R_2) <= 1/6*1/7 = 1/42|R2|≤16⋅17=142
abs(R_3) <= 1/24*1/10 = 1/240 < 0.01|R3|≤124⋅110=1240<0.01
So to have an error smaller than 0.010.01 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∫10e−x3dx≅1−14+114−160=420−105+30−7420=338420=169210
