What is the Maclaurin series expansion for the integral from 0 to X ʃte^t dt?

Question

Given that the Maclaurin series for f(t)=te^t is Ʃ(n)/n!, which should be right

if you could explain or give easy steps on how you did this so i can understand it when it comes time to test.

Answer 1

The Maclaurin series for e^t is

∞
Ʃ(t^n)/n!
n=0

so the Maclaurin series for te^t is

∞
Ʃ(t^{n+1})/n!
n=0

Integrating this with respect to t gives

∞
Ʃ(t^{n+2})/[n!(n+2)]
n=0

At the upper limit of integration, X, this sum is

∞
Ʃ(X^{n+2})/[n!(n+2)]
n=0

and at the lower limit of integration the sum is zero, so the integral is given by

∞
Ʃ(X^{n+2})/[n!(n+2)]
n=0

Answer 2

te^t = ∑ t^(n+1) / n!
∫ te^t dt = ∑ t^(n+2) / [(n+2)n!]

Applying the limits 0 to X, we get
∑ X^(n+2) / [(n+2)*n!]