# Thread: Integral of e^(2x-1) as an infinite series

1. ## Integral of e^(2x-1) as an infinite series

I am trying to figure out what I am even being asked to do here. I'm guessing that I'm supposed to integrate the Mclaurin Series? If so, is my answer at the bottom left correct?

2. ## Re: Integral of e^(2x-1) as an infinite series

When you are at this point:

$\displaystyle \int e^{2x-1}\,dx=\sum_{n=0}^{\infty}\frac{(2x-1)^{n+1}}{2(n+1)!}$

You might then observe that we can write:

$\displaystyle \int e^{2x-1}\,dx=\frac{1}{2}\left(\sum_{n=1}^{\infty}\frac{( 2x-1)^{n}}{n!}\right)=\frac{1}{2}\left(e^{2x-1}-1\right)$

3. ## Re: Integral of e^(2x-1) as an infinite series

As an addendum, we should probably write:

$\displaystyle \int e^{2x-1}\,dx=\sum_{n=0}^{\infty}\frac{(2x-1)^{n+1}}{2(n+1)!}+C$

And then we'd end up with:

$\displaystyle \int e^{2x-1}\,dx=\frac{1}{2}\left(e^{2x-1}-1\right)+C$

And it we define:

$\displaystyle C_1=C-\frac{1}{2}$

We could write:

$\displaystyle \int e^{2x-1}\,dx=\frac{1}{2}e^{2x-1}+C_1$