Problem
$$I(x) = \int_{1}^x \frac{e^t – 1}{t}$$
Find $I'( \sqrt{x} )$.
Solution
We know that $F'(x) = f(x)$ by the fundamental theorem of calculus so $$I'(x) = \frac{e^t -1}{t}$$ And so $$I'( \sqrt{x}) = \frac{ e^{\sqrt{x}} -1 }{ \sqrt{x}}$$
Problem
Find the fourth taylor polynomial of $I(x)$ around $0$ (Guess this makes it a Maclaurin series?)
Taylor series: $$ \sum_{n=o}^\infty \frac{ f^{(n)} (a) }{n!} (x-a)^n$$
So as we know the first derivative, we can compute the first Taylor expansion:
$$\frac{e^t-1}{t} (x)^n$$
Do I just continue like this? Am I doing this whole problem right? I'm just not feeling sure at all!
Best Answer
Here is a start. Using the power series of $e^t$, we have
$$ I(x) = \int_{1}^x \frac{e^t - 1}{t} = \int_{1}^x \sum_{k=1}^{\infty}\frac{t^{k-1}}{k!} dt = \sum_{k=1}^{\infty}\frac{1}{k!}\int_{1}^{x}{t^{k-1}} dt. $$
I think you can finish the problem.
Note: You only need a fourth degree polynomial.