Question

Now, lets evaluate the same integral using power series. first, find the power series for the function f(x = \frac{32}{x^2+4}. then, integrate it from 0 to 2, and call it s. s should be an infinite series. what are the first few terms of s ?

Answer 1

No idea what the previous part of the problem is, but you have

$f(x)=\dfrac{32}{x^2+4}=\dfrac8{1-\left(-\frac{x^2}4\right)}=\displaystyle8\sum_{n\ge0}\left(-\frac{x^2}4\right)^n$
$f(x)=\displaystyle8\sum_{n\ge0}\left(-\dfrac14\right)^nx^{2n}$

which is valid for $\left|-\dfrac{x^2}4\right|$, or $|x|$. So the integral from 0 to 2 is

$\displaystyle\int_0^2f(x)\,\mathrm dx=\int_0^28\sum_{n\ge0}\left(-\frac14\right)^nx^{2n}\,\mathrm dx$
$=\displaystyle8\sum_{n\ge0}\left(-\frac14\right)^n\int_0^2x^{2n}\,\mathrm dx$

Note that since the power series only converges on the interval if $x$ is strictly less than 2, which means we have to treat this as an improper integral.

$=\displaystyle8\sum_{n\ge0}\left(-\frac14\right)^n\lim_{t\to2^-}\int_0^tx^{2n}\,\mathrm dx$[/tex]
$=\displaystyle8\sum_{n\ge0}\left(-\frac14\right)^n\lim_{t\to2^-}\frac{x^{2n+1}}{2n+1}\bigg|_{x=0}^{x=t}$
$=\displaystyle8\sum_{n\ge0}\frac{(-1)^n}{2^{2n}(2n+1)}\lim_{t\to2^-}t^{2n+1}$
$=\displaystyle16\sum_{n\ge0}\frac{(-1)^n}{2n+1}$
$=16-\dfrac{16}3+\dfrac{16}5-\dfrac{16}7+\dfrac{16}9+\cdots$