# Integral Problem help!

## Homework Statement

Evaluate the integral...

int(0 to 2) of 40/(x^2 + 4)dx

a: Your answer should be in the form  kpi , where  k  is an integer. What is the value of  k ?

b: Now, lets evaluate the same integral using power series. First, find the power series for the function  f(x) = (40)/(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 ?

c: The answers to part (a) and (b) are equal (why?). Hence, if you divide your infinite series from (b) by  k  (the answer to (a)), you have found an estimate for the value of  pi  in terms of an infinite series. Approximate the value of  pi  by the first 5 terms.

d: What is the upper bound for your error of your estimate if you use the first 10 terms? (Use the alternating series estimation.)

## The Attempt at a Solution

k is 5, this is correct, I know how to get k...

the first term of the series is 20, I know how to get that, but then the rest of my series goes wacky.

Here's my equation for the series

the sum of (-1)^n * ((2)^(2n + 1)/(10^(n - 1) * (2n + 1))

I can show how I got that in case anyone wants, but I think it's fairly obvious.

And at the end of that, using exact integral notation, there'd be the same equation but instead of that first 2 there'd be a 0, but that simplifies to 0 so it doesn't matter.

When I use n = 0, I get 20 like I'm supposed to, but when I use n = 1, i get -2.6667...answer is -6.6667

my next term is .64, should be getting 4...then -.182857, should get a -2.85714285714...then .0568888889, should get a 2.22222222222...

For c and d of course I'm gonna be off, but I got 3.5694 and 3.64722E-6 respectively. The correct answers are 3.33968253968 and 0.190476190476.

Help?

OK. So, first of all, your series is close, but not correct. Since $$\frac{40}{x^2+4}$$ is the same as $$\frac{10}{1+\frac{x^2}{4}}$$, for all x in the interval of convergence,
$${\int_0^2 f(x) dx} = 10 {\int_0^2 {\sum_n \left(\frac{-1}{4}\right)^n x^{2n} } dx} = 10{\sum_n \left(\frac{-1}{4}\right)^n \frac{2^{2n+1} }{2n+1}}$$