# Integral Problem help

• the7joker7
In summary, the conversation is about evaluating an integral and finding an estimate for the value of pi using the first few terms of an infinite series. The answer for part (a) should be in the form of kpi, where k is an integer. For part (b), the power series for the function f(x) = 40/(x^2 + 4) is needed, and the first few terms of the resulting infinite series should be found. Part (c) involves using the answers from parts (a) and (b) to approximate the value of pi by the first 5 terms of the infinite series. Lastly, part (d) asks for the upper bound of the error when using the first 10 terms of

## 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, let's 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 going to 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,

$${\int_0^2 f(x) dx} = 10 {\int_0^2 {\sum_n \left(\frac{-1}{4}\right)^n x^{2n} } dx} = 10{\sum_n \left(\frac{-1}{4}\right)^n \frac{2^{2n+1} }{2n+1}}$$

This should fix your problem. :)