Question

Find a power series representation for the antiderivative of the integral below

integral from 0 to pi of 8sin(x) / x dx

Answer 1

\[\int_0^\pi8{\sin x\over x}{\rm d}x\]

Answer 2

is that it?

Answer 3

yes

Answer 4

\[\sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-...\]
\[\frac{\sin(x)}{x}=1-\frac{x^2}{3!}+\frac{x^4}{5!}\] at least to three terms

Answer 5

integrate term by term, then compute

Answer 6

so I can't substitute the summation equivalent for f(x) = sin(x) ?

Answer 7

Integrating term by term I got \[\sum_{n=0}^{\Pi} \frac{ (8)(-1)^{2n} (x)^{2n+1} }{ (2n+1)(2n+1)! }\]

Answer 8

but it's incorrect

Answer 9

It's not inccorect ... I swear

Answer 10

okay you have a number to compute, not an anti derivative
i assume you are going to compute this as a number, so once you have the power series for \(\frac{\sin(x)}{x}\) and integrate term by term, you have to pick some point at which to stop to evaluate

Answer 11

the power series you get is not the sum from \(0\) to \(\pi\), it is from \(0\) to \(\infty\)

Answer 12

i think you have confused two things, one is the power series, one is how to evaluate the integral

Answer 13

i stopped at 3 terms
\[\frac{\sin(x)}{x}=1-\frac{x^2}{3!}+\frac{x^4}{5!}\] then
\[\int \frac{\sin(x)}{x}=1-\frac{x^3}{18}+\frac{x^5}{600}\] evaluate at \(\pi\) and \(0\) and subtract

Answer 14

damn that was wrong \[\int \frac{\sin(x)}{x}dx=x-\frac{x^3}{18}+\frac{x^5}{600}\]

Answer 15

i see! i didn't integrate from 0 to pi

Answer 16

yeah you have to compute some number
i.e. you are going to make an approximation

Answer 17

so would the summation be \[\sum_{n=0}^{\infty} \frac{ \pi ^{2n+1} }{(2n+1)(2n+1)! }\]

Answer 18

i think you are missing a \((-1)^n\) term

Answer 19

\[\sum_{n=0}^{\infty} \frac{(-1)^{n+1} \pi ^{2n+1} }{(2n+1)(2n+1)! }\] might work

Answer 20

no that wont work

Answer 21

maybe just
\[\sum_{n=0}^{\infty} \frac{(-1)^{n} \pi ^{2n+1} }{(2n+1)(2n+1)! }\]

Answer 22

yeah, that is the one

Answer 23

yeah you're right about the (-1)^n. i'll try it and let you know.

Answer 24

all of that times 8 right?

Answer 25

i checked, it is right
take a look
http://www.wolframalpha.com/input/?i= \sum_{n%3D0}^{\infty}+\frac{%28-1%29^{n}+\pi+^{2n%2B1}+}{%282n%2B1%29%282n%2B1%29!+}
compare to
http://www.wolframalpha.com/input/?i=intgral+0+to+pi+sin%28x%29%2Fx

Answer 26

yes, times 8

Answer 27

yes! it's correct!

Answer 28

thank you so much!