Question

An Integral A Day:
To stay fresh over the summer, I'll be doing an integral a day. I figured it would be good motivation if I had some other people working on it as well, that way I'd be more likely to stay at it.
Here's today:
\[Int_0^{3}x \sin(\frac{n \pi x}{3})dx~~~~\text{n is an integer}\]

Answer 1

sorry: \[\int_0^3 x \sin(\frac{n \pi x}{3}) dx\]

Answer 2

i would say it equals \[-9\cos(n \pi)/n \pi \]
maybe i made some calculation mistakes, :)

Answer 3

That's what I got as well, but I evaluated it a little further:
\[\frac{9(-1)^n}{n\pi} \text{for n} \neq 0 \\ 0 ~~\text{for n} = 0\]

Answer 4

ya that's look better

Answer 5

slight correction: \[ \frac{9(-1)^{n+1}}{n\pi} \text{for n} \neq 0 \\ 0 ~~\text{for n} = 0\]

Answer 6

\[\frac{-9}{n \pi}\cos(n \pi)+\frac{9}{n^2 \pi^2}\sin(n \pi)\]

Answer 7

right cinar, but you can evaluate sin and cos at integer multiples of pi to simplify it.

Answer 8

sin nPi is = 0

Answer 9

\[\frac00\]

Answer 10

only if n is 0, but if n was zero, you wouldn't be doing the integration anyway

Answer 11

I'll let you know I tried - and failed. Haha! :) Good job

Answer 12

Maybe I should post a complete solution after a half hour or so?

Answer 13

\[\int_0^3 x \sin(\frac{n \pi x}{3})dx \\ \text{integration by parts} \\ = \frac{-3 x cos\frac{n \pi x}{3})}{n \pi}|_0^3 + \int_0^3 \frac{3}{n\pi}cos(\frac{n \pi x}{3}) dx\\ = \frac{-9 cos(n\pi)}{n \pi}+\frac{9 sin(\frac{n\pi x}{3})}{n^2\pi^2}| _0^3\\ = \frac{-9 cos(n\pi)}{n \pi}+\frac{9 sin(n\pi)}{n^2\pi^2}\\ \text{but} sin(n\pi) = 0~~and~~cos(n\pi) = (-1)^n~~so\\ = \frac{9(-1)^{n+1}}{n \pi} \]