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Man, it's been so long since I did this. I am trying to do this:

NB: limits are $-\pi$ and 0, but I can't get the minus in the limits. If anybody knows how do to that please let me know, the $\pi$ symbol jumps into the integrand when I type minus before it...

$$ \frac{1}{\pi}\int^0_{-\pi} -t dt $$

I move the minus out of the integral: $$ -\frac{1}{\pi}\int^0_{-\pi} t dt $$

I do the integration:

$$ -\frac{1}{\pi} \cdot \left[\frac{t^2}{2}\right]^0_{-\pi} $$

I insert the limits, the final limit is zero, so the second term is removed:

$$ -\frac{1}{\pi} \cdot \frac{(-\pi)^2}{2} $$

$$ -\frac{1}{\pi} \cdot \frac{\pi^2}{2} $$

$$ -\frac{\pi}{2} $$

But this is incorrect. It should be:

$$ +\frac{\pi}{2} $$

So my question is: Do I have to change the signing when I do definite integrals on the left side of the y axis?

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  • $\begingroup$ @scott thanks, curly brackets. Got it! $\endgroup$ – DrOnline Apr 28 '13 at 23:34
  • $\begingroup$ No problem DrOnline! $\endgroup$ – Scott H. Apr 28 '13 at 23:35
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Note that you evaluate $$-\frac{1}{\pi}\left(\dfrac{t^2}{2}\right)\Big|_{-\pi}^0$$

Which is $$-\frac 1{\pi}\left[\left(\dfrac{0^2}{2}\right) - \left(\dfrac{(-\pi)^2}{2}\right)\right] = -\frac 1{\pi}\left[0 - \left(\dfrac{(-\pi)^2}{2}\right)\right] -\frac 1{\pi}\cdot -\frac{\pi^2}{2} = \frac{\pi}{2}$$

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  • $\begingroup$ Wow, I was doing F(a)-F(b) when it should be like you do F(b)-F(a) $\endgroup$ – DrOnline Apr 28 '13 at 23:29
  • $\begingroup$ Yes, that's all it was...just a simple mix-up in the order of evaluating at your bounds. Otherwise, you were "spot on" $\endgroup$ – Namaste Apr 28 '13 at 23:30
  • $\begingroup$ Thanks a lot. I can't vote up until I have 15 rep, so I'm gonna vote you up when I get to that, bookmarking for now. Thanks again! $\endgroup$ – DrOnline Apr 28 '13 at 23:32
  • $\begingroup$ You're welcome, DrOnline! $\endgroup$ – Namaste Apr 28 '13 at 23:32
  • $\begingroup$ Done! I was waiting for the timer to go out, one has to wait like 5 minutes after receiving the answer to accept it :D $\endgroup$ – DrOnline Apr 28 '13 at 23:33

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