anonymous
  • anonymous
calculate the integral
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
\[\int\limits_{-\infty}^{\infty}dz/z ^{2}+25\]
anonymous
  • anonymous
So?
TuringTest
  • TuringTest
I'm working on it did you try a trig sub\[z=5\tan\theta\]?

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anonymous
  • anonymous
Well we have a table ofontegrals at the back of the book and it shows how to find the diffeential. it is a spectial differential
anonymous
  • anonymous
(1/a) arctan(x/a)+c
anonymous
  • anonymous
a=5 over here
anonymous
  • anonymous
so i guess it wld be (1/5)arctan(x/5)
anonymous
  • anonymous
whoops replace the x with z
TuringTest
  • TuringTest
yeah, ok forgot about that so now it's an improper integral and needs to be split up as far as I know\[\lim_{n \rightarrow -\infty}\frac1 5\tan^{-1}(\frac z5) |_{n}^{0}+\lim_{n \rightarrow \infty}\frac1 5\tan^{-1}(\frac z5) |_{0}^{n}\]but I'm not sure how to do these limits, so there is probably another method... Who else has an idea?
anonymous
  • anonymous
well we plug in no?
anonymous
  • anonymous
so tan ^(-1)(0)=0
anonymous
  • anonymous
so we are left with (1/5)tan^(-1)(b/5)
anonymous
  • anonymous
turing u showed that n was positive both times but the first one shldbe negative and second time should be positive no?
TuringTest
  • TuringTest
no because they are different limits the first will go to -infinity, the second to positive so n doesn't have to be positive or negative, it makes no difference\[\lim_{n\rightarrow -\infty}\frac1 5\tan^{-1}(\frac z5) |_{n}^{0}+\lim_{n\rightarrow \infty}\frac1 5\tan^{-1}(\frac z5) |_{0}^{n}\]\[=\frac 15[\lim_{n\rightarrow -\infty}\tan^{-1}(\frac n5)+\lim_{n\rightarrow \infty}\tan^{-1}(\frac n5)] \]now each limit is pi/2 so the answer will be \[\frac\pi 5\]but my problem is I get\[\frac1 5(-\frac\pi 2+\frac\pi 2)=0\]somehow they are supposed to add though apparently, I must be missing something.
anonymous
  • anonymous
no one equation goes to negative infinity and the other to positive I think you solve the 2 parts of the equation sepaprately
TuringTest
  • TuringTest
the correct answer is pi/5, so it seems this is set up right I have done the limits separately, if that is what you mean. I just seem to be missing a negative sign somehwere
anonymous
  • anonymous
ok I will review it and try to see what went wrong Thanks:D I appreciate your help :D seriously i know it took long to type it out:D
TuringTest
  • TuringTest
anytime, sorry I am a bit preoccupied so I can't devote all my attention to finding the missing minus sign good luck!
anonymous
  • anonymous
k wait i have a question
anonymous
  • anonymous
\[\int\limits_{-\infty}^{0}f(x) +\int\limits_{0}^{\infty}f(x) = 2\int\limits_{0}^{\infty}f(x)\]
anonymous
  • anonymous
is this true?
anonymous
  • anonymous
Turing are you there?
TuringTest
  • TuringTest
sorry, hey I figure it out the integrand is even so we can rewrite this as\[2\int_{0}^{\infty}\frac{dz}{z^2+25}=\frac2 5\lim_{n \rightarrow \infty}\tan^{-1}\frac n 5=\frac{\pi}{5}\]...oh but you just wrote that nice job :D
TuringTest
  • TuringTest
...just because the integrand is even this is true
anonymous
  • anonymous
ohhhh thanks turing you r AWESOME :D

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