anonymous
  • anonymous
Determine whether each integral is convergent or divergent. Evaluate 1) ∫ from -infinity to infinity (xe^-x^2) dx answer comes out to be 0 but I need step by step guidance 2) ∫ from -3 to 6 (x^-4) dx
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
Answer for #2 is divergent so I assume answer comes out to be infinity? How do I arrive to that answer? Thanks.
anonymous
  • anonymous
your function is \[\frac{1}{x^4}\] which is not defined at x = 0
anonymous
  • anonymous
so you have to compute \[\lim_{t\to 0^+}\int_t^6 \frac{1}{x^4}dx\]

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anonymous
  • anonymous
anti derivative of \[x^{-4}\] is \[-\frac{1}{3}x^{-3}=-\frac{1}{x^3}\] so you need \[\lim_{t\to 0^+}-\frac{1}{x^3}\] which does not exist
anonymous
  • anonymous
So integral of 1/x^4 then evaluate from t to 6 then plug in t for 0? May I ask why 0 is + and not just left as 0? OR is there a difference?
anonymous
  • anonymous
I see you didn't evaluate from t to 6. Was it not necessary?
anonymous
  • anonymous
you have to break up your integral in to two parts, from -3 to 0 and from 0 to 6 since the function is not defined at 0, you need the limit as t goes to zero actually twice, from the left and from the right
anonymous
  • anonymous
it is true that you can evaluate at \(\frac{1}{x^3}\) at \(x=6\) this is a finite number for sure, but i don't care because the limit as t goes to zero is infinite, so we will not get a number out of it

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