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Divergent or Convergent?

  1. Mar 23, 2014 #1
    1. The problem statement, all variables and given/known data

    Determine whether the integral is divergent or convergent. If it is convergent, evaluate it.

    ∫ from negative infinity to infinity of (x^8*e^-x^9)
    3. The attempt at a solution
    The answer is diverged to infinity. But I got that by guessing. Can someone explain to me why, please?

    100% lost on this problem.
    I changed the interval to 2∫ from 0 to infinity of (x^8*e^-x^9). I dont even know if I needed to do that or not.
     
    Last edited: Mar 23, 2014
  2. jcsd
  3. Mar 23, 2014 #2
    I'm not so sure that you can change the interval, actually -- you can only do that for even functions, and I think the $$e^{-x^{9}}$$ messes that up. Have you been able to do the anti-derivative of the function? If so, what did you get? If not, u-substitution looks like a good option.
     
  4. Mar 23, 2014 #3
    u=-x^9
    du=-9x^8

    -1/9∫from -inf to inf of (x^8e^u)
     
  5. Mar 23, 2014 #4
    Close -- with the exponential function you copy down the original function, then do the derivative of the exponent. So you would keep the e^-x^9, and then multiply that by the derivative of ##-x^{9}##

    Edit: Sorry, saw (I think you edited) your other stuff. Yes, you would have u be -x^9, and then that would be the correct du.
     
  6. Mar 23, 2014 #5
    I edited my above post. Could I also do it like that?
     
  7. Mar 23, 2014 #6
    and then how do i actually take integral from -inf to inf?
     
  8. Mar 23, 2014 #7
    Yes, that looks right. Make sure you replace ##x^{8}## with du in the integral though.
     
  9. Mar 23, 2014 #8
    okay, so would the integral = [1/(9x^9)e^-x^9)] if I did that correctly?
     
  10. Mar 23, 2014 #9
    in bounds from -inf to inf. This was the major part I was confused on. How to I do this bounds part to let me know it is diverging to infinity?
     
  11. Mar 24, 2014 #10
    No, I think you have an error in substituting. If you have ##u = -x^{9}##, then ##du = -9x^{8}dx##, so substituting into your integral you should get:$$-\frac{1}{9}\int e^{u}du$$
     
  12. Mar 24, 2014 #11
    but this is what I got that integral equaled. like after taking the antiderivative and subsituting back in u.
    [1/(9x^9)e^-x^9)]
     
  13. Mar 24, 2014 #12
    Well, first you want to integrate and get everything back in terms of x. Then you essentially just plug in the infinities as you would with other bounds, and then if it converges you'll get a finite answer, and an infinite answer otherwise.
     
  14. Mar 24, 2014 #13
    No, that isn't the right antiderivative though. You're missing a negative, and the ##x^{9}## shouldn't be there in the front.
     
  15. Mar 24, 2014 #14
    but thats my problem. when i plug in infinity, wouldnt it give me infinity, and then subtract from when i plug in -inf, and that gives me -inf and dont those cancel to equal 0?
     
  16. Mar 24, 2014 #15
    ok okay, the right antiderviative is -1/9e^-x^9
     
  17. Mar 24, 2014 #16
    i got it. thank you so much jackarms.
     
  18. Mar 24, 2014 #17
    Well, make sure you have the right antiderivative, to start. And no, not necessarily. ##\infinity - \infinity## is not zero -- it's indeterminate. Improper integrals like this one are really limits. Really you should evaluate it like this:
    $$\stackrel{lim}{a \rightarrow -\infty}\int^{0}_{a}f(x)dx + \stackrel{lim}{b \rightarrow \infty}\int^{b}_{0}f(x)dx$$
     
  19. Mar 24, 2014 #18
    Ah, okay. Glad you could work it out.
     
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