Quantcast

Got Homework?

Connect with other students for help. It's a free community.

  • across
    MIT Grad Student
    Online now
  • laura*
    Helped 1,000 students
    Online now
  • Hero
    College Math Guru
    Online now

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

klimenkov

Why does the formula \[\int x^n\,dx=\frac{x^{n+1}}{n+1}+C\]give the wrong result if we put \(n=-1\) ?

  • one year ago
  • one year ago

  • This Question is Closed
  1. SandeepReddy
    Best Response
    You've already chosen the best response.
    Medals 0

    AS you only stated the formula, substituting n = -1 in the formula on RHS, gives undefined form

    • one year ago
  2. klimenkov
    Best Response
    You've already chosen the best response.
    Medals 0

    This not the answer I want to hear. Why does this formula give the right result if we put any \(n\) except \(n=-1\) ? May be there is any trick for \(n=-1\) ?

    • one year ago
  3. PeterPan
    Best Response
    You've already chosen the best response.
    Medals 3

    Because the antiderivative of x^{-1} is ln(x)

    • one year ago
  4. klimenkov
    Best Response
    You've already chosen the best response.
    Medals 0

    \[\int \frac1x\,dx=\frac{x^0}{0}+C=\frac10+C\]What is wrong with it?

    • one year ago
  5. PeterPan
    Best Response
    You've already chosen the best response.
    Medals 3

    Simply, 1/0 does not exist :)

    • one year ago
  6. klimenkov
    Best Response
    You've already chosen the best response.
    Medals 0

    \[\int \frac1x\,dx=\frac{x^0}{0}+C=\frac10+C=\ln x+C\]\[\frac10=\ln x\]

    • one year ago
  7. PeterPan
    Best Response
    You've already chosen the best response.
    Medals 3

    Maybe it's best to recall that we have this formula \[\int x^n\,dx=\frac{x^{n+1}}{n+1}+C\]BECAUSE of this formula \[\Large \frac{d}{dx}x^n=nx^{n-1}\]

    • one year ago
  8. PeterPan
    Best Response
    You've already chosen the best response.
    Medals 3

    Or, more relevantly \[\Large \frac{d}{dx}\frac{x^{n+1}}{n+1}=x^n\]

    • one year ago
  9. PeterPan
    Best Response
    You've already chosen the best response.
    Medals 3

    But you see that it no longer applies for n = -1

    • one year ago
  10. PeterPan
    Best Response
    You've already chosen the best response.
    Medals 3

    Because \[\Large \frac{d}{dx}\frac{x^{-1+1}}{-1+1}=\frac{d}{dx}\frac10\]and 1/0 never existed to begin with :/

    • one year ago
  11. klimenkov
    Best Response
    You've already chosen the best response.
    Medals 0

    Ok. It is time to tell you the truth.

    • one year ago
  12. PeterPan
    Best Response
    You've already chosen the best response.
    Medals 3

    I do love the truth :)

    • one year ago
  13. PeterPan
    Best Response
    You've already chosen the best response.
    Medals 3

    Noting of course, that there are many ways to explain this.

    • one year ago
  14. klimenkov
    Best Response
    You've already chosen the best response.
    Medals 0

    Try to compute\[\frac{d}{dx}\left(\frac{x^{n+1}-1}{n+1}\right)\]

    • one year ago
  15. PeterPan
    Best Response
    You've already chosen the best response.
    Medals 3

    Well, of course, it's still \[x^n\]

    • one year ago
  16. klimenkov
    Best Response
    You've already chosen the best response.
    Medals 0

    But now if you put \(n=-1\) what will you get?

    • one year ago
  17. PeterPan
    Best Response
    You've already chosen the best response.
    Medals 3

    Indeterminate.

    • one year ago
  18. klimenkov
    Best Response
    You've already chosen the best response.
    Medals 0

    Don't mind on indeterminate. Just compute this like a simple limit.

    • one year ago
  19. PeterPan
    Best Response
    You've already chosen the best response.
    Medals 3

    0/0

    • one year ago
  20. klimenkov
    Best Response
    You've already chosen the best response.
    Medals 0

    \(\frac{\sin n}{n}\) is \(\frac00\) too, but we all know that it is equal to 1. What is that limit?

    • one year ago
  21. PeterPan
    Best Response
    You've already chosen the best response.
    Medals 3

    I really don't know.

    • one year ago
  22. klimenkov
    Best Response
    You've already chosen the best response.
    Medals 0

    Do you know the L'hopital's rule?

    • one year ago
  23. PeterPan
    Best Response
    You've already chosen the best response.
    Medals 3

    No...

    • one year ago
  24. klimenkov
    Best Response
    You've already chosen the best response.
    Medals 0

    \[\lim_{n\rightarrow-1}\frac{x^{n+1}-1}{n+1}\]Anybody here know how to compute this limit? @ParthKohli

    • one year ago
  25. klimenkov
    Best Response
    You've already chosen the best response.
    Medals 0

    @PeterPan Do you know what is Wolfram Alpha?

    • one year ago
  26. PeterPan
    Best Response
    You've already chosen the best response.
    Medals 3

    That's why Google is here. :)

    • one year ago
  27. PeterPan
    Best Response
    You've already chosen the best response.
    Medals 3

    It seems its limit is ln(x)

    • one year ago
  28. klimenkov
    Best Response
    You've already chosen the best response.
    Medals 0

    Yes.

    • one year ago
  29. PeterPan
    Best Response
    You've already chosen the best response.
    Medals 3

    Your point?

    • one year ago
  30. klimenkov
    Best Response
    You've already chosen the best response.
    Medals 0

    Sorry, cant understand what means "Your point?".

    • one year ago
  31. PeterPan
    Best Response
    You've already chosen the best response.
    Medals 3

    Never mind. :)

    • one year ago
  32. PeterPan
    Best Response
    You've already chosen the best response.
    Medals 3

    You did say you were looking for a specific answer, so there was preference :P

    • one year ago
  33. PeterPan
    Best Response
    You've already chosen the best response.
    Medals 3

    Besides, you gave a specific example, which is not valid for proofs :) It has to be true for all constants. I think I'll stand by my original statements, interesting though this was...

    • one year ago
  34. klimenkov
    Best Response
    You've already chosen the best response.
    Medals 0

    Actually, the right expression is \[\int x^n\,dx=\frac{x^{n+1}-1}{n+1}+C\]

    • one year ago
  35. klimenkov
    Best Response
    You've already chosen the best response.
    Medals 0

    Now we can include \(n=-1\) also.

    • one year ago
  36. PeterPan
    Best Response
    You've already chosen the best response.
    Medals 3

    No you can't. It has to be \[\Large \int x^k\,dx=\lim_{k\rightarrow n}\frac{x^{n+1}-1}{n+1}+C\]

    • one year ago
  37. PeterPan
    Best Response
    You've already chosen the best response.
    Medals 3

    \[\Large \int x^k\,dx=\lim_{n\rightarrow k}\frac{x^{n+1}-1}{n+1}+C\]

    • one year ago
  38. PeterPan
    Best Response
    You've already chosen the best response.
    Medals 3

    Without the limit, it still won't make sense.

    • one year ago
  39. klimenkov
    Best Response
    You've already chosen the best response.
    Medals 0

    Very nice. But forget using limits. It is restrictions that really does not play a big role.

    • one year ago
  40. PeterPan
    Best Response
    You've already chosen the best response.
    Medals 3

    Finally. But I find it easier to just affirm that the modified power rule \[\Large \frac{d}{dx}\frac{x^{n+1}}{n+1}=x^n\]does not hold for n=-1 And just proving that the derivative of ln(x) is 1/x,

    • one year ago
  41. shubhamsrg
    Best Response
    You've already chosen the best response.
    Medals 3

    the diff is there since 1/x is not continuous at x=0 x^n always passes through (0,0) if it is not equal to -1, but for n=-1, the shape of graph is completely different. And, integration is simply the area under the graph. I know my logic is very wayward and not very strong, but still.. a point worth mentioning. :P

    • one year ago
  42. PeterPan
    Best Response
    You've already chosen the best response.
    Medals 3

    ^^

    • one year ago
    • Attachments:

See more questions >>>

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.