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klimenkov

  • 3 years ago

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

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  1. SandeepReddy
    • 3 years ago
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    AS you only stated the formula, substituting n = -1 in the formula on RHS, gives undefined form

  2. klimenkov
    • 3 years ago
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    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 \(n=-1\) ?

  3. PeterPan
    • 3 years ago
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    Because the antiderivative of x^{-1} is ln(x)

  4. klimenkov
    • 3 years ago
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    \[\int \frac1x\,dx=\frac{x^0}{0}+C=\frac10+C\]What is wrong with it?

  5. PeterPan
    • 3 years ago
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    Simply, 1/0 does not exist :)

  6. klimenkov
    • 3 years ago
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    \[\int \frac1x\,dx=\frac{x^0}{0}+C=\frac10+C=\ln x+C\]\[\frac10=\ln x\]

  7. PeterPan
    • 3 years ago
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    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 \[\Large \frac{d}{dx}x^n=nx^{n-1}\]

  8. PeterPan
    • 3 years ago
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    Or, more relevantly \[\Large \frac{d}{dx}\frac{x^{n+1}}{n+1}=x^n\]

  9. PeterPan
    • 3 years ago
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    But you see that it no longer applies for n = -1

  10. PeterPan
    • 3 years ago
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    Because \[\Large \frac{d}{dx}\frac{x^{-1+1}}{-1+1}=\frac{d}{dx}\frac10\]and 1/0 never existed to begin with :/

  11. klimenkov
    • 3 years ago
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    Ok. It is time to tell you the truth.

  12. PeterPan
    • 3 years ago
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    I do love the truth :)

  13. PeterPan
    • 3 years ago
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    Noting of course, that there are many ways to explain this.

  14. klimenkov
    • 3 years ago
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    Try to compute\[\frac{d}{dx}\left(\frac{x^{n+1}-1}{n+1}\right)\]

  15. PeterPan
    • 3 years ago
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    Well, of course, it's still \[x^n\]

  16. klimenkov
    • 3 years ago
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    But now if you put \(n=-1\) what will you get?

  17. PeterPan
    • 3 years ago
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    Indeterminate.

  18. klimenkov
    • 3 years ago
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    Don't mind on indeterminate. Just compute this like a simple limit.

  19. PeterPan
    • 3 years ago
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    0/0

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

  21. PeterPan
    • 3 years ago
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    I really don't know.

  22. klimenkov
    • 3 years ago
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    Do you know the L'hopital's rule?

  23. PeterPan
    • 3 years ago
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    No...

  24. klimenkov
    • 3 years ago
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    \[\lim_{n\rightarrow-1}\frac{x^{n+1}-1}{n+1}\]Anybody here know how to compute this limit? @ParthKohli

  25. klimenkov
    • 3 years ago
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    @PeterPan Do you know what is Wolfram Alpha?

  26. PeterPan
    • 3 years ago
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    That's why Google is here. :)

  27. PeterPan
    • 3 years ago
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    It seems its limit is ln(x)

  28. klimenkov
    • 3 years ago
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    Yes.

  29. PeterPan
    • 3 years ago
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    Your point?

  30. klimenkov
    • 3 years ago
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    Sorry, cant understand what means "Your point?".

  31. PeterPan
    • 3 years ago
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    Never mind. :)

  32. PeterPan
    • 3 years ago
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    You did say you were looking for a specific answer, so there was preference :P

  33. PeterPan
    • 3 years ago
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    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...

  34. klimenkov
    • 3 years ago
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    Actually, the right expression is \[\int x^n\,dx=\frac{x^{n+1}-1}{n+1}+C\]

  35. klimenkov
    • 3 years ago
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    Now we can include \(n=-1\) also.

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

  37. PeterPan
    • 3 years ago
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    \[\Large \int x^k\,dx=\lim_{n\rightarrow k}\frac{x^{n+1}-1}{n+1}+C\]

  38. PeterPan
    • 3 years ago
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    Without the limit, it still won't make sense.

  39. klimenkov
    • 3 years ago
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    Very nice. But forget using limits. It is restrictions that really does not play a big role.

  40. PeterPan
    • 3 years ago
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    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,

  41. shubhamsrg
    • 3 years ago
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    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

  42. PeterPan
    • 3 years ago
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    ^^

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