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 one year ago
Why does the formula \[\int x^n\,dx=\frac{x^{n+1}}{n+1}+C\]give the wrong result if we put \(n=1\) ?
 one year 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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SandeepReddy
 one year ago
Best ResponseYou've already chosen the best response.0AS you only stated the formula, substituting n = 1 in the formula on RHS, gives undefined form

klimenkov
 one year ago
Best ResponseYou've already chosen the best response.0This 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 \(n=1\) ?

PeterPan
 one year ago
Best ResponseYou've already chosen the best response.3Because the antiderivative of x^{1} is ln(x)

klimenkov
 one year ago
Best ResponseYou've already chosen the best response.0\[\int \frac1x\,dx=\frac{x^0}{0}+C=\frac10+C\]What is wrong with it?

PeterPan
 one year ago
Best ResponseYou've already chosen the best response.3Simply, 1/0 does not exist :)

klimenkov
 one year ago
Best ResponseYou've already chosen the best response.0\[\int \frac1x\,dx=\frac{x^0}{0}+C=\frac10+C=\ln x+C\]\[\frac10=\ln x\]

PeterPan
 one year ago
Best ResponseYou've already chosen the best response.3Maybe 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 \[\Large \frac{d}{dx}x^n=nx^{n1}\]

PeterPan
 one year ago
Best ResponseYou've already chosen the best response.3Or, more relevantly \[\Large \frac{d}{dx}\frac{x^{n+1}}{n+1}=x^n\]

PeterPan
 one year ago
Best ResponseYou've already chosen the best response.3But you see that it no longer applies for n = 1

PeterPan
 one year ago
Best ResponseYou've already chosen the best response.3Because \[\Large \frac{d}{dx}\frac{x^{1+1}}{1+1}=\frac{d}{dx}\frac10\]and 1/0 never existed to begin with :/

klimenkov
 one year ago
Best ResponseYou've already chosen the best response.0Ok. It is time to tell you the truth.

PeterPan
 one year ago
Best ResponseYou've already chosen the best response.3I do love the truth :)

PeterPan
 one year ago
Best ResponseYou've already chosen the best response.3Noting of course, that there are many ways to explain this.

klimenkov
 one year ago
Best ResponseYou've already chosen the best response.0Try to compute\[\frac{d}{dx}\left(\frac{x^{n+1}1}{n+1}\right)\]

PeterPan
 one year ago
Best ResponseYou've already chosen the best response.3Well, of course, it's still \[x^n\]

klimenkov
 one year ago
Best ResponseYou've already chosen the best response.0But now if you put \(n=1\) what will you get?

klimenkov
 one year ago
Best ResponseYou've already chosen the best response.0Don't mind on indeterminate. Just compute this like a simple limit.

klimenkov
 one year ago
Best ResponseYou've already chosen the best response.0\(\frac{\sin n}{n}\) is \(\frac00\) too, but we all know that it is equal to 1. What is that limit?

klimenkov
 one year ago
Best ResponseYou've already chosen the best response.0Do you know the L'hopital's rule?

klimenkov
 one year ago
Best ResponseYou've already chosen the best response.0\[\lim_{n\rightarrow1}\frac{x^{n+1}1}{n+1}\]Anybody here know how to compute this limit? @ParthKohli

klimenkov
 one year ago
Best ResponseYou've already chosen the best response.0@PeterPan Do you know what is Wolfram Alpha?

PeterPan
 one year ago
Best ResponseYou've already chosen the best response.3That's why Google is here. :)

PeterPan
 one year ago
Best ResponseYou've already chosen the best response.3It seems its limit is ln(x)

klimenkov
 one year ago
Best ResponseYou've already chosen the best response.0Sorry, cant understand what means "Your point?".

PeterPan
 one year ago
Best ResponseYou've already chosen the best response.3You did say you were looking for a specific answer, so there was preference :P

PeterPan
 one year ago
Best ResponseYou've already chosen the best response.3Besides, 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...

klimenkov
 one year ago
Best ResponseYou've already chosen the best response.0Actually, the right expression is \[\int x^n\,dx=\frac{x^{n+1}1}{n+1}+C\]

klimenkov
 one year ago
Best ResponseYou've already chosen the best response.0Now we can include \(n=1\) also.

PeterPan
 one year ago
Best ResponseYou've already chosen the best response.3No you can't. It has to be \[\Large \int x^k\,dx=\lim_{k\rightarrow n}\frac{x^{n+1}1}{n+1}+C\]

PeterPan
 one year ago
Best ResponseYou've already chosen the best response.3\[\Large \int x^k\,dx=\lim_{n\rightarrow k}\frac{x^{n+1}1}{n+1}+C\]

PeterPan
 one year ago
Best ResponseYou've already chosen the best response.3Without the limit, it still won't make sense.

klimenkov
 one year ago
Best ResponseYou've already chosen the best response.0Very nice. But forget using limits. It is restrictions that really does not play a big role.

PeterPan
 one year ago
Best ResponseYou've already chosen the best response.3Finally. 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,

shubhamsrg
 one year ago
Best ResponseYou've already chosen the best response.3the 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
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