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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 \(n=1\) ?
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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anonymous
 3 years 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
 3 years 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 \(n=1\) ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Because the antiderivative of x^{1} is ln(x)

klimenkov
 3 years 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?

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Maybe 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 \[\Large \frac{d}{dx}x^n=nx^{n1}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Or, more relevantly \[\Large \frac{d}{dx}\frac{x^{n+1}}{n+1}=x^n\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0But you see that it no longer applies for n = 1

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I do love the truth :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Noting of course, that there are many ways to explain this.

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Well, of course, it's still \[x^n\]

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

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

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0That's why Google is here. :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0It seems its limit is ln(x)

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0You did say you were looking for a specific answer, so there was preference :P

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Besides, 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
 3 years 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
 3 years ago
Best ResponseYou've already chosen the best response.0Now we can include \(n=1\) also.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0No you can't. It has to be \[\Large \int x^k\,dx=\lim_{k\rightarrow n}\frac{x^{n+1}1}{n+1}+C\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\Large \int x^k\,dx=\lim_{n\rightarrow k}\frac{x^{n+1}1}{n+1}+C\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Without the limit, it still won't make sense.

klimenkov
 3 years 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Finally. 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
 3 years 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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