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klimenkov
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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
klimenkov Group Title
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

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SandeepReddy Group TitleBest ResponseYou've already chosen the best response.0
AS you only stated the formula, substituting n = 1 in the formula on RHS, gives undefined form
 one year ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.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

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

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

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

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

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

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

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

PeterPan Group TitleBest ResponseYou've already chosen the best response.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

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

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

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

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

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

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

PeterPan Group TitleBest ResponseYou've already chosen the best response.3
Indeterminate.
 one year ago

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

klimenkov Group TitleBest 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?
 one year ago

PeterPan Group TitleBest ResponseYou've already chosen the best response.3
I really don't know.
 one year ago

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

klimenkov Group TitleBest 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
 one year ago

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

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

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

PeterPan Group TitleBest ResponseYou've already chosen the best response.3
Your point?
 one year ago

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

PeterPan Group TitleBest ResponseYou've already chosen the best response.3
Never mind. :)
 one year ago

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

PeterPan Group TitleBest ResponseYou've already chosen the best response.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

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

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

PeterPan Group TitleBest ResponseYou've already chosen the best response.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

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

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

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

PeterPan Group TitleBest ResponseYou've already chosen the best response.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

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.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
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