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

- chestercat

See more answers at brainly.com

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- anonymous

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

- klimenkov

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 \(n=-1\) ?

- anonymous

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- klimenkov

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

- anonymous

Simply, 1/0 does not exist :)

- klimenkov

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

- anonymous

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
\[\Large \frac{d}{dx}x^n=nx^{n-1}\]

- anonymous

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

- anonymous

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

- anonymous

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

- klimenkov

Ok. It is time to tell you the truth.

- anonymous

I do love the truth :)

- anonymous

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

- klimenkov

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

- anonymous

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

- klimenkov

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

- anonymous

Indeterminate.

- klimenkov

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

- anonymous

0/0

- klimenkov

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

- anonymous

I really don't know.

- klimenkov

Do you know the L'hopital's rule?

- anonymous

No...

- klimenkov

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

- klimenkov

@PeterPan
Do you know what is Wolfram Alpha?

- anonymous

That's why Google is here. :)

- anonymous

It seems its limit is ln(x)

- klimenkov

Yes.

- anonymous

Your point?

- klimenkov

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

- anonymous

Never mind. :)

- anonymous

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

- anonymous

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...

- klimenkov

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

- klimenkov

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

- anonymous

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

- anonymous

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

- anonymous

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

- klimenkov

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

- anonymous

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,

- shubhamsrg

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

- anonymous

^^

Looking for something else?

Not the answer you are looking for? Search for more explanations.