Improper Integral: 1/x2 | Converges to -1?

[andrey21]

Decide if the following improper integral converges or not

Integrate 1/x²

limits are 1 0

Now by simple integration we can see that:

[x^-2] = -x^-1

Substituting in 1 and 0:

-(1/1) - 0

So series converges to -1

Is this correct?

 

[Dick]

Can you explain in more detail how substituting 0 for x in -1/x gives 0?

 

[andrey21]

I don't understand why I have put that. With that in mind I don't know what to put there??

 

[Dick]

I don't understand why I have put that. With that in mind I don't know what to put there??

Look up the definition of improper integral. You don't put anything into -1/x. You have to find the limit as x->0 of -1/x.

 

[Mark44]

In this integral the integrand is undefined at one of the endpoints of integration. You need to use a limit to determine whether this integral is convergent.
$$\lim_{a \to 0^+}\int_a^b\frac{dx}{x^2}$$

 

[Dick]

As x tends to zero would the limit of -1/x be infinity??

Yes, it would. What does that tell you about the integral?

 

[andrey21]

Sorry I accidentally deleted my previous post, that means the integral itself tends to infinity and so is Divergent.

 

[Dick]

Sorry I accidentally deleted my previous post, that means the integral itself tends to infinity and so is Divergent.

Yes, it's divergent.

 

[andrey21]

Great thanks Dick, I do have a similar problem but the integral is:

1/x

So integrating gives:

[ln x]

So as x tends to zero the limit will be??

 

[Dick]

Great thanks Dick, I do have a similar problem but the integral is:

1/x

So integrating gives:

[ln x]

So as x tends to zero the limit will be??

You tell me. e^(-100) is pretty close to zero. What's ln(e^(-100))? Can you name some other numbers even closer to zero?

 

[andrey21]

Well ln and e cancel each other out so leave -100 correct?

 

[Dick]

Well ln and e cancel each other out so leave -100 correct?

Yes. So what do you say about the limit?

 

[andrey21]

So as x tends to zero the limit is -100??

 

[Dick]

So as x tends to zero the limit is -100??

You can't think of any numbers closer to zero than e^(-100)?

 

[andrey21]

Well e^(-101) , e^(-102) etc so could I say e^(-infinity)

so ln e^(-infinity) so limit is -infinity

 

[Dick]

Well e^(-101) , e^(-102) etc so could I say e^(-infinity)

so ln e^(-infinity) so limit is -infinity

That's better.

 

[andrey21]

Thank you dick for all your help