Solving Improper Integrals: Converging or Diverging?

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In summary, improper integrals are being studied in Calculus 2 class, specifically looking at convergence and divergence. The integral given is ∫(dt/(t+1)^2) on the interval from -1 to 5, and a u-substitution is used to simplify it. However, when taking the limit as "a" goes to -1, it results in an undefined value, indicating that the integral diverges. This means that there is no meaningful value for the integral, as it does not approach a finite number.
  • #1
Timebomb3750
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Just got into improper integrals, in my Calculus 2 class. We're looking to see if the integral converges or diverges.

Homework Statement


The integral given:
∫(dt/(t+1)^2) on the interval from -1 to 5

Homework Equations


uhhh...

The Attempt at a Solution


Took the limit as "a" goes to -1.

Did a simple u substitution with u=t+1, so that du=dt.

So, you're left with (du/u^2)

The integral of that is -(1/u) meaning -(1/(t+1))

Then I used the fundamental theorem of calculus by evaluating the integral from 5 to "a".

That looks like: -(1/(5+1)) - (1/(a+1))

Basically, I have no clue if I'm doing this right. According to the back of the book, it diverges. But I have no idea how to see that. Any help would be appreciated. Thanks.
 
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  • #2
You have
[tex]\int_{-1}^{5} \frac{dt}{(t+1)^{2}} = -\frac{1}{6} - [ \lim_{a \rightarrow -1} (-\frac{1}{a+1}) ] . [/tex]

What happens when you apply the limit?
 
  • #3
You mean apply the -1 into "a"? That would make it undefined because you'd be dividing by zero. But what does this tell me about convergence/divergence?
 
  • #4
And so that limit is undefined. Therefore, there is no meaningful value for the integral: that is what is meant by "divergence". We say the integral "converges" if the limit for its value approaches a finite number. If the limit for the integral does not approach a finite value, or does not even exist, the integral is said to "diverge". (It should also give the definitions of convergence and divergence of an integral in your textbook.)
 
  • #5
Wow. Thanks for clearing that up. I'm starting to understand it now, as I do more problems. Sometimes, I can't understand my textbook, nor my professor.
 

FAQ: Solving Improper Integrals: Converging or Diverging?

What is the definition of an improper integral?

An improper integral is an integral where one or both of the limits of integration are infinite, or where the integrand has a vertical asymptote within the bounds of integration.

How can I tell if an improper integral converges or diverges?

An improper integral converges if the limit of the integral as the limits of integration approach infinity or negative infinity exists and is finite. It diverges if the limit does not exist or is infinite.

What is the difference between a Type I and a Type II improper integral?

A Type I improper integral has one or both of the limits of integration as infinity, while a Type II improper integral has a vertical asymptote within the bounds of integration.

How do I solve for the convergence or divergence of an improper integral?

To solve for the convergence or divergence of an improper integral, you can use the limit comparison test, the integral comparison test, the p-test, or the limit comparison test with the p-test.

Can an improper integral be both convergent and divergent?

No, an improper integral can only be either convergent or divergent. It cannot be both at the same time.

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