1. ## Quick question about divergent and convergent

Here is a problem:

If you integrate that, you would get:

Now using the Fundamental theorem of calculus, F(b) - F(a) you would get an answer of 4ln(4) - 8.

But how is that answer even possible? If you used the formula F(b) - F(a), it would look something like this,

Wouldn't the log(0) give an indeterminate answer? Doesn't that automatically qualify the problem to be divergent?

Same thing with this problem:

Integrate that and you would get:

Use the fundamental theorem of calculus and you would get this:
2(sqrt4)ln(4)-4sqrt4-[(2sqrt0)ln(0)-4sqrt0]

This problem turns out to be divergent. But I'm not so sure if it is because of the ln(0) or what.

Please clarify how the answers are how they are for both of my questions. Thank you so much in advance.

2. These are examples of improper integrals. If the function is not defined at some point on your region of integration, you need to evaluate a limit.

In your first example, you need to rewrite this as

$\displaystyle \lim_{\epsilon \to 0}\int_{\epsilon}^{4}{\frac{\log{x}}{\sqrt{x}}\,dx } = \lim_{\epsilon \to 0}\left[2\sqrt{x}\log{x}-4\sqrt{x}\right]_{\epsilon}^{4}$

$\displaystyle = 2\sqrt{4}\log{4} - 4\sqrt{4} - \lim_{\epsilon \to 0}\left(2\sqrt{\epsilon}\log{\epsilon} - 4\sqrt{\epsilon}\right)$

Can you go from there?

In the second example, you need to break it up into two definite integrals.

3. Ah thank you so much for your help for the first one. I can finally solve it now.

For the second one, you said to break it up into two definite integrals.

Would this be it?

I chose those limits because I saw my book used it on an example problem. I'm not sure if the limits would be the same for every problem or it depends. Please clarify for me and thank you thus far.

4. No. The function is not defined where the denominator is zero, i.e. where $\displaystyle x = -1, 2$.

Since $\displaystyle x = 2$ is in your region of integration, that will be the point where you break the integral into two integrals.

So it should be $\displaystyle \int_0^2{\frac{1}{x^2 - x - 2}\,dx} + \int_2^3{\frac{1}{x^2 - x- 2}\,dx}$.

5. Originally Posted by Prove It
No. The function is not defined where the denominator is zero, i.e. where $\displaystyle x = -1, 2$.

Since $\displaystyle x = 2$ is in your region of integration, that will be the point where you break the integral into two integrals.

So it should be $\displaystyle \int_0^2{\frac{1}{x^2 - x - 2}\,dx} + \int_2^3{\frac{1}{x^2 - x- 2}\,dx}$.
So that is how you figure out. Thank you so much for your help!