# Integration, Limits, and e^(x)

1. Dec 5, 2011

### Krishan93

http://img827.imageshack.us/img827/4765/img0161u.jpg [Broken]

I don't know how to integrate e^(t^2) like that. Im thinking that it become e^(t^2+1) with some constant in front of the whole expression.

And then with A) I just don't know how to begin.

A kick in the right direction would help. Thanks.

Last edited by a moderator: May 5, 2017
2. Dec 5, 2011

### vela

Staff Emeritus
You can't integrate it to get a closed form expression, so you need to figure out a different way of evaluating those limits. You'll likely want to make use of the fundamental theorem of calculus.

3. Dec 5, 2011

### Krishan93

Yeah...I'm still stumped.
I don't know how to do this one, let alone apply the fundamental theorem.

How should one come about the solution to these problems?

4. Dec 5, 2011

### vela

Staff Emeritus
What do you get if you try applying L'Hopital's rule to the first one?

5. Dec 5, 2011

### Krishan93

How do you know to use Lhopitals rule?
Whats the rule when you're dealing with infinity-I already know the 0/0, just not infinity.

Deriving top and bottom expression should yield
xe^(x^2)+F(x)
x^2(e^(x^2))(2x) <-- (chain rule, right?)

6. Dec 5, 2011

### vela

Staff Emeritus
L'Hopital's rule can be used when you have indeterminate forms. 0/0 is one such form; ∞/∞ is another.

The denominator would be correct without the x2 in front. When you differentiate the exponential, you get the exponential, and then the chain rule gives you the factor of 2x. You end up with
$$\frac{xe^{x^2}+F(x)}{2xe^{x^2}} = \frac{1}{2} + \frac{F(x)}{2xe^{x^2}}$$When you try to take the limit now, the second term is again ∞/∞, so you need to apply L'Hopital's rule on it.

7. Dec 5, 2011

### SammyS

Staff Emeritus
Assuming you're working with the (a), the derivative of the denominator is incorrect.

$\displaystyle \frac{d}{dx}e^{x^2}=2xe^{x^2}$

8. Dec 5, 2011

### dextercioby

You can show that F(x) diverges when x grows to +infinity, hence you're allowed to use differentiation (required by L'Ho^pital's rule) when computing those limits.