Integration, Limits, and e^(x)

[Krishan93]

http://img827.imageshack.us/img827/4765/img0161u.jpg

I don't know how to integrate e^(t^2) like that. I am 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.

 

[vela]

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.

 

[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?

 

[vela]

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

 

[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?)

 

[vela]

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.

 

[SammyS]

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?)

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

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

 

[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.