# Evaluating a limit with an integral in it?

• Mar 31st 2010, 07:27 PM
lauren72
Evaluating a limit with an integral in it?
• Mar 31st 2010, 07:45 PM
Anonymous1
Wolfram Mathematica Online Integrator

yikes.

It may be a change of variables type of limit. You know, something like $\displaystyle \lim_{h\rightarrow 0} \frac{1}{h} \int f dx = \lim_{y\rightarrow \infty} y \int f dx.$

Do you have any notes on this?
• Mar 31st 2010, 07:46 PM
lauren72
*sigh* No. And I already tried using wolfram. But I think I might have an idea using the second fundamental theorem of calculus... maybe.
• Mar 31st 2010, 07:48 PM
Anonymous1
Are you taking Real Analysis? Or calc?
• Mar 31st 2010, 07:49 PM
SethP
I'd evaluate the integral, substitute 2 and 2+h for the t values, and multiply that by 1/h. Then I'd evaluate the limit.
• Mar 31st 2010, 07:50 PM
lauren72
Quote:

Originally Posted by Anonymous1
Are you taking Real Analysis? Or calc?

Calculus 1
• Mar 31st 2010, 07:51 PM
Anonymous1
Quote:

Originally Posted by SethP
I'd evaluate the integral, substitute 2 and 2+h for the t values, and multiply that by 1/h. Then I'd evaluate the limit.

*edit* You would, if you COULD.
• Mar 31st 2010, 07:52 PM
lauren72
Actually I think I got it figured out... I'm not sure how I did it, but I got the right answer.

Thanks anyway, guys.
• Mar 31st 2010, 07:52 PM
SethP
Oh, yeah. It's been a while since I've done limits; my brain sort of gets rid of stuff I haven't thought of in a while.
• Mar 31st 2010, 07:54 PM
Anonymous1
Quote:

Originally Posted by lauren72
Actually I think I got it figured out... I'm not sure how I did it, but I got the right answer.

Thanks anyway, guys.

(Clapping)

You used the $\displaystyle 2^{nd}$ $\displaystyle FTC?$
• Mar 31st 2010, 07:58 PM
lauren72
Yup. And got sqrt(15)
• Mar 31st 2010, 08:23 PM
Random Variable
Let $\displaystyle F(x)$ be an antiderivative of $\displaystyle \sqrt{7+t^{3}}$

then $\displaystyle \lim_{h \to 0} \ \frac{1}{h} \int^{2+h}_{2} \sqrt{7+t^{3}} = \lim_{h \to 0} \frac{F(2+h)-F(2)}{h} = F'(2)= \sqrt{7+2^{3}} = \sqrt{15}$