I have NO idea how to do this:

http://online2.byu.edu/webwork2_file...02c95fb4d1.png

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- Mar 31st 2010, 07:27 PMlauren72Evaluating a limit with an integral in it?
I have NO idea how to do this:

http://online2.byu.edu/webwork2_file...02c95fb4d1.png - Mar 31st 2010, 07:45 PMAnonymous1
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 PMlauren72
*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 PMAnonymous1
Are you taking Real Analysis? Or calc?

- Mar 31st 2010, 07:49 PMSethP
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 PMlauren72
- Mar 31st 2010, 07:51 PMAnonymous1
- Mar 31st 2010, 07:52 PMlauren72
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 PMSethP
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 PMAnonymous1
- Mar 31st 2010, 07:58 PMlauren72
Yup. And got sqrt(15)

- Mar 31st 2010, 08:23 PMRandom Variable
Let $\displaystyle F(x) $ be an antiderivative of $\displaystyle \sqrt{7+t^{3}} $

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

so your answer is correct