# Thread: Evaluating a limit with an integral in it?

1. ## Evaluating a limit with an integral in it?

2. Wolfram Mathematica Online Integrator

yikes.

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

Do you have any notes on this?

3. *sigh* No. And I already tried using wolfram. But I think I might have an idea using the second fundamental theorem of calculus... maybe.

4. Are you taking Real Analysis? Or calc?

5. 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.

6. Originally Posted by Anonymous1
Are you taking Real Analysis? Or calc?
Calculus 1

7. 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.

8. 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.

9. 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.

10. 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.

You used the $2^{nd}$ $FTC?$

11. Yup. And got sqrt(15)

12. Let $F(x)$ be an antiderivative of $\sqrt{7+t^{3}}$

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

so your answer is correct