1. ## complicated integral

Consider the function.

Find F''(2)

[IMG]file:///C:/Users/rsanford/AppData/Local/Temp/moz-screenshot.jpg[/IMG][IMG]file:///C:/Users/rsanford/AppData/Local/Temp/moz-screenshot-1.jpg[/IMG]

2. Originally Posted by rsanford
Consider the function.

Find F''(2)

[IMG]file:///C:/Users/rsanford/AppData/Local/Temp/moz-screenshot.jpg[/IMG][IMG]file:///C:/Users/rsanford/AppData/Local/Temp/moz-screenshot-1.jpg[/IMG]
these are just using the second fundamental theorem of calculus: in a nutshell it says:

if $f(x) = \int_c^{g(x)}f(t)~dt$, then $f'(x) = f(g(x)) \cdot g'(x)$

where $c$ is any constant and $g(x)$ is some function of $x$.

Use this to find $f'(x)$. then to get $f''(x)$, just take the regular derivative of $f'(x)$

Can you continue?

3. yes! Thank you! I just didn't see that.

4. I'm sorry.. I'm having trouble taking the integral of
$

\int_1^{x^2} \frac{\sqrt{5+x^4}}{x}

$

5. Originally Posted by rsanford
I'm sorry.. I'm having trouble taking the integral of
$

\int_1^{x^2} \frac{\sqrt{5+x^4}}{x}

$
Make a trig substitution with $5\tan(\theta)$

6. Originally Posted by rsanford
I'm sorry.. I'm having trouble taking the integral of
$

\int_1^{x^2} \frac{\sqrt{5+x^4}}{x}

$
the point of my first post was showing that you do not have to do any integrals here. just use the fundamental theorem of calculus that i posted. do you understand what the theorem says? all you need for this problem is derivatives, and that's only because we are going to the second derivative. DO NOT EVALUATE ANY INTEGRALS.

Originally Posted by Mathstud28
Make a trig substitution with $5\tan(\theta)$
that would not work...

7. Originally Posted by Jhevon
that would not work...
I havent worked it but letting $x^2=5tan(\theta)$?

o wait I see why..hmm...well then that makes that harder

8. Originally Posted by Mathstud28
I havent worked it but letting $x^2=5tan(\theta)$?

o wait I see why..hmm...well then that makes that harder
yes, the fourth power is the problem. thank God we don't have to do this integral