FTC & Integral Homework: Find F''(2)

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Homework Statement


F(x)= \int_1^x f(t)dt

f(t)= \int_1^{t^2} \frac{\sqrt{1+u^4}}{u} du

Find F''(2)

Homework Equations


Just learned the FTC part 1 and 2.


The Attempt at a Solution


Pretty lost on how to do this...
 
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You can write f(x) in terms of x: f(x) = \int^{x^2}_1 \frac{\sqrt{1+u^4}}{u} \ du. Now what can you say, by the FTC, about how F'(x) is related to f(x)? And how is this related to what you can do with the expression for f(x) given above?
 
F'(x) = f(x) for all x in (1,x^2) ?

And for the other part I am not sure... f(x) would equal F(x^2) - F(1) where F is the the anti derivative.
 
You have F'(x) = f(x). You can see that F''(x) is f'(x), right? How would you get the latter from what you are given?
 
Take the derivative of f(x).
 
Alright I think understand. I will look over it some more. Thanks for your help.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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