How do I find F''(2)

Question

F(x) = \int_{1}^{x} f(t)\, dt

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

I have been stuck on this problem forever how do I find the second derivative at x=2? Any help would be appreciated.

Answer 1

F(x) = ∫_{(1 to x)} f(t)dt

By the Fundamental Theorem of Calculus, F'(x) = f(x). So, F"(x) = f'(x).

f(t) = ∫_{(1 to t}²₎ [ √(4 + u²) / u ]du

f'(t) = [√(4 + t⁴) / t²] (t²)' = 2√(4 + t⁴) / t

So, f'(2) = 2√20 / 2 = 2√5

Therefore, F"(2) = f'(2) = 2√5