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The problem

For $$F(x) = \int_0^x \cos\left( t^2 \right)\,dt$$ find $$F(0), F'(0), F''(x)$$ and $$\lim_{x\rightarrow0} \frac{F(x)}{x} $$

What I've done

Not much to be frank, though not for a lack of trying. I could find $F(0)$ (since that's quite trivial) and I have differentiated similar integrals before. What I did for those is expand the integral to $F(a) - F(b)$ (where $F(x)$ is its anti-derivative and $a$ the upper bound, $b$ the lower) and derive that. Unfortunately $cos\ t^2$ does not lend itself to finding its anti-derivative easily (or at all it seems).

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    Use the *first fundamental theorem of integral calculus*. – 2017-02-13
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    With appropriate smoothness assumptions, if $F(x) = \int_0^x f(t) dt$ then $F'(x) = f(x)$. The rest are standard differentiation. – 2017-02-13
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    do you mean $cos(t^2)$? – 2017-02-13
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    @Dr.SonnhardGraubner Yes, I do. Sorry, I'm used to $cos^2t = cos(t)^2$ so I thought I was being clear. – 2017-02-13

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We use the Fundamental Theorem of Calculus

If $f$ is continuous on $[a, b]$ and defining $$ F(x) = \int_a^x \! f(t) \, dt $$ for $x \in [a, b]$, then $F'(x) = f(x)$ for $x \in (a,b)$.

Thus the first derivative of $F(x)$ is $f(x)$, the second is $f'(x)$, and so forth.

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    Okay so $F'(x) = cos(x^2)$? – 2017-02-13
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    @ejbs Yep. Or $\cos(x^2)$... your notation "$\cos t^2$ is a bit unclear, but just replace $t$ by $x$ with whatever is inside of the integral – 2017-02-13
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    Gosh I was trying to be clear this time but I misplaced the parenthesis haha. But alright, I'll accept that. I only knew that that 'worked' for indefinite integrals, I'll need to read up on that. Thank you. – 2017-02-13