# Solving this odd mix of integrals/derivatives?

I've got a homework assignment that looks something like this: (numbers changed)

$$F(x) =\int_1^x f(t) \, \mathrm{d} t$$ $$f(t) =\int_x^{t^2} \frac{\sqrt{7+u^4}}{u} \, \mathrm{d} u$$

Find $F\;''(1)$

I changed the numbers because I don't want the answer. This seems simple but I can't wrap my head around it. Any ideas?

Edit: I entered this in completely wrong for some reason. Will the same answer still apply?

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What do you know about the Fundamental Theorem of Calculus? – mixedmath Oct 30 '11 at 5:06

With respect to your modified question, things are a little stickier. One thing that might help you in getting the answer is to consider the following representation of $f(t)$:

$$f(t)=\int_x^{t^2} \frac{\sqrt{7+u^4}}{u} \, \mathrm du=\int_p^{t^2} \frac{\sqrt{7+u^4}}{u} \, \mathrm du-\int_p^{x} \frac{\sqrt{7+u^4}}{u} \, \mathrm du$$

where $p$ is some constant. (Why this is justified is something you'll have to explain.) We then have

\begin{align*}F(x)=\int_1^x f(t) \, \mathrm dt&=\int_1^x \left(\int_p^{t^2} \frac{\sqrt{7+u^4}}{u} \, \mathrm du-\int_p^{x} \frac{\sqrt{7+u^4}}{u} \, \mathrm du\right) \, \mathrm dt\\&=\int_1^x \int_p^{t^2} \frac{\sqrt{7+u^4}}{u} \, \mathrm du\,\mathrm dt-\left(\int_p^{x} \frac{\sqrt{7+u^4}}{u} \, \mathrm du\right)\left(\int_1^x\mathrm dt\right)\end{align*}

(You'll also have to explain how I got that last bit.) Differentiating the second term requires that you use the product rule in addition to the Fundamental Theorem; differentiating the first term will require the careful use of the chain rule...

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Interesting. Looks pretty intense, to be honest. :( – Yep Oct 30 '11 at 19:08

By the fundamental theorem of calculus, if $\displaystyle F(x)=\int_a^x f(t)\ dt$, then $F'(x)=f(x)$. So here $$F'(x)=\frac{\sqrt{7+x^4}}{x}.$$ You can take the second derivative to find $F''(x)$, and then plug in $x=1$.

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I was wondering if you could explain how did you find that: $F'(x)=\frac{\sqrt{7+x^4}}{x}$? Thanks. – NoChance Oct 30 '11 at 5:29
@Emmad Sure, the $t$ in my first statement could have just as well been a $u$, so $F(x)=\int_a^x f(u)\ du$ implies $F'(x)=f(x)$. Write $f(u)=\frac{\sqrt{7+u^4}}{u}$, so $F(x) =\int_1^x \frac{\sqrt{7+u^4}}{u} \, \mathrm{d} u=\int_1^x f(u)\ du$. Then $F'(x)=f(x)=\frac{\sqrt{7+x^4}}{x}$. – yunone Oct 30 '11 at 5:32
Thank you for your clear and quick reply. Very clever. – NoChance Oct 30 '11 at 5:37
Whoops, can you check my original question again? I entered it in wrongly. – Yep Oct 30 '11 at 14:05