Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
What do you know about the Fundamental Theorem of Calculus? – mixedmath Oct 30 '11 at 5:06
up vote 3 down vote accepted

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...

share|cite|improve this answer
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 $x=1$.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.