Take the 2-minute tour ×
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.

Find $ F''(1)$ if

$$ F(x) = \int_1^x f(t)dt $$ $$ f(t) = \int_1^{2t} \sqrt{1+u^3} du $$

My work:

From the looks of it, it looks like the Fundamental Theorem of Calculus, twice.

From FTC... $F'(x) = f(x) $

On the second equation, derive both sides and use chain rule:

$f'(t)= d/dt*[ \int_1^{2t} \sqrt{1+u^3} du ] $

Using $z=2t$

$f'(t)=d/dz*[\int_1^{z} \sqrt{1+u^3} du]*dz/dt $

$f'(t) = 2 * \sqrt{1+(2t)^3} $

Just by gut feeling, I want to plug in $t=1$ and solve. Does $f'(t)$ represent $F''(x)$ ? $t$ and $x$ are different variable. Am I plugging in $1$ too early?

share|improve this question
1  
I didn't check every detail, but it looks like you are heading in the right direction. And yes $F''(1) = f'(1)$. This follows from your first application of the FTC. –  Code-Guru Dec 7 '12 at 2:00
add comment

1 Answer

up vote 2 down vote accepted

You are not plugging in $1$ too early; $x, t, u$ are defined as separate variables, but just as $F'(x)=f(x)$, we can similarly say $f'(x)=2\sqrt{1+8x^3}$ by FTC and chain rule.

It may seem odd, since $x$ and $t$ are defined to be distinct variables, but what happens when you replace $t$ with $x$ in the second line? The math is correct; the issue here is not getting jumbled up by the variable notation.

share|improve this answer
add comment

Your Answer

 
discard

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.