Second integral problem

[Kevrad]

Compute f'(1) where
f(x)=(integral from 3x down to x)sqrt[8+t^3]dt

It says leave it in whatever your calculator calculates.

 

[daon2]

Compute f'(1) where
f(x)=(integral from 3x down to x)sqrt[8+t^3]dt

It says leave it in whatever your calculator calculates.

So what have you tried? What does your book or notes say?

 

[Kevrad]

it doesnt say much about there being a 3x its kind of a challenge problem and i would like to know how to do it. should i not change the f(x) value since it is deriving an integral?

 

[daon2]

Here is the general idea:

Let $\displaystyle G(t)=\int g(t) dt$ (any antiderivative). Then by the fundamental theorem of calculus:

$\displaystyle \displaystyle \int_{p(x)}^{q(x)} g(t) dt = G(q(x))-G(p(x))$.

In your problem, we have $\displaystyle f(x) = G(q(x))-G(p(x))$, and from the last problem you posted you should know what $\displaystyle G'$ is. Now use the chain rule to take the derivative, and plug in 1.

 

[Kevrad]

Thanks for the help you are awesome i will let you know what i get!

 

[Kevrad]

answer

So i used your equation can you check and see if i got the right outcome
my answer is 3*sqrt[8+(3x)^3]-sqrt[8+x^3]

 

[daon2]

Yes, looks good

 

[Kevrad]

so now i just use the f'(1) and plug 1 into this equation?

 

[HallsofIvy]

Since you are not actually asked to find f but only its derivative, you could use Lagrange's formula:
$\displaystyle \frac{d}{dx}\int_{p(x)}^{q(x)} F(x,t)dt= \frac{dq}{dx}F(x, q(x))- \frac{dp}{dx}F(x, p(x))+ \int_{p(x)}^{q(x)}\frac{\partial F}{\partial x}dt$

Here, p(x)= x and q(x)= 3x while $\displaystyle F(x,t)= \sqrt{8+ t^3}$ does not depend on x so
$\displaystyle f'(x)= 3\sqrt{8+ 27x^3}- \sqrt{8+ x^3}$.