Second integral problem

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Kevrad

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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.
 
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Kevrad said:
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.
Click to expand...

So what have you tried? What does your book or notes say?
 
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?
 
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.
 
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]
 
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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
\(\displaystyle f'(x)= 3\sqrt{8+ 27x^3}- \sqrt{8+ x^3}\).
 
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