Question

Calculus question, medal for help and explanation. Much appreciated I will post below. thanks

Answer 1

\[ f(t)=\int\limits_{1}^{t^2}(\sqrt{7+u^6}du/u \]
and \[ F(x)=\int\limits_{1}^{x}f(t)dt, \]
Find F''(2).

Thanks and id like to understand it as well

Answer 2

I also dont have the answer so il try my answer and if someone could confirm thatd be good to. so F'(x)=f(x) so \[f(x)=\int\limits_{1}^{x^2}(\sqrt{7+u^6du/u}\]
so F''(x)= d/dx of that integral, which would be sqrt(7+x^2)^6/x^2 times 2x and evaluated at x=2

Answer 3

OK, first let's make clearance about your problem. I would like to understand the f(t)

\[F(x) = \int\limits_{1}^{x}f(t)dt\]
So F'(x) = f(t) with t from 1--> x. And F''(x) = f'(t)

But f'(t) can be written as\[f' (t) = \frac{\sqrt{7+u^6}}{u}\] with u from 1 --> t^2

That's almost done. Now just do some simplification to the final result.

Step 1: substitute u=1 and u=t^2 in order to calculate f'(t)
\[f \prime (t) = \frac{\sqrt{7+t^{12}}}{t^2} - \sqrt{8}\]
Step 2: substitute t =1 and t=x in order to calculate F''(x) = f'(t), below I already simplify the expression for you.
\[F \prime \prime (x) = \frac{\sqrt{7+x^{12}}}{x^2} - \sqrt{8}\]

Step 3: F''(2) is ?
\[F \prime \prime (2) = \frac{\sqrt{4103}}{4}-\sqrt{8} \approx 13.185\]

Answer 4

Where did your - sqrt come from

Answer 5

You mean -sqrt(8) ? This was taken from definition of definite integrals

\[F(x) = \int\limits_{a}^{b} f(t)dt = F(b) - F(a)\]

Answer 6

Okay and do you not multiple by 2x because of the x^2 and chain rule

Answer 7

So do you agree from the steps: F''(x) = f'(t) with t = 1-->x ? With substitute t = x then the chain rule cannot apply.

Answer 8

More clearly, let me explain by :
\[F \prime (x) = f(t)\]
Take derivative of d/dx for both sides
\[\frac{d}{dx}F'(x) = \frac{d}{dx}f(t) = f'(t) . \frac{dt}{dx}\]
Here is the chain rule application, but because t = x at boundary then dt/dx = 1

Answer 9

Alright makes sense, thanks