Find $F''(2): $$F(x)=∫_x^2 f(t)\,dt$$

[Starrrrr]

Suppose that $$F(x)=∫_x^2 f(t)\,dt \\ f(t)=∫_{t^2}^1 \sqrt{5+u^6}/u\, du$$ Find $F′′(2)$?

 

[I like Serena]

Suppose that $$F(x)=∫_x^2 f(t)\,dt \\ f(t)=∫_{t^2}^1 \sqrt{5+u^6}/u\, du$$ Find $F′′(2)$?

Hi Eletronic! Welcome to MHB! ;)

Let's define:
$$\hat f(t)=\int f(t)\,dt, \\s(u)=\sqrt{5+u^6}/u, \\\hat s(u)=\int s(u)\,du$$
Then:
$$F(x)=\int_x^2 f(t)\,dt =\hat f(2)-\hat f(x) \\
F'(x)=-\hat f'(x) = -f(x) \\
F''(x)=-f'(x) \\
f(t)=\int_{t^2}^1 s(u)\, du = \hat s(1) - \hat s(t^2) \\
f'(t)=-\hat s'(t^2)\cdot 2t = -s(t^2)\cdot 2t
$$
So:
$$F''(x)=-s(x^2)\cdot 2x$$

How could we continue? (Wondering)

 

[Starrrrr]

Hi Eletronic! Welcome to MHB! ;)

Let's define:
$$\hat f(t)=\int f(t)\,dt, \\s(u)=\sqrt{5+u^6}/u, \\\hat s(u)=\int s(u)\,du$$
Then:
$$F(x)=\int_x^2 f(t)\,dt =\hat f(2)-\hat f(x) \\
F'(x)=-\hat f'(x) = -f(x) \\
F''(x)=-f'(x) \\
f(t)=\int_{t^2}^1 s(u)\, du = \hat s(1) - \hat s(t^2) \\
f'(t)=-\hat s'(t^2)\cdot 2t = -s(t^2)\cdot 2t
$$
So:
$$F''(x)=-s(x^2)\cdot 2x$$

How could we continue? (Wondering)

Where did you get the s? ,I'm kinda confused.

 

[I like Serena]

Where did you get the s? ,I'm kinda confused.

The second given integral is $f(t)=∫_{t^2}^1 \sqrt{5+u^6}/u\, du$, isn't it?
We're defining $s(u)$ to be the integrand for ease of reference.

 

[Starrrrr]

The second given integral is $f(t)=∫_{t^2}^1 \sqrt{5+u^6}/u\, du$, isn't it?
We're defining $s(u)$ to be the integrand for ease of reference.

Yes it is and ok.

 

[Starrrrr]

The second given integral is $f(t)=∫_{t^2}^1 \sqrt{5+u^6}/u\, du$, isn't it?
We're defining $s(u)$ to be the integrand for ease of reference.

Do I just sub in x=2 into the second integral?

 

[I like Serena]

Do I just sub in x=2 into the second integral?

Not on the second integral...

We need to dub x=2 in the final expression for F''(x).
And as part of that we need to sub $u=x^2=4$ into $s(u)'$. (Thinking)