Question

Suppose that F(x)=∫1 to x f(t)dt, where
f(t)=∫1 to t^2 sqrt(6+(u^4))/ (u) du

Find F″(2).

Answer 1

Woah...

Answer 2

Let me write this out.

Answer 3

Here's a pic of the problem if you want to see it

Answer 4

Hmm this is a cool problem :3 thinking...

Answer 5

Like wouldn't just be we have F'(x)=f(x) by fundamental theorem of calculus
then differentiate that again F''(x)=f'(x)
then find f'(t) using the other equation thingy

Answer 6

you know also by fundamental theorem thingy

Answer 7

I don't know, sorry. I'm taking Calc 1 this semester, so, I'm not great with integrals yet.

Answer 8

do you understand what I mean @lawls

Answer 9

i'm in calc 1 too lol

Answer 10

you can find what from \[F(x)=\int\limits_{1}^{x}f(t) dt \text{ that } F'(x)=f(x) \text{ and from that } F''(x)=f'(x) \\ \text{ then you can also find } f'(x) \text{ from } f(x)=\int\limits_{1}^{x^2}\frac{\sqrt{6+u^4}}{u} du\]

Answer 11

Ooo neato :O

Answer 12

i'm not sure if this is right but is f'(x) = sqrt (6 + (x^2)^4) / x^2 * 2x ?

Answer 13

that looks good to me

Answer 14

and that is also F''(x) since F''(x)=f'(x)

Answer 15

ok. now i just plug in 2?

Answer 16

yeah

Answer 17

i got 16.18641406

Answer 18

thank you!!

Answer 19

or exact form sqrt(262)