Question

g(x)=integral of f(t^2)

Answer 1

do you know the fundamental theorem of calculus?

Answer 2

yes i do

Answer 3

i have up to now y=4(x-1)+g(1)
I am not able to find g(1)

Answer 4

First find \(g(x)\).

Answer 5

g(x)=integral of f(t^2)2tdt from 0 to x=F(x^2)-F(0). No idea how to find the function F.

Answer 6

You are not quite correct there.

Answer 7

Or rather, you are not going about it the right way.

Answer 8

I don't really know how that is wrong!

Answer 9

We don't want to introduce \(F(x)\) in the equation since \(F(x) = g(x) + c\) which introduces more uncertainty into this problem.

Answer 10

\[
g(x) = \int_0^xf(t^2)2t\;dt
\]First consider \[
h(t) = f(t^2)2t
\]Giving us: \[
g(x) = \int_0^xh(t)dt
\]

Answer 11

We know now \[
g'(x) = h(x) = f(x^2)2x
\]

Answer 12

this brings me back to H(x)-H(0).

Answer 13

f(x)-f(0).

Answer 14

Thus \[
g'(1) = f(1)\cdot 2(1)
\]

Answer 15

f(1)-f(0)=4-2

Answer 16

g(1)=2

Answer 17

yes, the slope of the tangent line is 4
But g(1)=? I think my previous steps are wrong.

Answer 18

Anyone? Please help. I am lost.

Answer 19

are you sure that g'(x) is f(x)2x wio? I think it's just f(x)

Answer 20

There is one thing I'm trying to work out.

Answer 21

@TuringTest of that, I'm sure. Even if you do that \(u=t^2\) sub, you will have to change your bounds to be from \(0\) to \(x^2\).

Answer 22

oh yeah, sorry

Answer 23

Finding \(g(1)\) is a bit tricky.

Answer 24

Have you found a way though :O?

Answer 25

I spent 3h on this.

Answer 26

Hmmm \[
g(x) - g(1) = \int_1^x f(t^2)2t\:dt
\]

Answer 27

yep, that;s what I'm staring at lol

Answer 28

I've also looked at that for a long time! No idea how to solve. I'm losing all hopes.

Answer 29

Well, since the problem assumes that \(f(t)\) is any function, why not try some sample functions?

Answer 30

Try \(f(t) = 1\) and \(f(t) = 1\) and see what you come up with.

Answer 31

I mean \(f(t) = 1\).

Answer 32

Crap, I mean \(f(t)=t\).

Answer 33

Actually, scratch that, all we need is to come up with a sample function that meets our conditions.
Those conditions being: \(f(0)=4\) and \(f(1) = 2\).

Answer 34

So try \(f(t) = 4- 2t\)

Answer 35

you mean f(t^2)=4-2t^2?