# Thread: I need HELP!

1. ## I need HELP!

I have a problem that reads; Find f(t) if f '(t)=t, and f(0)=3

2. Originally Posted by 10164064
I have a problem that reads; Find f(t) if f '(t)=t, and f(0)=3
Note that $f(t)=\int f'(t)\,dt$

Integrating, we get

$\int t \,dt = \frac{1}{2}t^2+C$

However, you're given an initial condition. Use it to find the value of C.

$\because f(0)=3, \ 3=\frac{1}{2}(0)^2+C\implies \color{blue}\boxed{C=3}$

Thus, $\color{blue}\boxed{f(t)=\frac{1}{2}t^2+3}$

Hope that this makes sense!

--Chris

3. Originally Posted by 10164064
I have a problem that reads; Find f(t) if f '(t)=t, and f(0)=3
$f'(t)=t$, means that $f(t)=\int_0^t u \ du +c$. Do the integral then put $t=0$ into the resulting expression for $f(x)$ and solve for $c$.

RonL

4. ## got it

YES thank you that makes sense

5. Originally Posted by 10164064
I have a problem that reads; Find f(t) if f '(t)=t, and f(0)=3
Haha, maybe this will be fun

Let $f'(t)=at+b$

So then

$\int{f'(t)}dt=f(t)=\frac{at^2}{2}+bt+C$

Now seeing that

$f(0)=3\Rightarrow{C=3}$

So we have that

$f(t)=\frac{at^2}{2}+bt+3$

...you guys should thank me for doing all the heavy lifting around here.

6. Originally Posted by Mathstud28
Haha, maybe this will be fun

Let $f'(t)=at+b$

So then

$\int{f'(t)}dt=f(t)=\frac{at^2}{2}+bt+C$

Now seeing that

$f(0)=3\Rightarrow{C=3}$

So we have that

$f(t)=\frac{at^2}{2}+bt+3$

...you guys should thank me for doing all the heavy lifting around here.
That is not the question asked, and is a pointless extension why not $f'(x)=P_n(x)$ where $P_n(x)$ is an $n$-th order polynomial in x?

Also, the idea is to give the minimum help to allow the original poster to complete the solution themselves.

RonL

7. Originally Posted by CaptainBlack
That is not the question asked, and is a pointless extension why not $f'(x)=P_n(x)$ where $P_n(x)$ is an $n$-th order polynomial in x?

Also, the idea is to give the minimum help to allow the original poster to complete the solution themselves.

RonL
Let $f'(x)=P_n(x)$ where....just kidding