# Thread: evaluating infinite integral

1. ## evaluating infinite integral

$f'(x)=x(1-f(x))$

Evaluate $\int_0^{\infty} x(1-f(x))$

f(0)= 10

2. Originally Posted by yoman360
$f'(x)=x(1-f(x))$

Evaluate $\int_0^{\infty} x(1-f(x))$
f(0)=10
heres my attempt

$\lim_{b \to \infty} \int_0^b x(1-f(x))$

then after that i don't know what to do.

3. the integral equals $f(b)-f(0)=f(b)-10,$ and in order to have convergence, we require that $\lim_{b\to\infty}f(b)$ exists.

4. $f'(x)=x(1-f(x))\implies \ln\left|\frac{1}{f(x)-1}\right|=\frac{x^2}{2}+C$

5. Originally Posted by Drexel28
$f'(x)=x(1-f(x))\implies \ln\left|\frac{1}{f(x)-1}\right|=\frac{x^2}{2}+C$
Good job seeing that! But why isn't it $-\ln\left|{1-f(x)}\right|$, instead of $\ln\left|\frac{1}{f(x)-1}\right|$?

6. Originally Posted by mathemagister
Good job seeing that! But isn't it $\ln\left|\frac{1}{1-f(x)}\right|$, not $\ln\left|\frac{1}{f(x)-1}\right|$
Isn't $|-1|=1$

7. Sorry, that's not what I meant. Please refer to my modified post. Thanks!

8. Originally Posted by mathemagister
Sorry, that's not what I meant. Please refer to my modified post. Thanks!
Isn't $-\ln|1-f(x)|=-\ln|f(x)-1|=\ln\left(\frac{1}{|f(x)-1|}\right)=\ln\left(\frac{|1|}{|f(x)-1|}\right)=\ln\left|\frac{1}{f(x)-1}\right|$?

9. Originally Posted by Drexel28
Isn't $-\ln|1-f(x)|=-\ln|f(x)-1|=\ln\left(\frac{1}{|f(x)-1|}\right)=\ln\left(\frac{|1|}{|f(x)-1|}\right)=\ln\left|\frac{1}{f(x)-1}\right|$?
Haha! Wow, I'm really not into logarithms today... Thanks for clearing that up. Conventionally, are both forms equally simplified?

10. Originally Posted by mathemagister
Haha! Wow, I'm really not into logarithms today... Thanks for clearing that up. Conventionally, are both forms equally simplified?
Not being a conventionalist, I wouldn't know. I sometimes put it in a convoluted form to make the OP work. It's medicine for their soul.

11. Originally Posted by Drexel28
$f'(x)=x(1-f(x))\implies \ln\left|\frac{1}{f(x)-1}\right|=\frac{x^2}{2}+C$
I don't understand how you got that

12. Originally Posted by yoman360
I don't understand how you got that
Are you acquainted with the idea of separable ODEs?

13. Originally Posted by Drexel28
Are you acquainted with the idea of separable ODEs?
I learned separable equations. Is that the same as separable ODEs?

14. Originally Posted by yoman360
I learned separable equations. Is that the same as separable ODEs?
Don't ask me.

$f'(x)=x(1-f(x))\implies \frac{f'(x)}{1-f(x)}=x$. Integrate both sides with respect to $x$.

15. Originally Posted by Drexel28
Don't ask me.

$f'(x)=x(1-f(x))\implies \frac{f'(x)}{1-f(x)}=x$. Integrate both sides with respect to $x$.
I actually see how you got that now

f'(x)=x(1-f(x))

dy/dx=x(1-y) where y=f(x)

$\int \frac{1}{1-y} dy = \int x dx$

$-ln|1-y| = \frac{1}{2} x^2 + C$

$-ln|1-f(x)| = \frac{1}{2} x^2 + C$

then using the initial condition f(0) = 10 I can find C

$-ln|1-f(0)| = \frac{1}{2} (0)^2 + C$
$-ln|9|= C$

thanks for the help i think i can finish the problem from here but if i get stuck i will post it.