evaluating infinite integral

**yoman360** · March 27th 2010, 12:49 PM

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

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

f(0)= 10

**yoman360** · March 27th 2010, 01:23 PM

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.

**Krizalid** · March 27th 2010, 01:29 PM

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.

**Drexel28** · March 27th 2010, 01:31 PM

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

**mathemagister** · March 27th 2010, 01:35 PM

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 $\ln\left|\frac{1}{f(x)-1}\right|$ ?

**Drexel28** · March 27th 2010, 01:37 PM

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 $|-1|=1$

**mathemagister** · March 27th 2010, 01:43 PM

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

**Drexel28** · March 27th 2010, 01:46 PM

Originally Posted by mathemagister

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

Isn't $-\ln|1-f(x)|=-\ln|f(x)-1|=\ln\left(\frac{1}{|f(x)-1|}\right)=\ln\left(\frac{|1|}{|f(x)-1|}\right)=\ln\left|\frac{1}{f(x)-1}\right|$ ?

**mathemagister** · March 27th 2010, 02:12 PM

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?

**Drexel28** · March 27th 2010, 02:14 PM

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.

**yoman360** · March 27th 2010, 02:36 PM

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

**Drexel28** · March 27th 2010, 02:38 PM

Originally Posted by yoman360

I don't understand how you got that

Are you acquainted with the idea of separable ODEs?

**yoman360** · March 27th 2010, 02:52 PM

Originally Posted by Drexel28

Are you acquainted with the idea of separable ODEs?

I learned separable equations. Is that the same as separable ODEs?

**Drexel28** · March 27th 2010, 02:53 PM

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 $x$ .

**yoman360** · March 27th 2010, 03:00 PM

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.

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