1. Improper Integrals

I'm stuck on this integral and it's due online tonight, thanks a lot guys!

The Integral from 0 to infinity of: 1700te^(-0.5t)

2. Originally Posted by amaya
I'm stuck on this integral and it's due online tonight, thanks a lot guys!

The Integral from 0 to infinity of: 1700te^(-0.5t)
since this is to be handed in, we can't give you a full solution. instead, try to see if you can get by on hints.

you want to find $1700 \int_0^\infty te^{-0.5t}~dt = 1700 \lim_{N \to \infty} \int_0^N te^{-0.5t}~dt$

now, the standard way to approach this integral is to use integration by parts. using $u = t$ and $dv = e^{-0.5t}$

or, as an alternative, you could find $\frac d{dt} -te^{-0.5t} = \cdots$

integrate both sides and solve for the required integral

3. improper integrals..same problem!

Originally Posted by Jhevon
since this is to be handed in, we can't give you a full solution. instead, try to see if you can get by on hints.

you want to find $1700 \int_0^\infty te^{-0.5t}~dt = 1700 \lim_{N \to \infty} \int_0^N te^{-0.5t}~dt$

now, the standard way to approach this integral is to use integration by parts. using $u = t$ and $dv = e^{-0.5t}$

or, as an alternative, you could find $\frac d{dt} -te^{-0.5t} = \cdots$

integrate both sides and solve for the required integral

Ok thanks. I feel like I should know this and I hope I'm not just missing something obvious.
I'm doing integration by parts but I'm having trouble figuring out the integral of e^0.5t (this willl give me v). I know the integral of e^t will just be e^t but I'm not sure what to do when we have the -0.5 in front of the t.

Thanks a lot.

4. Originally Posted by amaya
Ok thanks. I feel like I should know this and I hope I'm not just missing something obvious.
I'm doing integration by parts but I'm having trouble figuring out the integral of e^0.5t (this willl give me v). I know the integral of e^t will just be e^t but I'm not sure what to do when we have the -0.5 in front of the t.

Thanks a lot.
in general: $\int e^{kt}~dt = \frac 1{k}e^{kt} + C$ for $k \in \mathbb{Z},~k \ne 0$

5. Originally Posted by amaya
The Integral from 0 to infinity of: 1700te^(-0.5t)
Let's just take what really matters: $\int_0^\infty te^{-t/2}\,dt.$

Substitute $u=\frac t2,$ the integral becomes

$4\int_0^\infty {ue^{ - u} \,du} = 4\cdot\Gamma (2) = 4.$