Finding the integral of t e^(-lambda t) from t=0 to t=infinity

[stunner5000pt]

\int_{0}^{infty} t e^{-\lambda t} \lambda dt\frac{1}{\lambda e^{\lambda t}} (1 + t \lambda) \right]_{0}^{\infty} = \frac{1}{\lambda}

is this corect??

for hte infinity part we don't have to worry since the 1/exponential term goes to zero as does the t/exponential term. For t = 0 however t/exponentail is zero and what is left is 1/lambda

 

[StatusX]

You're missing a factor of lambda. It's helpful if you think of t as being a time, so that lambda is a frequency and the integral needs to have dimensions of time^2.

 

[stunner5000pt]

You're missing a factor of lambda. It's helpful if you think of t as being a time, so that lambda is a frequency and the integral needs to have dimensions of time^2.

how am i missing it??

did i not do everything right??

see the whole work

\frac{-1}{\lambda e^{\lambda t}} (1 + t \lambda) \right]_{0}^{\infty} = \lim_{t \rightarrow \infty} \left(\frac{-1}{\lambda e^{\lambda t}} \right) - \left( \frac{-1}{\lambda}\right) - 0 \right) = \frac{1}{\lambda}

where does the extra fact oro lambda come from??

 

[StatusX]

The problem is before the first step you've shown, since the dimensions are wrong there as well. How did you take the antiderivative?

 

[stunner5000pt]

are u talkin about hte negative sign i was missing??

i did integration by parts..

 

[StatusX]

No, you're missing a factor of lambda in the antiderivative. Go back through it step through step.

 

[stunner5000pt]

since we are on this what about
\int_{T}^{\infty} e^{-\lambda t} \lambda dt = -e^{-\lambda t} |_{T}^{\infty} = 0 + e^{-\lambda T}

am i missing a fact or of lambda here too.. WHY??

 

[stunner5000pt]

No, you're missing a factor of lambda in the antiderivative. Go back through it step through step.

i typed out the question wrong...

it should be e^\lambda t * lambda

 

[StatusX]

I'm confused. Your antiderivative in the first step is right except for a factor of lambda. If you meant for there not to be a t multiplying the exponential, then your original answer is completely wrong. So which is it?

 

[stunner5000pt]

Ok let's restart

\int_{0}^{\infty} t e^{-\lambda t} \lambda dt

and the integral is
\frac{-1}{\lambda e^{\lambda t}} (1 + t \lambda) \right]_{0}^{\infty} = \frac{1}{\lambda}

good?

 

[StatusX]

Yea, that's right.