how do i integrate t*e^(kt)dt from 0 to infinity?

[BobLoblaw]

k = -.000121

[DavidG]

hey mate,

just to let you know, whether or not this can be achieved is based on the value of k, k must be negative for the integral to be finite,

however assume k is < 0, i.e. k = -b ; b>0

Then,

int (te^(kt) dt ; t = 0 to inf) = int(te^(-bt) dt ; t = 0 to inf)

Employ Integration by parts int(uv') = uv - int(vu')

Here let,
u = t --> u' = 1
v' = e^(-bt) --> v = (-1/b)e^(-bt)

Thus,

int(te^(-bt) dt ; t = 0 to inf) = [ (-t/b)e^(-bt) ; t = 0 to inf ] - int( (-1/b)e^(-bt) (1) dt ; t = 0 to inf)
= (0 - 0) + (1/b)int(e^(-bt) dt ; t = 0 to inf)
= (1/b) [ (-1/b)e^(-bt) ; t = 0..int ]
= (1/b) [ (-1/b) - 0]
= -1/b^2

Hope this helps,

David

[DavidG]

hey mate,

just to let you know, whether or not this can be achieved is based on the value of k, k must be negative for the integral to be finite,

however assume k is < 0, i.e. k = -b ; b>0

Then,

int (te^(kt) dt ; t = 0 to inf) = int(te^(-bt) dt ; t = 0 to inf)

Employ Integration by parts int(uv') = uv - int(vu')

Here let,
u = t --> u' = 1
v' = e^(-bt) --> v = (-1/b)e^(-bt)

Thus,

int(te^(-bt) dt ; t = 0 to inf) = [ (-t/b)e^(-bt) ; t = 0 to inf ] - int( (-1/b)e^(-bt) (1) dt ; t = 0 to inf)
= (0 - 0) + (1/b)int(e^(-bt) dt ; t = 0 to inf)
= (1/b) [ (-1/b)e^(-bt) ; t = 0..int ]
= (1/b) [ (-1/b) - 0]
= -1/b^2

Hope this helps,

David

[5aboob]

by parts:

u= t ---> du = dt

dv = e^(kt) dt ----> v= 1/k e^kt

uv - int v du = t/k e^kt - int 1/k e^kt dt

= t/k e^kt - 1/k^2 e^kt + c

for the given limits the answer is infinity