- Jan 17, 2006
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Sorry, I really hate to ask for calculus help here, but I have 30+ attempts on this problem and emailing the professor didn't really help.
The Problem:
A particle that moves along a straight line has velocity v(t)=t^2 * e^-3t meters per second after t seconds. How many meters will it travel during the first t seconds?
Here is my work so far, the S is the integral sign:
u=t^2 dv=e^-3t
du=2t v= -1/3 e^-3t
when I do uv-Svdu I get:
2t * -1/3 e^-3t - S(2t*-1/3 e^-3t)
Bringing the terms out front I get:
2t * -1/3 e^-3t + 2/3 S(t*e^-3t)
I integrate by parts again, but this time letting:
u=t dv=e^-3t
du=1 v=-1/3 e^-3t
So my formula becomes:
2t * -1/3 e^-3t + 2/3 (t*-1/3 * e^-3t - S (-1/3 e^-3t))
Bring out the terms once again:
2t * -1/3 e^-3t + 2/3 (-t/3 e^-3t + 1/3 S (e^-3t)
Integrate the last part and I get:
2t * -1/3 e^-3t + 2/3 (-t/3 e^-3t + 1/3(-1/3 e^-3t))
So my final answer becomes:
-2t/3 * e^-3t + 2/3 (-t/3 e^-3t - 1/9 e^-3t)
Thanks!
The Problem:
A particle that moves along a straight line has velocity v(t)=t^2 * e^-3t meters per second after t seconds. How many meters will it travel during the first t seconds?
Here is my work so far, the S is the integral sign:
u=t^2 dv=e^-3t
du=2t v= -1/3 e^-3t
when I do uv-Svdu I get:
2t * -1/3 e^-3t - S(2t*-1/3 e^-3t)
Bringing the terms out front I get:
2t * -1/3 e^-3t + 2/3 S(t*e^-3t)
I integrate by parts again, but this time letting:
u=t dv=e^-3t
du=1 v=-1/3 e^-3t
So my formula becomes:
2t * -1/3 e^-3t + 2/3 (t*-1/3 * e^-3t - S (-1/3 e^-3t))
Bring out the terms once again:
2t * -1/3 e^-3t + 2/3 (-t/3 e^-3t + 1/3 S (e^-3t)
Integrate the last part and I get:
2t * -1/3 e^-3t + 2/3 (-t/3 e^-3t + 1/3(-1/3 e^-3t))
So my final answer becomes:
-2t/3 * e^-3t + 2/3 (-t/3 e^-3t - 1/9 e^-3t)
Thanks!
