Please help! Integration by parts?

Question

A particle that moves along a straight line has velocity v(t)=t^(2)e^(-2t) meters per second after t seconds. How many meters will it travel during the first t seconds?

Thanks, but I think that this is more than an integration problem. Simply integrating does not give the correct answer. Maybe if I knew more about the relationship between position and velocity?

Answer 1

v(t) = t^2 e^(-2t) = (t e^(-t))^2, so v(t) is always non-negative. Consequently the distance travelled in the first t seconds really is just ∫(0 to t) v(r) dr
= ∫(0 to t) r^2 e^(-2r) dr
= [r^2 (e^(-2r) / (-2)][0 to t] - ∫(0 to t) (2r)(e^(-2r)/(-2)) dr
= -t^2 e^(-2t) + ∫(0 to t) r e^(-2r) dr
= -t^2 e^(-2t) + [r(e^(-2r)/(-2))][0 to t] - ∫(0 to t) e^(-2r)/(-2) dr
= -t^2 e^(-2t) - te^(-2t) / 2 + (1/2) [e^(-2r)/(-2)][0 to t]
= -t^2 e^(-2t) - te^(-2t) / 2 - (1/4) (e^(-2t) - 1)
= 1/4 - e^(-2t) (t^2 + t/2 + 1/4)
= 1/4 (1 - e^(-2t)(4t^2 + 2t + 1).

Note that if v(t) was sometimes positive and sometimes negative, you'd have to split the integral up into parts to get the distance travelled (distance = ∫ |v(t)| dt). However, the displacement (i.e. final position) will still be just ∫ v(t) dt.

The relationship between position and velocity is precisely that velocity is the derivative of position. In fact, that's the definition of velocity. Both are signed quantities (or more generally, vector quantities; but at this level we normally deal with 1 dimension only, so we only need positive and negative values to indicate the direction).

Answer 2

1 2 many times

Answer 3

I'm not sure of the problem, but I can do the integration by parts.

Integral (t^2 e^(-2t)) dt

Let u = t^2. dv = e^(-2t)
du = 2t dt. v = (-1/2) e^(-2t)

t^2 e^(-2t) - Integral (2t (-1/2) e^(-2t) dt)
t^2 e^(-2t) + Integral (t e^(-2t) dt)

Doing integration by parts again,

Let u = t. dv = e^(-2t) dt
du = dt v = (-1/2) e^(-2t)

t^2 e^(-2t) + (-1/2)t e^(-2t) - Integral ( (-1/2) e^(-2t) ) dt

t^2 e^(-2t) - (1/2) t e^(-2t) + Integral ( (1/2) e^(-2t) ) dt
t^2 e^(-2t) - (1/2) t e^(-2t) + (1/2) (-1/2)e^(-2t) + C
t^2 e^(-2t) - (1/2) t e^(-2t) - (1/4)e^(-2t) + C

Answer 4

if going down on earth, 32ft. per second per second.
if linear, you will have to know the value of t ( time )
these are hypothetical, for they do not include variables such as co-efficient of lift, drag, etc.

Answer 5

use tabular integration, it's a lot easier than integration by parts.

http://marauder.millersville.edu/~bikenaga/calc/parts/partspf.html