Velocity, time, distance problem

[muddyjch]

Integration by parts velocity, time, distance problem

Homework Statement

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? This problem comes from an online homework covering sections on Substitution in indefinite integrals and integration by parts

Homework Equations

The Attempt at a Solution

It is just the integral right? s(t) = ∫ t^2 e^(-2t) dt

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

Integrating by parts, we have s(t) = (-1/2) e^(-2t) • t^2 - (1/2)(2) ∫ t e^(-2t) dt

Again, integrating by parts:

Let U = t dV = e^(-2t) dt
dU = dt V = (-1/2) e^(-2t)

s(t) = (-1/2) t^2 e^(-2t) - [(-1/2) t e^(-2t) - (-1/2) ∫ e^(-2t) dt] = (-1/2) t^2 e^(-2t) + (1/2) t e^(-2t) + (1/4) e^(-2t) = (-1/2) e^(-2t) ( t^2 - t + 1/2)

Do I plug in at t=0, s=0 to get my constant. That would give me C=.25, so s(t)= (-1/2)e^(-2t)*(t^2-t+1/2)+.25 but this is not the right answer?

 

[LCKurtz]

Homework Statement

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? This problem comes from an online homework covering sections on Substitution in indefinite integrals and integration by parts

Homework Equations

The Attempt at a Solution

It is just the integral right? s(t) = ∫ t^2 e^(-2t) dt

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

Integrating by parts, we have s(t) = (-1/2) e^(-2t) • t^2 - (1/2)(2) ∫ t e^(-2t) dt

Wrong sign on the second term. See if that fixes it.