This is a calc 2 problem?

Question

A particle that moves along a straight line has velocity

v(t) = t^2 e^{-3 t}

meters per second after t seconds. How many meters will it travel during the first t seconds?

Answer 1

Distance during the first 3 seconds = integral v(t) dt from 0 to t

= integral t^2 e^{-3 t} dt from 0 to t

Integral can be computed using integration by parts.

Take t^2 as first function and e^(-3t) as second function.

Integral = first function * integral of second - integral of (differential of first * integral of second)

Integral = t^2 * e^(-3t)/(-3) - intgral [2t * e^(-3t)/(-3)]

= -(1/3)t^2e^(-3t) + 2/3 * integral [t * e^(-3t)]

= -(1/3)t^2e^(-3t) + 2/3 * [t * e^(-3t)/(-3) - integral e^(-3t)/(-3)] (again integrating by parts taking t as first function and e^(-3t) as second)

= -(1/3)t^2e^(-3t) - 2/9*t * e^(-3t) + 2/9 * integral e^(-3t)

= -1/3 * t^2e^(-3t) - 2/9 * t * e^(-3t) + 2/9 * e^(-3t)/(-3)

= -1/3 * t^2e^(-3t) - 2/9 * t * e^(-3t) - 2/27 * e^(-3t)

Taking limit from 0 to t

Distance = [-1/3 * t^2e^(-3t) - 2/9 * t * e^(-3t) - 2/27 * e^(-3t)] - [0 + 0 - 2/27]

= -1/3 * t^2e^(-3t) - 2/9 * t * e^(-3t) - 2/27 * e^(-3t) + 2/27

Ans: -1/3 * t^2e^(-3t) - 2/9 * t * e^(-3t) - 2/27 * e^(-3t) + 2/27

Answer 2

in other words, the question asks for the integral of v(t) from 0 to t.

d(t) = ∫v(t)

∫t² e^(-3t) dt

integrate by parts

∫u dv = uv - ∫v du

let u = t² ; dv = e^(-3t)

du = 2t dt ; v = (-1/3)e^(-3t)

substitute:

(-1/3)t² e^(-3t) - ∫(-1/3)(2)t e^(-3t) dt

(-1/3)t² e^(-3t) + ∫(2/3)t e^(-3t) dt

integrate by parts again

let u = (2/3)t ; dv = e^(-3t)

du = (2/3)dt ; v = (-1/3)e^(-3t)

(-1/3)t² e^(-3t) + (2/3)(-1/3)t e^(-3t) - ∫(-1/3)(2/3) e^(-3t) dt

(-1/3)t² e^(-3t) - (2/9)t e^(-3t) + ∫(2/9)e^(-3t) dt

(-1/3)t² e^(-3t) - (2/9)t e^(-3t) + (2/9)(-1/3)e^(-3t) + C

(-1/3)t² e^(-3t) - (2/9)t e^(-3t) - (2/27)e^(-3t) + C

so d(t) = (-1/3)t² e^(-3t) - (2/9)t e^(-3t) - (2/27)e^(-3t) + C

so during the first t seconds, it will travel:

d(t) = (-1/3)t² e^(-3t) - (2/9)t e^(-3t) - (2/27)e^(-3t) - (2/27) + C

where C is the intial position

Edit: i'm sure my integration is 100% correct but i'm not sure if i should have put the (-2/27) + C there. Hope someone has a better explaination.