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 (from time =0 to time =t) ?

Answer 1

`\displaystyle -(1)/(2)\, t^(2) \, e^(-2t) - (1)/(2) t\, e^(-2 t) - (1)/(4) e^(-2 t) + (1)/(4)`.

Step-by-step explanation:

Distance travelled can be found by integrating speed, which is the absolute value (magnitude) of velocity. Since the value of velocity in this question is non-negative over the given range, the expression for speed would be the same as that for velocity:
`v(t) = t^(2)\, e^(-2 t)`.

The distance travelled during the given time would be equal to the definite integral:

`\displaystyle \int \limits_(0)^(t) t^(2) e^(-2 t)\, dt`.

To find the value of this definite integral, start by integrating
`v(t) = t^(2)\, e^(-2 t)` indefinitely using integration by parts.

In integration by parts:

`\displaystyle \int u\, v^(\prime) \, d t = u\, v - \int u^(\prime)\, v\, d t`,

Where:

• 
  `u` and
  `v` are two functions of
  `t`, and
• 
  `u^(\prime)` and
  `v^(\prime)` denote the first derivative of
  `u` and
  `v` with respect to
  `t`.

Using the LIATE rule, set
`u = t^(2)` (algebraic) and
`v^(\prime) = e^(-2 t)` (exponential,) such that
`u^(\prime) = 2\, t` and
`v = (-1/2)\, e^(-2 t)` (the constant of integration is omitted here.) Hence:

`\begin{aligned} \int t^(2) \, e^(-2 t)\, dt &= t^(2)\, \left(-(1)/(2)\, e^(-2 t)\right) - \int 2\, t\left(-(1)/(2)\, e^(-2 t)\right)\, d t \\ &= -(1)/(2)\, t^(2)\, e^(-2 t) + \int t\, e^(-2 t)\, d t\end{aligned}`.

Similarly, apply integration by parts to integrate
`t\, e^(-2 t)`. Set
`u = t` (algebraic) and
`v^(\prime) = e^(-2 t)` (exponential,) such that
`u^(\prime) = 1` and
`v = (-1/2)\, e^(-2 t)`. Using integration by parts:

`\begin{aligned} \int t \, e^(-2 t)\, dt &= t \, \left(-(1)/(2)\, e^(-2 t)\right) - \int \left(-(1)/(2)\, e^(-2 t)\right)\, d t \\ &= -(1)/(2)\, t\, e^(-2 t) +(1)/(2)\int e^(-2 t)\, d t \\ &= -(1)/(2)\, t \, e^(-2 t) + (1)/(2)\left(-(1)/(2)\, e^(-2 t)\right) + C \\ &= -(1)/(2)\, t \, e^(-2 t) -(1)/(4)\, e^(-2 t) + C\end{aligned}`,

Where
`C` is a constant of integration.

Substitute this result into the previous one:

`\begin{aligned} \int t^(2) \, e^(-2 t)\, dt &= t^(2)\, \left(-(1)/(2)\, e^(-2 t)\right) - \int 2\, t\left(-(1)/(2)\, e^(-2 t)\right)\, d t \\ &= -(1)/(2)\, t^(2)\, e^(-2 t) + \int t\, e^(-2 t)\, d t \\ &= -(1)/(2)\, t^(2)\, e^(-2 t) + \left( -(1)/(2)\, t \, e^(-2 t) -(1)/(4)\, e^(-2 t) + C\right) \\ &= -(1)/(2)\, t^(2)\, e^(-2 t) -(1)/(2)\, t \, e^(-2 t) -(1)/(4)\, e^(-2 t) + C\end{aligned}`.

Evaluate the definite integral for displacement using this result:

`\begin{aligned} &\int \limits_(0)^(t) t^(2) e^(-2 t)\, dt\\ =\; & \left[-(1)/(2)\, t^(2)\, e^(-2 t) -(1)/(2)\, t \, e^(-2 t) -(1)/(4)\, e^(-2 t)\right]^(t)_(0) \\ =\; & \left(-(1)/(2)\, t^(2)\, e^(-2 t) -(1)/(2)\, t \, e^(-2 t) -(1)/(4)\, e^(-2 t)\right) - \left(-(1)/(4)\right) \end{aligned}`.