Suppose a particle moves along a straight line with velocity v(t) = t^2 e^{-2 t} meters per second after t seconds.

It travels ______ meters during the first t seconds.

To find the distance traveled by the particle during the first \( t \) seconds, we need to integrate the velocity function \( v(t) = t^2 e^{-2t} \) with respect to time over the interval from \( 0 \) to \( t \).

The formula for the distance traveled is given by:

\[
\text{Distance} = \int_0^t v(\tau) \, d\tau = \int_0^t \tau^2 e^{-2\tau} \, d\tau
\]

Let's calculate the integral of \( \tau^2 e^{-2\tau} \) from \( 0 \) to \( t \). The integral can be solved using integration by parts.

### Step 1: Set up integration by parts
We apply the formula for integration by parts:
\[
\int u \, dv = uv - \int v \, du
\]
Let:
- \( u = \tau^2 \) and \( dv = e^{-2\tau} d\tau \)

Now, we differentiate \( u \) and integrate \( dv \):
- \( du = 2\tau d\tau \)
- \( v = -\frac{1}{2} e^{-2\tau} \) (since \( \int e^{-2\tau} \, d\tau = -\frac{1}{2} e^{-2\tau} \))

### Step 2: Apply integration by parts

\[
\int_0^t \tau^2 e^{-2\tau} \, d\tau = \left[ \tau^2 \left( -\frac{1}{2} e^{-2\tau} \right) \right]_0^t + \int_0^t 2\tau \left( -\frac{1}{2} e^{-2\tau} \right) \, d\tau
\]

Let's now compute this step by step:

1. First term:
\[
\left[ -\frac{1}{2} \tau^2 e^{-2\tau} \right]_0^t = -\frac{1}{2} t^2 e^{-2t} - 0 = -\frac{1}{2} t^2 e^{-2t}
\]

2. The remaining integral:
\[
\int_0^t \tau e^{-2\tau} \, d\tau
\]
This also requires integration by parts. Let:
- \( u = \tau \) and \( dv = e^{-2\tau} d\tau \)

Differentiate \( u \) and integrate \( dv \):
- \( du = d\tau \)
- \( v = -\frac{1}{2} e^{-2\tau} \)

Using integration by parts again:

\[
\int_0^t \tau e^{-2\tau} \, d\tau = \left[ \tau \left( -\frac{1}{2} e^{-2\tau} \right) \right]_0^t + \int_0^t \left( -\frac{1}{2} e^{-2\tau} \right) \, d\tau
\]

Evaluate this step:
\[
\left[ -\frac{1}{2} \tau e^{-2\tau} \right]_0^t = -\frac{1}{2} t e^{-2t}
\]
and
\[
\int_0^t \left( -\frac{1}{2} e^{-2\tau} \right) d\tau = \left[ \frac{1}{4} e^{-2\tau} \right]_0^t = \frac{1}{4} \left( e^{-2t} - 1 \right)
\]

### Step 3: Combine all parts

Putting everything together, the distance traveled is:

\[
\int_0^t \tau^2 e^{-2\tau} \, d\tau = -\frac{1}{2} t^2 e^{-2t} + \left( -t e^{-2t} + \frac{1}{4} \left( 1 - e^{-2t} \right) \right)
\]

Thus, the distance traveled during the first \( t \) seconds is:

\[
\boxed{D(t) = -\frac{1}{2} t^2 e^{-2t} - t e^{-2t} + \frac{1}{4} \left( 1 - e^{-2t} \right)}
\]

Would you like further details or clarification on any part of this process?

---

Here are five questions related to the concept:

1. How does integration by parts help in solving velocity integrals?
2. What is the physical significance of integrating velocity over time?
3. Can you explain why an exponential decay function appears in many velocity functions?
4. How do you find the total distance if the velocity function changes direction?
5. What is the difference between displacement and total distance?

**Tip**: When solving integrals involving products of polynomials and exponentials, integration by parts is usually an effective technique.

Suppose a particle moves along a straight line with velocity v(t) = t^2 e^{-2 t} meters per second after t seconds. It travels ______ meters during the first t seconds.

Learn how to calculate the distance traveled by a particle with velocity v(t) = t^2 e^{-2t} meters per second during the first t seconds. This problem involves integration by parts and exponential decay, suitable for high school and college-level calculus students.

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