To find how many meters the particle travels during the first $t$ seconds, we need to calculate the definite integral of the velocity function $v(t)$ over the interval $[0, t]$. This gives the displacement:

$s(t) = \int_0^t v(x) \, dx = \int_0^t x^2 e^{-3x} \, dx$

Step 1: Set up the integral

The integral to evaluate is:

$\int_0^t x^2 e^{-3x} \, dx$

Step 2: Use integration by parts

We use integration by parts, where: $\int u \, dv = uv - \int v \, du$ Let:

• $u = x^2$ (so $du = 2x \, dx$)
• $dv = e^{-3x} \, dx$ (so $v = \frac{e^{-3x}}{-3} = -\frac{1}{3} e^{-3x}$)

Substitute into the formula:

$\int x^2 e^{-3x} \, dx = -\frac{x^2}{3} e^{-3x} + \int \frac{2x}{3} e^{-3x} \, dx$

Step 3: Repeat integration by parts for the remaining integral

For $\int \frac{2x}{3} e^{-3x} \, dx$, let:

• $u = x$ (so $du = dx$)
• $dv = e^{-3x} \, dx$ (so $v = -\frac{1}{3} e^{-3x}$)

Substitute again:

$\int x e^{-3x} \, dx = -\frac{x}{3} e^{-3x} + \int \frac{1}{3} e^{-3x} \, dx$

The final term is a straightforward integral:

$\int e^{-3x} \, dx = \frac{e^{-3x}}{-3} = -\frac{1}{3} e^{-3x}$

Thus: $\int x e^{-3x} \, dx = -\frac{x}{3} e^{-3x} - \frac{1}{9} e^{-3x}$

Step 4: Combine results

Now substitute back into the original integral: $\int x^2 e^{-3x} \, dx = -\frac{x^2}{3} e^{-3x} + \frac{2}{3} \left( -\frac{x}{3} e^{-3x} - \frac{1}{9} e^{-3x} \right)$

Simplify: $\int x^2 e^{-3x} \, dx = -\frac{x^2}{3} e^{-3x} - \frac{2x}{9} e^{-3x} - \frac{2}{27} e^{-3x}$

Factor out $e^{-3x}$: $\int x^2 e^{-3x} \, dx = e^{-3x} \left( -\frac{x^2}{3} - \frac{2x}{9} - \frac{2}{27} \right)$

Step 5: Apply limits

Evaluate the definite integral over $[0, t]$: $s(t) = \left[ e^{-3x} \left( -\frac{x^2}{3} - \frac{2x}{9} - \frac{2}{27} \right) \right]_0^t$

At $x = 0$, the expression becomes 0 because $e^{-3 \cdot 0} = 1$, but all terms involving $x$ vanish.

Thus: $s(t) = e^{-3t} \left( -\frac{t^2}{3} - \frac{2t}{9} - \frac{2}{27} \right) - 0$

Simplify: $s(t) = e^{-3t} \left( -\frac{t^2}{3} - \frac{2t}{9} - \frac{2}{27} \right)$

This is the displacement of the particle during the first $t$ seconds.

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Would you like to see a numerical example or further details? 😊

Related Questions:

1. How is integration by parts applied to more complex functions?
2. Can you explain the physical significance of definite integrals in kinematics?
3. What if the velocity function were sinusoidal instead of exponential?
4. How does one compute such integrals numerically if they are hard to solve analytically?
5. What happens to the particle’s displacement as $t \to \infty$?

Tip:

When solving integrals involving products of polynomials and exponentials, integration by parts is usually the go-to method!