Let's go through the steps to solve this problem related to the Laplace transform of $f(t) = t + 6$.

(a) Set up the integral for the Laplace transform:

The Laplace transform of a function $f(t)$ is given by:

$F(s) = \mathcal{L}\{f(t)\} = \int_0^{\infty} f(t) e^{-st} \, dt$

For $f(t) = t + 6$, the integral becomes:

$F(s) = \int_0^{\infty} (t + 6)e^{-st} \, dt$

Here:

• $A = 0$ (lower limit of the integral),
• $B = \infty$ (upper limit of the integral).

(b) Find the antiderivative of the integrand:

To solve the integral, we need to break it up:

$F(s) = \int_0^{\infty} t e^{-st} \, dt + 6 \int_0^{\infty} e^{-st} \, dt$

The antiderivative of each part is:

1. For $\int t e^{-st} \, dt$, we can use integration by parts. $\int t e^{-st} \, dt = \frac{1}{s^2}$
2. For $\int e^{-st} \, dt$, the result is: $\int e^{-st} \, dt = \frac{1}{s}$

Thus, the antiderivative becomes: [ F(s) = \left[ \frac{1}{s^2} \right]_0^\infty + 6 \left[ \frac{1}{s} \right]_0^\infty ]

(c) Evaluate limits to compute the Laplace transform:

Evaluating the limits:

• As $t \to \infty$, $e^{-st} \to 0$ (for $s > 0$),
• At $t = 0$, we substitute into the antiderivative.

After calculating the limits, the Laplace transform is:

$F(s) = \frac{1}{s^2} + \frac{6}{s}, \quad \text{for } s > 0$

(d) Domain of $F(s)$:

For the Laplace transform to exist, $s$ must be greater than 0, i.e., the domain is:

$s > 0$

Let me know if you want more details or clarifications on any part!

Here are 5 related questions:

1. How is the Laplace transform of a function $f(t)$ derived from first principles?
2. How does the property of linearity apply to Laplace transforms?
3. What are the common uses of Laplace transforms in solving differential equations?
4. How can we apply the Laplace transform to solve initial value problems?
5. How does the behavior of $F(s)$ change as $s \to \infty$?

Tip: Always remember that the Laplace transform converts a time-domain function into a function of a complex variable $s$, which simplifies solving differential equations.