1. **State the problem:** Find the Laplace transform $F(s)$ of the function $f(t) = e^{4t} + 4$ using the definition of the Laplace transform. 2. **Recall the definition:** The Laplace transform of a function $f(t)$ is given by $$F(s) = \mathcal{L}\{f(t)\} = \int_0^\infty e^{-st} f(t) \, dt$$ where $s$ is a complex variable and the integral converges for certain values of $s$. 3. **Apply linearity:** Since $f(t) = e^{4t} + 4$, we use linearity of the Laplace transform: $$F(s) = \mathcal{L}\{e^{4t}\} + \mathcal{L}\{4\}$$ 4. **Find each transform:** - For $\mathcal{L}\{e^{4t}\}$, use the formula $\mathcal{L}\{e^{at}\} = \frac{1}{s - a}$ for $s > a$. - For $\mathcal{L}\{4\}$, since 4 is a constant, $\mathcal{L}\{c\} = \frac{c}{s}$ for $s > 0$. 5. **Calculate each term:** $$\mathcal{L}\{e^{4t}\} = \frac{1}{s - 4} \quad \text{for } s > 4$$ $$\mathcal{L}\{4\} = \frac{4}{s} \quad \text{for } s > 0$$ 6. **Combine results:** $$F(s) = \frac{1}{s - 4} + \frac{4}{s}$$ 7. **Determine domain:** The Laplace transform exists where both terms converge, so $$s > 4$$ **Final answer:** $$F(s) = \frac{1}{s - 4} + \frac{4}{s} \quad \text{for } s > 4$$