Para obter a Transformada de Laplace de \( f(t) = 1 \), utilizamos a fórmula: \[ \mathcal{L}\{f(t)\} = \int_{0}^{\infty} f(t)e^{-st} \, dt \] Substituindo \( f(t) = 1 \): \[ \mathcal{L}\{1\} = \int_{0}^{\infty} 1 \cdot e^{-st} \, dt \] Agora, resolvemos a integral: \[ \mathcal{L}\{1\} = \int_{0}^{\infty} e^{-st} \, dt \] A integral de \( e^{-st} \) é: \[ \int e^{-st} \, dt = -\frac{1}{s} e^{-st} \] Avalia-se de \( 0 \) a \( \infty \): \[ \left[-\frac{1}{s} e^{-st}\right]_{0}^{\infty} = 0 - \left(-\frac{1}{s} e^{0}\right) = \frac{1}{s} \] Portanto, a Transformada de Laplace de \( f(t) = 1 \) é: \[ \mathcal{L}\{1\} = \frac{1}{s}, \quad \text{para } s > 0 \]