Question

The Laplace transform of \(e^{-at}u(t)\) is

• A. \(\dfrac{1}{s+a}\)
• B. \(\dfrac{1}{s-a}\)
• C. \(\dfrac{s}{s+a}\)
• D. \(\dfrac{a}{s+a}\)

Hint:

Some standard Laplace transforms: \[ \boxed{ \mathcal{L}\{e^{-at}u(t)\} = \frac{1}{s+a} } \] \[ \boxed{ \mathcal{L}\{e^{at}u(t)\} = \frac{1}{s-a} } \]

Answer 1

The Correct Option is A

Step 1: Recall the Laplace transform formula. For \(a>0\), \[ \mathcal{L}\{e^{-at}u(t)\} = \int_{0}^{\infty} e^{-at}e^{-st}\,dt. \]

Step 2: Evaluate the integral. \[ \mathcal{L}\{e^{-at}u(t)\} = \int_{0}^{\infty} e^{-(s+a)t}\,dt = \left[ -\frac{e^{-(s+a)t}}{s+a} \right]_{0}^{\infty} = \frac{1}{s+a}. \] Hence, \[ \boxed{\frac{1}{s+a}} \] Therefore, \[ \boxed{(A)} \] is the correct answer.