Question

Use the Laplace transform integral, X(s)=∫₀[infinity]x(t)e⁻ˢᵗdt, to prove that the Laplace transform of x(t)=e⁻ᵃᵗ is X(s)= 1/s+a

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Answer 1

To prove the Laplace transform, substitute x(t) = e^(-at) into the Laplace transform integral and solve for X(s). By evaluating the integral, we find X(s) = 1/(s+a).

Step-by-step explanation:

To prove that the Laplace transform of x(t) = e^(-at) is X(s) = 1/(s+a), we can use the definition of the Laplace transform integral. The Laplace transform of x(t) is represented by the integral X(s) = ∫₀[∞]x(t)e^(-st)dt. In this case, we substitute x(t) = e^(-at) into the integral and solve for X(s).

We have X(s) = ∫₀[∞]e^(-at)e^(-st)dt. By combining the exponents, we have X(s) = ∫₀[∞]e^(-(s+a)t)dt. Now, we can evaluate this integral.

Using the integral property ∫e^(-kt)dt = -1/k * e^(-kt) + C, where k = s + a in this case, we find X(s) = -1/(s+a) * e^(-(s+a)t)]₀[∞] = 1/(s+a).