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(Solved): If \( f(t) \) is continuous for \( t \geq 0 \), the "Laplace transform" of \( f \) is the function ...



If \( f(t) \) is continuous for \( t \geq 0 \), the Laplace transform of \( f \) is the function \( F \) defined by
\[
F(s)

If \( f(t) \) is continuous for \( t \geq 0 \), the "Laplace transform" of \( f \) is the function \( F \) defined by \[ F(s)=\int_{0}^{\infty} f(t) e^{-s t} d t \] and the domain of \( F \) is the set consisting of all number \( s \) for which the integral converges. (a) Find the Laplace transform of \( f(t)=1 \). (Make sure you can state the domain of \( F \) if we ask for it later!) \( F(s)= \) (b) Find the Laplace transform of \( f(t)=e^{t} \). (Make sure you can state the domain of \( F \) if we ask for it later!) \( F(s)= \) (c) Find the Laplace transform of \( f(t)=t \). (Make sure you can state the domain of \( F \) if we ask for it later!) \( F(s)= \)


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Given that the Laplace transform of f is the function F defined by F(s)=?0?f(t)e?stdt the domain of F is the set consisting of all number s for which
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