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Math Help - L.Transform of Piecewise Function

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    L.Transform of Piecewise Function

    The function f(t) is defined as follows:

    <br />
f(t) = \begin{cases} t(t+1) & \ 0 < t < 2 \\ 1 & \ 2 \leq t \end{cases}

    Find  \mathcal{L} (f(t)) .
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    Rewrite using the unit step function u(t)

    f(t) = u(t)(t)(t-1) + u(t-2)[1- t(t-1)]


    use L{u(t-a)f(t-a)} = e^(-as)F(s)

    where F(s) = L{f(t)}

    note 1-t(t-1) = -t^2+t +1 = -(t-2)^2 -4(t-2) - 3
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    Quote Originally Posted by MrJack1990 View Post
    The function f(t) is defined as follows:

    <br />
f(t) = \begin{cases} t(t+1) & \ 0 < t < 2 \\ 1 & \ 2 \leq t \end{cases}

    Find  \mathcal{L} (f(t)) .
    If you are good at integration by parts, you can also do this from the definition of the LT:

    \mathcal{L}f (s)=\int_0^{\infty} f(x) e^{-st}\; dt=<br />
\int_0^{2} t(t+1) e^{-st}\; dt + \int_2^{\infty} e^{-st}\; dt

    CB
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