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Math Help - Laplace Transforms

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    Laplace Transforms

    Given the following:
    f(t) = 1 for 0 <= t <= 1
    f(t) = -1 for 1 < t <= 2
    f(t) = 0 elsewhere

    Find the Laplace Transform

    Solution thus far:
    The problem needs to be broken into two separate functions, one each for the time when f(t) = 1 and f(t) = -1.

    For 0 <= t <= 1

    L(s) = Integral (0 to 1) of e**-st f(t) dt

    L(s) = (1-e**-s)/s

    For 1 < t <= 2

    L(s) = Integral (1 to 2) of e**-st f(t) dt

    L(s) = (e**-2s - e**-s)/s

    Right?

    The solution in the back of the book (Advanced Engineering Mathematics, Kreyszig) says [(1 - e**-s) **2]/s.

    I need help understanding how to go from my answer (or correcting my answer) to the final answer. I'm missing something. Thanks in advance.
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    Quote Originally Posted by dsprice View Post
    Given the following:
    f(t) = 1 for 0 <= t <= 1
    f(t) = -1 for 1 < t <= 2
    f(t) = 0 elsewhere

    Find the Laplace Transform

    Solution thus far:
    The problem needs to be broken into two separate functions, one each for the time when f(t) = 1 and f(t) = -1.

    For 0 <= t <= 1

    L(s) = Integral (0 to 1) of e**-st f(t) dt

    L(s) = (1-e**-s)/s

    For 1 < t <= 2

    L(s) = Integral (1 to 2) of e**-st f(t) dt

    L(s) = (e**-2s - e**-s)/s

    Right?

    The solution in the back of the book (Advanced Engineering Mathematics, Kreyszig) says [(1 - e**-s) **2]/s.

    I need help understanding how to go from my answer (or correcting my answer) to the final answer. I'm missing something. Thanks in advance.
    \mathcal{L}(f(t))=\int_{0}^{\infty}e^{-st}f(t)dt=
    \int_{0}^{1}e^{-st}dt-\int_{1}^2 e^{-st}dt

    -\frac{1}{s}(e^{-s}-1)+\frac{1}{s}(e^{-2s}-e^{-s})=\frac{1}{s}(e^{-2s}-2e^{-2}+1)=\frac{1}{s}(e^{-s}-1)^2
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