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given the piecewise function f(t)= { t 0<= t< 1 { 0 t>=1 the step unit function would look like \[f(t)=t-t\mathcal{U}(t-1)\] so would the laplace of that be the same as the laplace of the piecewise function? (obvious answer would be yes assuming i didnt write the unit step function wrong, but i calculated the laplace using the piecewise function rather than a step function and a completely different answer than i was suppose to)
answer is yes although calculating using the step function is much quicker if you know the laplace transform of unit functions than integrating the piece wise functions
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