(adsbygoogle = window.adsbygoogle || []).push({}); [SOLVED] A Funky Integral

Let

[tex]f(t) =\left\{\begin{array}{lll}

0 &\mbox{ if }t < 0 \\

1 &\mbox{ if }0 \leq t \leq 1 \\

0 &\mbox{ if }t > 1

\end{array}\right[/tex]

I need a simplified version of [itex]g(t)[/itex] shown below:

[tex]g(t) = \int_{-\infty}^\infty f(u) \cdot f(t - u) \, du[/tex]

I divide the integral into the following sum:

[tex]g(t) = \int_{-\infty}^0 f(u) \cdot f(t - u) \, du + \int_{0}^1 f(u) \cdot f(t - u) \, du + \int_{1}^\infty f(u) \cdot f(t - u) \, du[/tex]

[itex]f(u)[/itex] in the first and third term is 0 according to the definition of [itex]f[/itex]. Hence

[tex]g(t) = \int_0^1 f(t - u) \, du[/tex]

Is this right? What do I do know? Should I consider the cases where [itex]t < 0[/itex], [itex]0 \le t \le 1[/itex] and [itex]t > 0[/itex]? Or should that be [itex]t - u[/itex]?

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# Homework Help: A Funky Integral

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