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I need some help with the following integration/use of fundamental theorem of calculus:

$\displaystyle x(t) = \int_{0}^{t} \exp (-2s)a(s) \ ds$, where

$a(x) = \left\{ \begin{array}{lr} -1 & : t \in [0,1]\\ 1& : t \in (1, \infty) \end{array} \right.$

How should I simplify the integral?

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first integrate inner part and show us what you get – dato datuashvili Jan 15 '13 at 6:22
up vote 3 down vote accepted

Since $\displaystyle \exp \left \{-\int\limits_{0}^{s} 2 \ du\right \}=e^{-2s}, $ then $$x(t)=\int\limits_{0}^{t}{e^{-2s}a(s) \ ds}=\begin{cases} -\int\limits_{0}^{t}{e^{-2s} \ ds}: & t \in [0,1]\\ -\int\limits_{0}^{1}{e^{-2s} \ ds}+\int\limits_{1}^{t}{e^{-2s} \ ds}: & t \in (1, \infty) \end{cases}$$

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Show that left and right limits at $t=1$ are equal: $$\lim\limits_{t\to{1-0}}x(t)=\lim\limits_{t\to{1+0}}x(t)=-\int\limits_{0}^{1}{e‌​^{-2s} \ ds}=\dfrac{1}{2}-\dfrac{1}{2e^2}$$ – M. Strochyk Jan 15 '13 at 6:47
Why would $x$ not be continuous at $t=1$? – AD. Jan 15 '13 at 6:58

Compute. If $0\le t\le 1$, we want $$\int_0^t -e^{-2s}\,ds.$$ For $t\gt 1$, we want $$\int_0^1 -e^{-2s}\,ds +\int_1^t e^{-2s}\,ds.$$ The integration is straightforward.

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