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 Calculus Calculus Math Forum

 May 7th, 2019, 09:00 AM #1 Senior Member   Joined: Oct 2015 From: Greece Posts: 137 Thanks: 8 Problem with a simple integration! Things I know: $\displaystyle \delta ( \tau -1 ) = 1, \tau=1$ else $\displaystyle \delta ( \tau -1 ) = 0, \tau \neq 1$ and $\displaystyle \int \delta (\tau -1) = u(\tau -1)$ where $\displaystyle u ( \tau -1 ) = 1, \tau>1$ else $\displaystyle u ( \tau -1 ) = 0, \tau<1$ Integrate to Solve: $\displaystyle e^{-2t} \int_{1}^{t}e^{\tau} + e^{2 \tau} \delta ( \tau -1 )d \tau$ Solution: $\displaystyle e^{-2t} \cdot [ \int_{1}^{t}e^{\tau}d \tau + \int_{1}^{t}e^{2 \tau} \delta ( \tau -1 )d \tau ] = e^{-2t} \cdot [ e^{\tau}|_{1}^{t} + \int_{1}^{2}e^{2 \tau} \delta ( \tau -1 )d \tau + \int_{2}^{t}e^{2 \tau} \delta ( \tau -1 )d \tau]$ I know that $\displaystyle \int_{2}^{t}e^{2 \tau} \delta ( \tau -1 )d \tau = 0$ because $\displaystyle \delta ( \tau -1 ) = 0$ for $\displaystyle \tau \neq 1$ but what about $\displaystyle \int_{1}^{2}e^{2 \tau} \delta ( \tau -1 )d \tau$ can I say that $\displaystyle \delta ( \tau -1 ) = 1$ ? Or that 2 on the upper limit of the integral is messing this up? Tags integration, problem, simple Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post too_old Calculus 4 October 24th, 2017 07:52 AM fobbz Calculus 2 February 22nd, 2013 07:30 PM OriaG Calculus 4 September 18th, 2012 11:04 AM RMXByker Calculus 3 April 17th, 2010 09:44 PM raa4 Calculus 2 December 5th, 2007 06:07 AM

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