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Homework Help: Piecewsie Integral

  1. Oct 5, 2005 #1
    Hi from a newbie.

    I have an apparently simple problem, but could use some guidance on how to do it.

    I have the following function:

    [tex]F(x)=\int^{x}_{0}f(t)dt,\ 0\leq x\leq3[/tex]

    where [tex]f(x)=\left\{\begin{array}{lll}1,&\mbox{ if }0\leq x<1\\2-x,&\mbox{ if }1\leq x<2\\0,&\mbox{ if }2\leq x\leq 3\end{array}\right[/tex]

    Is it as simple as:

    [tex]F(x)=\left\{\begin{array}{lll}x,&\mbox{ if }0\leq x<1\\2x-\frac{x^2}{2},&\mbox{ if }1\leq x<2\\0,&\mbox{ if }2\leq x\leq 3\end{array}\right[/tex]

    A plot of F(x) for the interval would show discontinuities at x=1 and x=2.

    Thanks for your help.

    Bailey
     
  2. jcsd
  3. Oct 5, 2005 #2

    Tom Mattson

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    No, it's a little more complicated than that. For instace when you integrated over the second interval you treated it as follows:

    [tex]\int_0^x(2-x)dx[/tex]

    However, [itex]F(x)[/itex] is not defined as [itex]2-x[/itex] for [itex]0\leq x<1[/itex], so it is not correct to start integrating that piece of the function at [itex]x=0[/itex]. There is a similar problem with the third interval.
     
  4. Oct 5, 2005 #3
    Okay...I can see I'm way off.

    What would be the proper steps to take to find F(x) knowing the integral is from 0 to x?

    Thanks for helping.

    Bailey
     
  5. Oct 5, 2005 #4

    Tom Mattson

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    The integral of your first piece is OK. You want to integrate it on the interval [0,x].

    But your second and third pieces need to be integrated over [1,x] and [2,x], respectively.
     
    Last edited: Oct 5, 2005
  6. Oct 5, 2005 #5

    Tom Mattson

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    Actually, the third piece is OK too, but only because it is zero.
     
  7. Oct 5, 2005 #6
    So, integrating for each piece over the respective intervals would yield a plot like the one attached, for 0 <= x <= 3.

    Is that correct, with the discontinuities at x = 1 and x = 2?
     

    Attached Files:

  8. Oct 6, 2005 #7
    Good morning all.

    Just want to find out if the graph above is correct.

    Bailey
     
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