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Simple yet tricky definite integration problem!

  1. Apr 12, 2013 #1
    I'm given:
    [itex]f(x)=\left\{\begin{matrix}
    1 & -1\leq x \leq 1\\
    0 & otherwise
    \end{matrix}\right.[/itex]

    The integral to evaluate is:
    [itex]\int_{-1}^1 f(x-t) dt[/itex]

    What integration techniques should i use to solve this problem?
    Could someone please provide the steps to solve this problem (as the answer only provides the solution, and not the steps)?
     
  2. jcsd
  3. Apr 12, 2013 #2
    I'm not sure if it really works, but let me just write it down:

    Do the change of variable u=x-t then the integral becomes [itex]\int\limits_{x - 1}^{x + 1} {f(u)du} [/itex].
    Now consider 4 cases:
    (i) [itex]x - 1 \le - 1 < 1 \le x + 1[/itex]
    (ii)[itex] - 1 \le x - 1 < 1 \le x + 1[/itex]
    (iii)[itex]x - 1 \le - 1 < x + 1 \le 1[/itex]
    (iv)[itex] - 1 \le x - 1 < x + 1 \le 1[/itex]
     
  4. Apr 12, 2013 #3

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Forum rules require you to show your work. Surely you can figure out most of this problem by yourself! Just think about what f(x-t) is for various x and t.
     
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