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Simple Integral

  1. Apr 10, 2012 #1
    1. The problem statement, all variables and given/known data

    Ok part of my math homework leads me to do this integral
    The integral from 0 to 7 of sqrt((2t-10)^2) dt

    2. Relevant equations



    3. The attempt at a solution

    The answer is 29, but I can only seem to get -21. I have it = integral of (2t-10)dt, so then I get t^2 - 10t from 0 to 7 = 49-70 = -21. Any help is appreciated. Thanks
     
  2. jcsd
  3. Apr 10, 2012 #2

    Ray Vickson

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    If f(t) = sqrt(2t-10)^2), what is f(0)? If g(t) = 2t-10, what is g(0)?

    RGV
     
  4. Apr 10, 2012 #3
    if y= sqrt(x^2)
    Can u simplify y?
     
  5. Apr 10, 2012 #4
    Thats agood question Vickson.
     
  6. Apr 10, 2012 #5
    Do i need an absolute value somewhere? If I don't simplify as I did and solve with substitution I still get (t-5)**2 from 0 to 7 which still yields -21. Where am I going wrong?
     
  7. Apr 10, 2012 #6
    Nvm that last part, I still used sqrt(x^2) = x
     
  8. Apr 10, 2012 #7

    Ray Vickson

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    In my first reply I asked you a question. You still have not answered it. (BTW: I asked for a very good reason!)

    RGV
     
  9. Apr 10, 2012 #8
    well one is -10 and the other is 10
     
  10. Apr 10, 2012 #9

    Ray Vickson

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    OK, so what can you learn from that? Do you, or do you not need to use absolute values?

    RGV
     
  11. Apr 10, 2012 #10
    i just ended up using x**2 = x*sqrt(x*2), so it keeps its sign and everything works out
     
  12. Apr 10, 2012 #11

    Ray Vickson

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    It sounds like you are trying to avoid the basic issue, which comes up over and over again and so is best dealt with now, once and for all. The point is:
    [tex] \text{for any real number }r, \; \sqrt{r^2} = |r|.[/tex]

    RGV
     
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