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Integration Proof with constant

  1. Dec 4, 2015 #1
    1. The problem statement, all variables and given/known data
    Prove that if f'(x) = g'(x) for all x in an interval (a,b) then f-g is constant on (a,b) then f-g is constant on (a,b) that is f(x) = g(x) + C

    2. Relevant equations
    Let C be a constant
    Let D be a constant
    3. The attempt at a solution
    f(x) = antiderivative(f'(x)) = f(x) + C
    g(x)= antiderivative(g'(x)) = g(x) + D

    f-g = f(x) + C - (g(x) + D)
    f-g = f(x) - g(x) + C - D

    but since f'(x) = g'(x) then f(x) = g(x) the only difference is their constant.

    then,

    f-g = f(x) - f(x) + C - D
    f-g = C - D

    Since C and D are constants

    then,

    f-g = constant

    if C = D

    then f-g = 0

    Note: I feel like I proved it but my notation is wrong since I cannot use f(x) = f(x) + C. I would like guidance for the proper notation to use. The possibility also exists my proof is completely wrong. I would like help

    Thanks in Advanced!
     
  2. jcsd
  3. Dec 4, 2015 #2

    Simon Bridge

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    You have not used the interval.
     
  4. Dec 4, 2015 #3
    How would I got about approaching this problem? I'm sorry perhaps I am lacking knowledge but I have only received a 30 minute lesson on integrals in my class, which was the last class of the semester.
     
  5. Dec 4, 2015 #4

    PeroK

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    Here's a suggestion to get you started. Let ##h(x) = f(x) - g(x)##. What can you say about ##h(x)##?
     
  6. Dec 4, 2015 #5
    h(x) is constant on (a,b). Should I be using the area beneath the curve formula?
     
  7. Dec 4, 2015 #6
    How did you deduce that ##h(x)## is constant on ##(a,b)##? PeroK has a very good suggestion, under what conditions would ##h(x)## be constant over an interval? Could this have something to do with its derivative?
     
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