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Math Help - Integration Theorem

  1. #1
    Senior Member slevvio's Avatar
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    Integration Theorem

    I have this in my notes:

    Define continuous functions on [a,b] by  f_+ (t) = max(f(t),0) and  f_- (t) = max(-f(t),0). Then  f(t) = f_+ (t) - f_- (t) and  |f(t)| = f_+ (t) + f_- (t). . The result follows from linearity and the triangle inequality by the following calculation:

     |\int_a^b f(t)dt | = | \int_a^b (f_+ (t) - f_- (t))dt | \le \int_a^b f_+(t)dt  + \int_a^b f_- (t) dt= \int_a^b |f(t)| dt .

    I understand that | \int_a^b (f_+ (t) - f_- (t))dt | \le |\int_a^b f_+(t)dt|  + |\int_a^b f_- (t) dt| but not why  | \int_a^b (f_+ (t) - f_- (t))dt | \le \int_a^b f_+(t)dt  + \int_a^b f_- (t) dt .

    Any help in this will be much appreciated thank you
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  2. #2
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    It uses the fact that \left|\int f(x)dx\right|\le \int |f(x)|dx.
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  3. #3
    Senior Member slevvio's Avatar
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    What I have written is the proof for that fact but its one of the steps that I don't understand
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  4. #4
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    Oh, of course, I completely misread.

    Do you see that if f(x) is non-negative on (a,b) then \left|\int_a^b f(x)dx\right|= \int_a^b f(x)dx= \int_a^b |f(x)| dx?

    Thats true because both f(x) and the integral are never negative so you can just "discard" the absolute values.

    Now, by its definition, f_+(x) is non-negative. By its definition, f_-(x) is non-positive so is -f_-(x) is non-negative.

    Clearly, f(x)= f_+(x)+ f_-(x) while |f(x)|= f_+(x)- f_-(x). If that's not clear, think about what happens for each x if (1) f(x)\ge 0, (2) f(x)< 0.

    Now, \int_a^b |f(x)|dx= \int_a^b (f_+(x)- f_-(x))dx= \int_a^b f_+(x)dx- \int_a^b f_-(x) dx while \left|\int_a^b f(x)dx\right|= \left|\int_a^b f^+(x)dx+ \int_a^b f_-(x)dx\right|.

    Finally, use the fact that, for numbers x and y, |x- y|\le |x|+ |y|.
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  5. #5
    Senior Member slevvio's Avatar
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    thank you very much
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