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Question about derivative of an integral

  1. Apr 16, 2015 #1
    What does it mean to say that ##\displaystyle\frac{d }{d x}\int f(x)dx = f(x)##? Does this somehow relate to the fundamental theorem of calculus? If so, how?
     
  2. jcsd
  3. Apr 16, 2015 #2
    What is the fundamental theorem according to you?
     
  4. Apr 16, 2015 #3

    jtbell

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    It means that if you find $$F(x) = \int {f(x)dx}$$ (i.e. the antiderivative, a.k.a. indefinite integral) and then you find $$\frac{d}{dx}F(x)$$ you get ##f(x)## back again.
     
  5. Apr 17, 2015 #4

    Mark44

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    Another way to look at it is that differentiation and antidifferentiation are inverse operations.
     
  6. Apr 17, 2015 #5

    HallsofIvy

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    The "Fundamental Theorem of Calculus" has two parts:
    1) If we define [itex]F(x)= \int_a^x f(x)dx[/itex] then [itex]dF/dx= f(x)[/itex].
    2) If f(x)= dF/dx then [itex]F(x)= \int f(x) dx+ [/itex] some constant.
     
  7. Apr 19, 2015 #6
    How are those two parts different? Also, isn't there a part that relates the definite integral to the antiderivative?
     
  8. Apr 20, 2015 #7

    SteamKing

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    In 1) above, F(x) is the antiderivative of f(x).

    The First Fundamental Theorem of Calculus is more often expressed as:

    [itex]\int_a^b f(x)dx = F(b) - F(a)[/itex], where F(x) is the antiderivative of f(x), as defined in part 2) above.
     
  9. Apr 20, 2015 #8

    WWGD

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    Actually, the FTC says that ## \frac {d}{dx} \int_0^x f(t)dt =f(x) ##
     
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