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Role of mean value theorem in fundamental theorem of calculus proof

  1. Jul 1, 2010 #1
    Hi, i've been watching the MIT lectures on single variable calculus, and whilst proving FTC, he mentions that we since we know that: <$> f'(x) = g'(x) </$>, then by MVT we know that <$> f(x) = g(x) + C </$>.

    I have tried searching for somewhere where this implication is spelled out for me, but I'm having trouble.

    As I understand MVT, it tells us that for two points on a function, there is a tangent, in that interval, parallel to the secant described by those points.


    A quick explanation would be really appreciated so I can fully appreciate this goodly resource!

    Thank you :)
     
  2. jcsd
  3. Jul 1, 2010 #2
    f'(x) = g'(x) then by CDT (Constant Difference Theorem) f(x)=g(x)+C.

    Here is something to know:

    "The Constant Difference Theorem has a simple geometric interpretation - it tells us that if f and g have the same derivative on an interval, then there is a constant k such that f(x)=g(x)+k for each x in the interval; that is, the graphs of f and g can be obtained from one another by a vertical translation" - Calculus (Seventh Edition) by Howard Anton, Irl Bivens and Stephen Davis.
     
  4. Jul 1, 2010 #3
    Analytically, the implication you are describing is a corollary of the fact that a function with derivative zero must be a constant. For a precise statement and the details, see http://en.wikipedia.org/wiki/Mean_value_theorem#A_simple_application". The corollary itself follows by defining a new function h to be the difference of f and g, i.e. h = f-g.
     
    Last edited by a moderator: Apr 25, 2017
  5. Jul 1, 2010 #4
    This is how it follows directly from MVT: For all x, f'(x) = g'(x) means for all x, (f-g)'(x) = 0. Consider the function (f-g)(x). Since it has a derivative equal to 0, it is differentiable (hence also continuous) everywhere. Consider for x and y with x < y, (f-g)(x) and (f-g)(y). By the mean value theorem, in (x,y) there exists z such that [(f-g)(x) - (f-g)(y)]/(x-y) = (f-g)'(z) = 0. Hence, (f-g)(x) = (f-g)(y) = C.

    EDIT: Well, I sort of jumped the gun on you, Office_Shredder -_-.
     
  6. Jul 1, 2010 #5

    Office_Shredder

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    I assume the question is more about how the MVT is used for this statement than why it's true.

    We can look at the function h(x)f(x)-g(x). It's derivative is zero, and I claim the MVT implies that means h(x) is constant. If you're not sure why, consider two points where h is unequal and ask what the MVT says about this situation
     
  7. Jul 1, 2010 #6
    Ah thank you for the extremely prompt and enlightening replies, I am still learning and what seems obvious to most is rather opaque to me. My mistake here, was I think, taking MVT at its boring face value, rather than the practical information it gives us from the derivative be it neg, pos or 0. Although I guess they're the same thing.

    Thanks again!
     
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