Homework Help: Role of mean value theorem in fundamental theorem of calculus proof

1. Jul 1, 2010

dylanbyte

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. Jul 1, 2010

njama

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.

3. Jul 1, 2010

snipez90

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
4. Jul 1, 2010

Tedjn

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 -_-.

5. Jul 1, 2010

Office_Shredder

Staff Emeritus
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

6. Jul 1, 2010

dylanbyte

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!