second fundamental calculus theorem

Dear All!

Proof ofthe second fundamental theorem

Please, how do we get in the sixth line that f(x) is right the one f(t), also, the sam f function as we have in the lines before?

In other words, how do we KNOW that $\displaystyle (F(x+\Delta x)-F(x))=f(x)$ The SAME f(x) we have in the right side before?

I understand that it is an DERIVATIVE, SURE! "An derivative! " But how do we know it is the same FUNCTION as we have f(t) on the right?

Many thanks!

Re: second fundamental calculus theorem

I get the error:

Forbidden

You don't have permission to access /~djk/18_01/chapter14/proof01.html on this server.

- Hollywood

Re: second fundamental calculus theorem

It is the "mean value theorem". As long as F is differentiable, there exist c between x and x+ h such that (F(x+h)- F(x))/h= F'(c)= f(c) because F is defined as the anti-derivative of f. As h goes to 0, since c is always between x and x+h, c goes to x and f(c) becomes f(x).

Re: second fundamental calculus theorem

Quote:

Originally Posted by

**HallsofIvy** It is the "mean value theorem". As long as F is differentiable, there exist c between x and x+ h such that (F(x+h)- F(x))/h= F'(c)= f(c) because F is defined as the anti-derivative of f.

In the link, the fact that F is the anti-derivative of f is what has to be proved. F is defined as the integral of f.

Re: second fundamental calculus theorem

Hello!

Thank U all very much, but I think we did not understand each other.

Sure it is clear that F'=f

My question is:

how do we know that this f from the left side of the eqation is the SAME f we have on the right side.

Proof ofthe second fundamental theorem

The functin on the right side we marked as "f".

Then, the derivative function on the left side we get we name AGAIN WITH THE SAME NAME! (f)

hOW can we know that f=f?

p.s. we could name the derivative on the left g8x) couln't we?

So, what woudl tell us then that g(x)=f(x)?

many thanks if someone has nerves for this-and sory if I am too intrusive!

Re: second fundamental calculus theorem

Quote:

Originally Posted by

**emakarov** In the link, the fact that F is the anti-derivative of f is what has to be proved. F is defined as the integral of f.

Ah! "F" is defined as the area under the curve. In that case, we **don't** know that "$\displaystyle F(x+ \Delta x)- F(x)= f(x)$", it is NOT, in general, true. What **is** true is that $\displaystyle \frac{F(x+ \Delta x)- F(x)}{\Delta x}= f(x_0)$ where $\displaystyle x_0$ is some number **between** $\displaystyle x$ and $\displaystyle x+ \Delta x$. That is true because of the "intermediate value property".

If we were to draw a horizontal line at $\displaystyle (x_M, f(x_M))$ where $\displaystyle f(x_M)$ is the **maximum** of f on the interval from x to $\displaystyle x+\Delta x$, we get a rectangle, of area $\displaystyle f(x_M)\Delta x$ which completely contains the area under the graph: $\displaystyle f(x_M)\Delta x> F(x+\Delta x)- F(x)$.

If we were to draw a horizontal line at $\displaystyle (x_m f(x_m))$ where $\displaystyle f(x_m)$ is the **minimum** of f on the interval from x to $\displaystyle x+ \Delta x$, we get a rectangle, of area $\displaystyle f(x_m)\Delta x$ which is completely **contained in** the area under the graph: $\displaystyle F(x+ \Delta x)- F(x)> f(x_m)\Delta x$.

(Using the intermediate value theorem, as well as guarenteeing that those "maximum" and "minimum" values **exist** requires that f be continuous. More generally, if f is not continuous at some points (but stays finite) we can work between points of discontinuity. The anti-derivative is the same as the 'area under the curve' (Riemann integral) as long as f has only a finite number of points of discontinuity and is bounded on the interval.)

Putting those together $\displaystyle f(x_m)\Delta x< F(x+ \Delta x)- F(x)< f(x_M)\Delta x$ which, since $\displaystyle \Delta x> 0$, is the same as $\displaystyle f(x_m)< \frac{F(x+ \Delta x)- F(x)}{\Delta x}< f(x_M$.

By the "intermediate value property then, there exist a value of x, $\displaystyle x_0$, between $\displaystyle x_m$ and $\displaystyle x_M$, such that $\displaystyle f(x_0)= \frac{F(x+ \Delta x)- F(x)}{\Delta x}$.

It is taking the **limit** as $\displaystyle \Delta x$ goes to 0 so that $\displaystyle x+ \Delta x$ goes to x, and $\displaystyle x_0$ is "trapped" between them, that gives F'(x)= f(x).