# First fundamental theorem of calculus

1. Oct 13, 2007

### ice109

how do i interpret this geometrically

$$F^\prime(x) = \lim_{h\to 0} \frac{F(x+h) - F(x)}{h} = f(x)$$
that the change in area with respect to change in "width" is the "height"? i don't think that's right.

2. Oct 13, 2007

### mathman

The expression on the left side before taking the limit is the tangent of the angle between the line (a), connecting the points (x,F(x)) and (x+h,F(x+h)), and the horizontal line (b), parallel to the x axis. As you know the limiting value is the slope of the tangent at (x,F(x)).

3. Oct 13, 2007

### ice109

anyone else?

4. Oct 13, 2007

### morphism

I guess a naive description is that F(x+h) - F(x) is the difference in area under the curve of f(t) from [a, x] and [a, x+h]. But if h is small, we would expect this new area to be between h*f(x) and h*f(x+h), i.e.
f(x) <= (F(x+h) - F(x))/h <= f(x+h)

Now take limits as h->0.. :tongue2:

Is that what you're looking for? If it's not too satisfying, then by further naivifying the situation, the extra area can be thought of as the area of the added "infinitesimal" rectangle, i.e. f(x)*h (for h "really, really" small). So in this case, F(x+h) - F(x) =~ f(x)*h

5. Oct 13, 2007

### ice109

yes this is all true and quite clear to me, the problem i'm having is with the "change with respect to" concept. rewriting
$$F^\prime$$
as
$$\frac{dF}{dx}$$
and then reading that as the change of F, which is the area, with respect to a change x is the value of f(x) at that value. it make no sense to me that the area function simply grows by adding up all the values of f(x), and it doesn't. but theres my problem.

6. Oct 13, 2007

### morphism

But F' is the "instantaneous change in area"... So by my last paragraph, this change is really the 'edge' of the area under the graph of f(t) on [a,x], i.e. f(x).

7. Oct 13, 2007

### ice109

the area of a dx slice is f(x)?

8. Oct 13, 2007

### morphism

Sure. Draw a picture.

9. Oct 13, 2007

### ice109

that can't be true

dF=f(x)dx

f(x) is not the area of dF;f(x)*dx is

10. Oct 13, 2007

### morphism

So dF/dx=f(x)...

11. Oct 13, 2007

### ice109

again we're going over the same things, i'll believe that f(x) divided by dx equal the height but that's not how the d/dx operator is read/understood.

12. Oct 14, 2007

### uart

I don't see what the confusion is about.

1. Draw the graph of a nice simple function and label it f(x).

2. Choose a particular value for x and draw a vertical line to f(x) there.

3. Choose a second point just a little larger than the first, call it x+h and draw a second vertical line there.

4. If F(x) represents the area (under the curve) to the left of the vertical line at x, and F(x+h) represents the area to the left of the second vertical line at x+h, then F(x+h)-F(x) represents the area between the two vertical lines. Draw a picture, it's a dead easy concept.

5. The area between the two vertical lines, F(x+h)-F(x), can clearly be approximated by the rectangle of height f(x) and width h. Surely you can see it from there.

Last edited: Oct 14, 2007
13. Oct 14, 2007

### ice109

maybe you didn't read the thread but i understand the dead easy concept that f(x)*dx is the area of the rectangle.

14. Oct 14, 2007

### uart

Ok I guess I don't understand what you're asking then.

You agree that it's easy to draw a diagram (as per the above steps) to illustrate that F(x+h) - F(x) tends to h * f(x) right. So if that's not the answer then I'm misunderstanding exactly what was the question?

15. Oct 14, 2007

### ice109

please tell me what you understand

$$\frac{dF}{dx}=f(x)$$
to mean

16. Oct 14, 2007

### uart

I understand it to mean exactly what you said in your first post. Namely that $$f(x) = \lim_{h\to 0} \frac{F(x+h) - F(x)}{h}$$. I thought you were just looking for a simple graphical demonstration of this fact.

17. Oct 14, 2007

### ice109

again :

i understand

$$\frac{dF}{dx}=f(x)$$

to mean the change of F(x) with respect to a change in x is f(x). this to means that if i add some value to x the change in AREA will be f(x+value) which is not correct.

18. Oct 14, 2007

### uart

No I don't know where you are getting that idea from. If you add a value (say h) to x then the change in area is h * f(x), not f(x+h).

19. Oct 14, 2007

### ice109

if i changed the d to $\Delta$ it becomes very obvious where i get that idea. plus that's how the operator is read.