First fundamental theorem of calculus

In summary, the conversation discusses the interpretation of the expression F'(x) = lim(h->0) (F(x+h) - F(x))/h = f(x) as the change in area with respect to change in "width" being the "height". The participants also mention the concept of the d/dx operator and its relationship to the height of the area under the curve of a function. One participant suggests drawing a diagram to understand the concept.
  • #1
ice109
1,714
6
how do i interpret this geometrically

[tex]F^\prime(x) = \lim_{h\to 0} \frac{F(x+h) - F(x)}{h} = f(x)
[/tex]
that the change in area with respect to change in "width" is the "height"? i don't think that's right.
 
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  • #2
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
anyone else?
 
  • #4
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.. :-p

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
morphism said:
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.. :-p

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

yes this is all true and quite clear to me, the problem I'm having is with the "change with respect to" concept. rewriting
[tex]
F^\prime
[/tex]
as
[tex]
\frac{dF}{dx}
[/tex]
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 there's my problem.
 
  • #6
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
morphism said:
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).

the area of a dx slice is f(x)?
 
  • #8
Sure. Draw a picture.
 
  • #9
morphism said:
Sure. Draw a picture.

that can't be true

dF=f(x)dx

f(x) is not the area of dF;f(x)*dx is
 
  • #10
So dF/dx=f(x)...
 
  • #11
morphism said:
So dF/dx=f(x)...

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
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:
  • #13
uart said:
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.
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
ice109 said:
maybe you didn't read the thread but i understand the dead easy concept that f(x)*dx is the area of the rectangle.

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
uart said:
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?

please tell me what you understand

[tex]\frac{dF}{dx}=f(x)[/tex]
to mean
 
  • #16
ice109 said:
please tell me what you understand

[tex]\frac{dF}{dx}=f(x)[/tex]
to mean

I understand it to mean exactly what you said in your first post. Namely that [tex]f(x) = \lim_{h\to 0} \frac{F(x+h) - F(x)}{h}[/tex]. I thought you were just looking for a simple graphical demonstration of this fact.
 
  • #17
again :

i understand

[tex]\frac{dF}{dx}=f(x)[/tex]

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
ice109 said:
again :

i understand

[tex]\frac{dF}{dx}=f(x)[/tex]

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.

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
uart said:
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).

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

FAQ: First fundamental theorem of calculus

What is the First Fundamental Theorem of Calculus?

The First Fundamental Theorem of Calculus is a mathematical theorem that establishes the relationship between differentiation and integration. It states that if a function f is continuous on the closed interval [a,b] and F is any function that satisfies F'(x) = f(x) for all x in [a,b], then the definite integral of f over the interval [a,b] can be evaluated as the difference between the values of F at the endpoints of the interval, i.e. ∫ab f(x) dx = F(b) - F(a).

How is the First Fundamental Theorem of Calculus different from the Second Fundamental Theorem of Calculus?

The First Fundamental Theorem of Calculus deals with the evaluation of definite integrals, while the Second Fundamental Theorem of Calculus deals with finding antiderivatives or indefinite integrals. In other words, the first theorem relates differentiation and integration, while the second theorem relates integration and differentiation.

What is the significance of the First Fundamental Theorem of Calculus?

The First Fundamental Theorem of Calculus plays a crucial role in the field of calculus as it provides a powerful tool for computing definite integrals. It also helps in understanding the relationship between the two fundamental operations of calculus - differentiation and integration.

What are the prerequisites for understanding the First Fundamental Theorem of Calculus?

To understand the First Fundamental Theorem of Calculus, one needs to have a solid understanding of basic calculus concepts such as limits, continuity, and the derivative. One must also be familiar with the properties of integrals and be able to solve simple integration problems.

Can the First Fundamental Theorem of Calculus be extended to multivariable calculus?

Yes, the First Fundamental Theorem of Calculus can be extended to multivariable calculus. The theorem states that if f is a continuous function on a closed, bounded region R in n-dimensional space and F is any function whose partial derivatives exist and are continuous on R, then the integral of f over R can be evaluated as the difference between the values of F at the endpoints of the region R. This is known as the Multivariable Fundamental Theorem of Calculus.

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