Part #1 of the Fundamental Theorem
You know that for a positive continious function, the area under the curve is represented by the following equation:
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NOTE: the reason the dummy-varable t is used in the above equation is so the function variable is not confused with the x used to define the upper bound in the examples to follow. Also, for your information, the graphs below use the same function f(x) = 2x as used in the main discussion of this section.
If you pick the lower bound a = 0 and the upper bound x, the
following graph shows an area A(x):
Let's say you add an incremental amount to x and call it h (a change in x). The upper bound now is (x+h) and the area (orange) is defined as A(x+h). The following graph shows this area:

If you subtract the first area from the second, you are left with a change in area A(x+h) - A(x).

The orange area above represents a change in area. An isolation of this area is shown below.

The Extreme Value Theorem states that a continious function on a closed interval always has an absolute maximum and an absolute minimum on that interval. The plots below show the area of the rectangle representing the minimum value (left hand sum in this example) and the Maximum value (right hand sum).

The area of both of these rectangles is a simple matter of taking the height times the width. In the first one, the area is at the minimum:
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In the second one the area is at the maximum:
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Therefore, the orange area [A(x+h) - A(x)] must be greater than the first and less than the second.
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Dividing through by h gives:
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As h goes to 0 (zero), the rectangle gets thinner and thinner and the height (the larger rectangle with M as a side is used here) gets closer and closer to a point chosen in the interval P(x). This point could be located at any point on the function within the interval, but for this example, the left hand side was chosen.
The movie below displays the plot above as h goes to zero. Notice in the upper right hand corner of the movie, the value for h displays.
As you can see, for all practical purposes, as h goes to zero, the area [A(x+h)-A(x)] divided by h gives you f(x).
Remember, when you divide the area of a rectangle by one of the sides you will get the other side. In this case, you start out dividing the area ( A(x+h) - A(x) ) by h and you get M. But at the very last moment, M turns into P(x) which is the same as f(x)!
The above inequality can now be placed in the following form:
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Since the LHS of the equation above is merely the derivative function, you are left with the remarkable fact that the derivative of the area function is the original function. In other words, integration and differentiation are inverse operations.
or
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In LiveMath you can "Undo" an integration by taking the derivative. The following two graphics show the concept in LiveMath using the same function f(x) = 2 x.
The anti-derivative of the function is:

Taking the derivative of the integral confirms the inverse operation:

This extraordinary concept that the derivative and integration are inverse operations that Newton and Leibniz discovered 300 years ago is what stands them apart from all the other mathematicians prior to that time.
In the Plug-in demonstration below, change the the function and watch the integral operation return that same function back to you. This profound experiment leads you to the fact that the derivative and integral are inexorably linked to each other.
The second part of the Fundamental Theorem states the following: given the first equation below (sometimes called the derivative form), the second can be derived and is the method that makes the theorem practical. It is sometimes called the anti-derivative form.

The derivation goes as follows:
The first equation above says that the area function A(x) is an anti-derivative of the function f(x). To form this relation, apply the integral to both sides of the equation.
But, you also know that the RHS is equal to:
Combining the two equations:
Remember that the area function A(x) represents the area under the curve from a to x, where x is any value between a and b. Now, if x = a then A = 0.

So it follows by substituting back into equation (3):
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isolating c gives you:
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and equation (3) becomes:
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by letting x = b you get:
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Since the area function A(x) is represented by the definite integral:
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You have the Fundamental Theorem Part #2:
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