It's a pity that most students studying Calculus take the definite integral for granted. I notice that many of them blindly substitute the limits of integral when finding areas without knowing the processes which are at work to make the substitution possible.
In this section, I will explain to the best as I can the Fundamental Theorem of Calculus, the reason why we can so easily find the areas under the graph.
THE FUNDAMENTAL THEOREM OF CALCULUS.2
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THE FUNDAMENTAL THEOREM OF CALCULUS.1
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VIDEO LECTURE
part 2
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LESSON
All high school calculus courses will most probably start a discussion on the integral calculus with the Riemann sums. While it does work by taking the sum of small little areas below the graph, we get a rather complicated definition of the definite integral as
While this method does work, it is impractical and in some case even impossible to use when evaluating complicated integrals such as
Thus begs the question, how do we progress in evaluating these integrals, and in turn finding the area under the graph, bearing in mind that they didn’t have calculators in those days. Nonetheless, from the mind of Newton and Leibniz came a new method.
Their method of calculating the definite integral of such functions seems paradoxical in first sight. The approach was to ask a more difficult question: Instead of considering a fixed area like the graph on the left, how do we calculate the variable area, denoted by Step 1. We need to establish the fact that
which says that the rate of change of area
Now
Using the graphical reasoning above, we can write
since Step 2. In the previous step, we have the equation
This allows us to find a formula for the area
To determine
Reintroducing the point
And so I present to you the Fundamental Theorem of Calculus: If
Out of convenience and convention, we can also write
though I must stress that this actually means from the antiderivative of All information presentated, less questions and exercises, is original content of Donny, with slight references to various books.
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where 
is any antiderivative of 
.
and 
, when the right side of the area is considered movable. 
where the area is a function of 
. Clearly 
and 
is the fixed area in the graph on the left. We need to find an explicit formula for 
and then determine the area by letting 
. This is a two-step process. You ready? Then let’s go!
with respect to a change in 
is equal to the length to the right of the region, or equivalently 
. Well this may or may not be obvious we shall prove it here. Recall differentiating by first principles,
is the area under the graph from 
to 
and 
is the area between 
and 
. And thus the expression 
is the area between 
and 
. By the mean value theorem, we see that this area is exactly equal to the area of a rectangle with the same base whose height is 
where 
lies between 
and 
.
is continuous. For further elaboration, 
is equivalent to 
. Since 
is caught between 
and 
, we also have 
, and in turn 
.
. Now, 
is one of the antiderivatives of 
. I emphasize ONE of the antiderivatives because we know from differentiating that is that we can differentiate a any constant and get the same result. So now say that 
is any antiderivative of 
and thus
, we let 
knowing that this will give us 
and so 
. Therefore
as we have shown on the graph, recognize that
is continuous on a closed interval 
, and if 
, then
. We can now reduce the problem of evaluating definite integrals of complicating functions to finding their antiderivatives. 
