If F'(x)=f(x) then the integral of f(x) from a to b is given by F(b)-F(a)

Fundamental Theorem of Calculus II

Let f be an integrable function on [a, b]. For x in [a, b], let F(x) be the integral of f(x) from a to x. Then F is continuous on [a, b]. If f is continuous at x0 in (a, b), then F is differentiable at x0 and F'(x0) = f(x0).

Note that FTC I (above w/u) does not imply FTC II and FTC II does not imply FTC I, and so are not equivalent definitions.
I'd like to try and provide a somewhat intuitive explanation of this theorem, since, if you're anything like me, when you first learn it, there seems to be nothing at all intuitive about it.

So, say we have some integrable function f defined on the real numbers. Then, if we assume f is positive, recall that the definite integral of f from a to b, written

     |  f(t) dt 
    \/ a

can be thought of as the area under f between a and b. OK, then consider the function F defined as follows:
F(x):= |  f(t) dt 
      \/ a
where x > a. Then F(x) can be thought of as giving us the "area so far" under f from a to x. We can see this graphically (where the "curve" is the graph of the function f):
|          .-.
|     ..--'   \    __.--- f
|    /*        '--'
|__.' *          *
|     *   F(x)   *
|     *          *
      a          x

Now the Fundamental Theorem of Calculus tells us that the derivative of F at x (let's call it F'(x) ), is equal to f(x). This does not seem obvious at first, but let's see if we can get an idea of why this is so. Remember that the derivative of F at x is defined to be the limit as h goes to zero of ((F(x+h)-F(x))/h); that is:
     f(x+h) -f(x)
lim  ------------
h->0      h

Now, consider the quantity ((F(x+h)-F(x))/h) for some small value of h. Now, F(x+h)-F(x) gives us the area under f between x and x+h:
    f(x)___..--'   * 
       *           *
       * F(x+h) -  *
       *   F(x)    *
       *           *
       x         x+h  
       <---- h ---->

But since h is small, we can approximate this area by the area of a rectangle with a height of f(x) and width h. So, we have the approximate equality, h*f(x) ~= F(x+h)-F(x). Dividing both sides by h, we have f(x) ~= (F(x+h)-F(x))/h. Now, if you make h smaller and smaller, then it seems like it would make sense that this approximation would get better and better. So, we haven't proven it, but I hope it's at least believable that when we take the limit as h goes to zero, the approximate equality above becomes actual equality, and we know (by definition) that the right hand side of the equality becomes F'(x), so f(x)=F'(x).

I feel like this makes it a little bit easier to understand what exactly is going on with the Fundamental Theorem (for proof, go to proof of the Fundamental Theorem of Calculus), but I'm open to any suggestions on how to make it more clear. /msg me!

Another way to better understand the FTC as the area under f(x) is to picture a line as f(x). We'll consider a simple line, y = x (can't get much simpler than that), for a better understanding of an elementary problem so the more complex ones will be better understood.

y = x

        4|   /
        3|  /
        2| /

Ok. We take the integral from a = 0 to b = 4:

∫ (x)dx = 

(1/2)x2]  =
(1/2)(42 - 02) =
(1/2)(16) = 8.

So 8 is the area under the graph. Now, let's look at it again. What does the area under the picture of the graph look like? Well, it looks like a triangle. So we are really just looking for the area of that triangle. Looking back to the days of basic geometry, we know the area of a triangle is (1/2)base*height. So, what is the base and the height? The base would be x-coordinate and the height would be the y-coordinate. Since we are going from x = 0 to x = 4, the length of the base is 4 - 0 or 4. Now, to find the height, we have to see where the range of f is when we go from x = 0 to x = 4. Going back to the equation of the line, when x = 0, y = 0, and when x = 4, y = 4. So, the height is also 4. Since base(or x) = 4 = height(or y), can't we say that the base*height is really just x2? So we see that the area of the triangle is (1/2)x2. When we put the value of x, 4, into the area equation, we see that the answer is, again, 8.

This is basically how the FTC works. On some AP or other standardized tests, they might give you a graph without the equation and have you find the definite integral. It will most likely be simple shapes such as triangles and rectangles (but remember, if the graph goes under the x-axis, that is negative area). With this better understanding of the FTC, these problems should be a given.

(This is, again, a very simplified exercise to show how the antiderivative works as the area under the graph. I'm sorry if I insulted any one, but I wanted to break it down severely for people that are really don't grasp the FTC.)

The Fundamental Theorem is "fundamental" because it links the two distinct branches of calculus, derivatives and integrals. As you can see above, it is split into two parts.

Here's what I mean by "take it-stick it-D it". Look at this example, first:

     /\ u(x)
  d  |
 --  |  f(t) dt 
 dx  |
    \/ a

Taking-sticking-D-ing is a phrase that was "patented" by my high school math teacher, Mr. Noeth. In this example, t is the variable of integration. To T-S-D, you take that end limit ("take it"), replace the variable of integration with it ("stick it"), and take the derivative WRT x of it ("D it"). So you have:

f(u(x)) * --

Note that you can use any variables, not just x and t.

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