Fundamental Theorem of Calculus(FTC):
Let f:[a,b]->R be a function that is continuous and integrable on [a,x] for every x in [a,b], and let c be such that a <= c <= b and define the function A:[a,b]->R as follows:

A(x) = Int(f,[c,x])

Where Int(f,J) is the definite integral of f over the subset J of R.
Then A(x) is differentiable on (a,b), and A' = f (where A' is the derivative of A).

Proof:
Let x be a point in [a,b] be given, and define the following quotient function q:

q(h) := ( A(x+h)-A(x) )/h

for non-zero real numbers h such that x+h is in [a,b].

A(x+h)-A(x) = Int(f,[c,x+h]) - Int(f,[c,x]) = Int(f,[x,x+h])

Adding and subtracting f(x) (I hate it when people use a trick and leave you to figure it out :), we get,

A(x+h)-A(x) = Int(f-f(x),[x,x+h])+Int(f(x),[x,x+h])

Note that above, f(x) refers to the constant function whose value is f(x) everywhere.
This yields,

q(h) = f(x) + (1/h)*Int(f-f(x),[x,x+h])

Now let some e > 0 be given. By continuity of f, we may assume that d > 0 is such that |f(t) - f(x)| < e for all t in [x-d,x+d]. Then for |h| < d, |f(t) - f(x)| < e for every t in [x,x+h]. Thus,

|(1/h)*Int(f-f(x),[x,x+h])| <= |1/h|*Int( | f-f(x) |,[x,x+h]) <= |1/h|*Int(e,[x,x+h]) <= |1/h|*h*e <= e.

But e was arbitrary, so this holds for any e. Thus, (1/h)*Int(f-f(x),[x,x+h]) converges to 0 at 0 (as h goes to 0), and q(h) = f(x) + (1/h)*Int(f-f(x),[x,x+h]) converges to f(x). But q(h) is exactly the derivative of A at x, so A'(x)=f(x). Since x was arbitrary, this holds for every x in [x,x+h].

Note: Any function with the form of A is called a "primitive" of f. Thus, the FTC says that every primitive of f is also an antiderivative of f.

I realize that the Int() notation for the definite integral is cumbersome. If there is a better alternative, I'd love to know of it... I'm exhausted :)

Log in or register to write something here or to contact authors.