State and prove Fundamental theorem of integral calculus?

Answer 1

This part is sometimes referred to as the First Fundamental Theorem of Calculus.

Let f be a continuous real-valued function defined on a closed interval [a, b]. Let F be the function defined, for all x in [a, b], by

Then, F is continuous on [a, b], differentiable on the open interval (a, b), and

for all x in (a, b).

Proof

For a given f(t), define the function F(x) as

For any two numbers x1 and x1 + Δx in [a, b], we have

and

Subtracting the two equations gives

It can be shown that (The sum of the areas of two adjacent regions is equal to the area of both regions combined.)

Manipulating this equation gives

Substituting the above into (1) results in

According to the mean value theorem for integration, there exists a c in [x1, x1 + Δx] such that

Substituting the above into (2) we get

Dividing both sides by Δx gives Notice that the expression on the left side of the equation is Newton's difference quotient for F at x1.

Take the limit as Δx → 0 on both sides of the equation.

The expression on the left side of the equation is the definition of the derivative of F at x1.

To find the other limit, we will use the squeeze theorem. The number c is in the interval [x1, x1 + Δx], so x1 ≤ c ≤ x1 + Δx.

Also, and

Therefore, according to the squeeze theorem,

Substituting into (3), we get

The function f is continuous at c, so the limit can be taken inside the function. Therefore, we get which completes the proof.