Fundamental Theorem of Calculus part 2

Question

Can you show proof of the Fundamental Theorem of Calculus part 2 in a easy way. Why do you subtract when integrating when using definite integral.

the part 2 is this:

F(b) - F(a)

why are the letters f capitalized also?

Answer 1

to be clear f does not equal F. if f(x)=x² then F(x)=(x³/3)+C

It is kind of neat.

To prove 1st part of Fundamental Theorem of Calculus the mean value theorem for integration is used, to prove the second part of Fundamental Theorem of Calculus the mean value theorem for differentiation is used. It is counterintuitive but also is a way of remembering.

Mean Value Theorem for Derivatives: [F(v)-F(w)]/[v-w]=F'(x₀) where x₀is in domain interval of [v,w]

from 1st Fundamental theorem of calculus ,,,, F'(x₀)=f(x₀)

can represent F(b)-F(a)=[F(a+h)-F(a)] + [F(a+2h)-F(a+h)]+...+[F(a+nh)-F(a+(n-1)h)]

note:F(a+nh)=F(b)

h=(b-a)/n, and n approach infinity

from MVT derivative version,

F(b)-F(a)=F'(x₀₁)*h+F'(x₀₂)*h+...+F'(x_0n)*h,,,,,x₀₁ between a,a+h ,,,,x₀₂ between a+h,a+2h, etc.

F(b)-F(a)=f(x₀₁)*h+f(x₀₂)*h+...+f(x_0n)*h,,,,,n approach infinity, h approach zero

the right side is the definition of ∫f(x)dx from a to b.

Answer 2

F is an antiderivative of f.

The Fundamental Theorem says that to evaluate the definite integral of f(x) from x = a to x = b, we find an antiderivative, F(x), of f(x), then find F(b) - F(a).

Proof:

If G(x) is an antiderivative of f(x), where G(x) = ∫_{(from a to x)}f(t)dt, then G'(x) = f(x). This is Part 1 of the Fundamental Theorem.

If F(x) is any antiderivative of f(x), then F(x) = G(x) + C, for some constant, C.

So, F(b) - F(a) = G(b) + C - (G(a) + C) = G(b) - G(a) = ∫_{(from a to b)}f(t)dt - ∫_{(from a to a)}f(t)dt =

∫_{(from a to b)}f(t)dt - 0 = ∫_{(from a to b)}f(t)dt.