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newton leibniz formula
If the function f(x) is continuous on [a, b] and the original function f(x) exists, then F(x) is integrable on [a, b], and

B (upper limit) ∫a (lower limit) f(x)dx=F(b)-F(a)

This is the Newton-Leibniz formula.

The significance of Newton-Leibniz formula lies in the connection between indefinite integral and definite integral, and also gives a perfect and satisfactory operation method of definite integral. The following is the whole process of proving this formula:

As we know, the definite integral of the function f(x) in the interval [a, b] is expressed as:

B (upper limit) ∫a (lower limit) f(x)dx

Now we take the upper limit of the integral interval as a variable, so we define a new function:

φ (x) = x (upper limit) ∫a (lower limit) f(x)dx

But here x has two meanings, one is the upper limit of the integral, and the other is the independent variable of the integrand function, but it is meaningless to take a fixed value for the independent variable of the integrand function in the definite integral. In order to express only the change of the upper limit of the integral, we change the independent variable of the integrand function to other letters such as t, so the meaning is clear:

φ (x) = x (upper limit) ∫a (lower limit) f(t)dt

Next, we will study the properties of this function φ (x):

1, define the function φ (x) = x (upper limit) ∫a (lower limit) f(t)dt, then φ' (x) = f (x).

It is proved that if the function φ (x) obtains the increment δ x, the corresponding function increments.

δ φ = φ (x+δ x)-φ (x) = x+δ x (upper limit) ∫a (lower limit) f(t)dt-x (upper limit) ∫a (lower limit) f(t)dt.

Obviously, x+δ x (upper limit) ∫a (lower limit) f(t)dt-x (upper limit) ∫a (lower limit) f(t)dt = x+δ x (upper limit) ∫x (lower limit) f(t)dt.

And δ φ = x+δ x (upper limit) ∫x (lower limit) f(t)dt=f(ξ)? Δ x (ξ is between x and x+Δ x, which can be deduced from the mean value theorem of definite integral.

You can also draw a picture yourself, the geometric meaning is very clear. )

When Δ x tends to 0, that is, Δ φ tends to 0 x, f (ξ) tends to x and f(ξ) tends to f(x), so there is lim Δ x→Δφ/Δ x = f (x).

It can be seen that this is also the definition of derivative, so we finally get φ' (x) = f (x).

2.b (upper limit) ∫a (lower limit) f (x) dx = f (b)-f (a), and f(x) is the original function of f(x).

Proof: We have proved that φ' (x) = f (x), so φ (x)+c = f (x).

But φ (a) = 0 (the integral interval becomes [a, a], so the area is 0), so f (a) = C.

So φ (x)+f (a) = f (x), when x=b, φ (b) = f (b)-f (a),

And φ (b) = b (upper limit) ∫a (lower limit) f(t)dt, so b (upper limit) ∫a (lower limit) f(t)dt=F(b)-F(a).

Re-writing T as X becomes the formula at the beginning, which is Newton-Leibniz formula.