Tuesday, May 9, 2017

Unizor - Definite Integrals - Newton-Leibniz Formula

Notes to a video lecture on http://www.unizor.com

Definite Integrals -
Newton-Leibniz Formula

Recall the definition of the definite integral:
abf(x) dx =
Σi∈[1,N] f(xiΔxi
where Δxi=xi−xi−1 represents partitioning of segment [a,b] into N parts, and it is assumed that the widest interval Δxi is shrinking to zero by length as N→∞.

First Fundamental Theorem
of Calculus

Consider a smooth function f(x) on segment [a,b] and any point t∈[a,b].
The following definite integral can be considered as a function of t:
F(t) = atf(x) dx
The First Fundamental Theorem of Calculus states that the derivative of function F(t) is function f(t):
FI(t) = f(t)


Since, by definition,
FI(t) =
Δt→0[F(t+Δt)−F(t)] /Δt
we have to prove that
limΔt→0 [at+Δtf(x) dx −
− atf(x) dx] /Δt = f(t)

From the properties of definite integrals we know that
atf(x) dx + tt+Δtf(x) dx =
at+Δtf(x) dx
from which follows that
at+Δtf(x) dx − atf(x) dx =
tt+Δtf(x) dx

Therefore, we have to prove that
limΔt→0 tt+Δtf(x) dx /Δt = f(t)

From properties of definite integrals we know that
Δt ≤ tt+Δtf(x) dx ≤ M·Δt
where m is minimum of function f(x) on segment [t,t+Δt] and M - its maximum on this segment.

This allows us to state that the expression
limΔt→0 tt+Δtf(x) dx /Δt
is bounded from below by m (minimum f(t) on segment [t,t+Δt]) and from above by M (maximum f(t) on segment [t,t+Δt]).

As Δt→0, our segment [t,t+Δt] shrinks to a point t. Since we assume sufficient "smoothness" of function f(t) (in this case we need just its continuity), both minimum and maximum of f(t) on segment [t,t+Δt] converge to the same value f(t).
That forces the limit above also to converge to f(t).

End of proof.

Since the derivative of function F(t) above (definite integral of function f(x) on a segment from a to t) equals to f(t), function F(t) can serve as an indefinite integral (antiderivative) of f(t).
Since we proved the existence and uniqueness of a definite (Riemann) integral for any continuous function, the theorem above has proven that for any continuous function there exists indefinite integral (its antiderivative).

Newton-Leibniz Formula

Let's assume that we want to find a definite (Riemann) integral abf(x) dx of some continuous function f(x) on segment [a,b].
Assume further that we know one particular function G(t) which is the indefinite integral (antiderivative) of f(t) (we deliberately decided to use different argument symbol t instead of x as an argument of function G(t) to have x only as a variable of integration).
That is, GI(t) = f(t).

Recall that there are many functions, derivative of which equal to f(t), but we know that all of them differ from each other only by an addition of a constant.

So, on one hand we have G(t) as one of the possible indefinite integrals (antiderivative) of f(t).
On the other hand, we have just proven that derivative of
F(t) = atf(x) dx
by t also equals to function f(t).
Therefore, we have two different functions, derivatives of both of which are the same function f(t):
GI(t) = f(t) and
FI(t) = [atf(x) dx]I = f(t)

Since two antiderivatives differ only by an addition of a constant, we conclude that
F(t) = atf(x) dx = G(t) + C
where C - some constant.

It's easy to find this constant. Since a definite integral on a null-segment [a,a] equals to zero, assign t=a in the above formula getting
aaf(x) dx = G(a) + C = 0
from which we get
C = −G(a)

The final formula for our definite integral, therefore, is
F(t) = atf(x) dx = G(t) − G(a)
In particular, for t=b, we obtain the Newton-Leibniz formula:
abf(x) dx = G(b) − G(a)
where G(t) - any function, derivative of which is f(t), in other words, an indefinite integral (antiderivative) of f(t).


To find a definite integral of some real function f(t) on segment t∈[a,b], it is sufficient to find any particular indefinite integral (antiderivative) G(t) of function f(t) and calculate the expression G(b)−G(a).
This establishes a connection between indefinite and definite integrals and justifies the usage of the same word integral for both.

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