## 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 =
lim
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)

Proof

Since, by definition,
FI(t) =
lim
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.

IMPORTANT:
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).

CONCLUSION

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.