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

__Definite Integrals -__

Newton-Leibniz Formula

Newton-Leibniz Formula

Recall the

**definition**of the

*definite integral*:

**∫**d_{a}^{b}f(x)**x =**

=lim=

**Σ**

_{i∈[1,N]}

*f(x*Δ_{i})·*x*_{i}where

**Δ**represents partitioning of segment [

*x*_{i}=x_{i}−x_{i−1}*] into*

**a,b***parts, and it is assumed that the widest interval*

**N****Δ**is shrinking to zero by length as

*x*_{i}*.*

**N→∞***First Fundamental Theorem*

of Calculus

of Calculus

Consider a smooth function

*on segment [*

**f(x)***] and any point*

**a,b***∈[*

**t***].*

**a,b**The following definite integral can be considered as a function of

*:*

**t**

**F(t) = ∫**d_{a}^{t}f(x)**x**The First Fundamental Theorem of Calculus states that the derivative of function

*is function*

**F(t)***:*

**f(t)**

**F**^{I}(t) = f(t)*Proof*

Since, by definition,

**F**

=lim^{I}(t) ==

_{Δt→0}[

**F(t+****Δ**]

*t)−F(t)*

**/****Δ**

*t*we have to prove that

*lim*

_{Δt→0}[

**∫***−*

**d**_{a}^{t+Δt}f(x)**x**−

*]*

**∫**d_{a}^{t}f(x)**x**

**/****Δ**=

*t*

**f(t)**From the properties of definite integrals we know that

*+*

**∫**d_{a}^{t}f(x)**x***=*

**∫**d_{t}^{t+Δt}f(x)**x**=

**∫**d_{a}^{t+Δt}f(x)**x**from which follows that

*−*

**∫**d_{a}^{t+Δt}f(x)**x***=*

**∫**d_{a}^{t}f(x)**x**=

**∫**d_{t}^{t+Δt}f(x)**x**Therefore, we have to prove that

*lim*

_{Δt→0}

**∫**d_{t}^{t+Δt}f(x)**x**

**/****Δ**=

*t*

**f(t)**From properties of definite integrals we know that

**m·****Δ**

*t ≤ ∫*_{t}^{t+Δt}f(x)*d*

**x**

**≤ M·****Δ**

*t*where

*is minimum of function*

**m***on segment [*

**f(x)**

**t,t+****Δ**] and

*t**- its maximum on this segment.*

**M**This allows us to state that the expression

*lim*

_{Δt→0}

**∫**d_{t}^{t+Δt}f(x)**x**

**/****Δ**

*t*is bounded from below by

*(minimum*

**m***on segment [*

**f(t)**

**t,t+****Δ**]) and from above by

*t**(maximum*

**M***on segment [*

**f(t)**

**t,t+****Δ**]).

*t*As

**Δ**, our segment [

*t→0*

**t,t+****Δ**] shrinks to a point

*t**. Since we assume sufficient "smoothness" of function*

**t***(in this case we need just its continuity), both minimum and maximum of*

**f(t)***on segment [*

**f(t)**

**t,t+****Δ**] converge to the same value

*t**.*

**f(t)**That forces the limit above also to converge to

*.*

**f(t)**End of proof.

IMPORTANT:

Since the derivative of function

*above (definite integral of function*

**F(t)***on a segment from*

**f(x)***to*

**a***) equals to*

**t***, function*

**f(t)***can serve as an indefinite integral (*

**F(t)***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

**∫**d_{a}^{b}f(x)**x***on segment [*

**f(x)***].*

**a,b**Assume further that we know one particular function

*which is the indefinite integral (antiderivative) of*

**G(t)***(we deliberately decided to use different argument symbol*

**f(t)***instead of*

**t***as an argument of function*

**x***to have*

**G(t)***only as a variable of integration).*

**x**That is,

*.*

**G**^{I}(t) = f(t)Recall that there are many functions, derivative of which equal to

*, but we know that all of them differ from each other only by an addition of a constant.*

**f(t)**So, on one hand we have

*as one of the possible indefinite integrals (antiderivative) of*

**G(t)***.*

**f(t)**On the other hand, we have just proven that derivative of

**F(t) = ∫**d_{a}^{t}f(x)**x**by

*also equals to function*

**t***.*

**f(t)**Therefore, we have two different functions, derivatives of both of which are the same function

*:*

**f(t)***and*

**G**^{I}(t) = f(t)*= [*

**F**^{I}(t)*]*

**∫**d_{a}^{t}f(x)**x**

^{I}= f(t)Since two antiderivatives differ only by an addition of a constant, we conclude that

**F(t) = ∫**d_{a}^{t}f(x)**x = G(t) + C**where

*- some constant.*

**C**It's easy to find this constant. Since a definite integral on a null-segment [

*] equals to zero, assign*

**a,a***in the above formula getting*

**t=a**

**∫**d_{a}^{a}f(x)**x = G(a) + C = 0**from which we get

**C = −G(a)**The final formula for our definite integral, therefore, is

**F(t) = ∫**d_{a}^{t}f(x)**x = G(t) − G(a)**In particular, for

*, we obtain the*

**t=b****Newton-Leibniz formula**:

**∫**d_{a}^{b}f(x)**x = G(b) − G(a)**where

*- any function, derivative of which is*

**G(t)***, in other words, an indefinite integral (antiderivative) of*

**f(t)***.*

**f(t)**CONCLUSION

To find a definite integral of some real function

*on segment*

**f(t)***∈[*

**t***], it is sufficient to find any particular indefinite integral (antiderivative)*

**a,b***of function*

**G(t)***and calculate the expression*

**f(t)***.*

**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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