Fundamental Theorem of Calculus/Second Part

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Theorem

Let $f$ be a real function which is continuous on the closed interval $\closedint a b$.

Then:

$(1): \quad f$ has a primitive on $\closedint a b$
$(2): \quad$ If $F$ is any primitive of $f$ on $\closedint a b$, then:
$\ds \int_a^b \map f t \rd t = \map F b - \map F a = \bigintlimits {\map F t} a b$


Proof 1

Let $G$ be defined on $\closedint a b$ by:

$\ds \map G x = \int_a^x \map f t \rd t$

We have:

$\ds \map G a = \int_a^a \map f t \rd t = 0$ from Integral on Zero Interval
$\ds \map G b = \int_a^b \map f t \rd t$ from the definition of $G$ above.

Therefore, we can compute:

\(\ds \int_a^b \map f t \rd t\) \(=\) \(\ds \int_a^a \map f t \rd t + \int_a^b \map f t \rd t\) Sum of Integrals on Adjacent Intervals for Continuous Functions
\(\ds \) \(=\) \(\ds \map G a + \map G b\) From the definition of $G$
\(\ds \) \(=\) \(\ds \map G b - \map G a\) $\map G a = 0$

By the first part of the theorem, $G$ is a primitive of $f$ on $\closedint a b$.

By Primitives which Differ by Constant‎, we have that any primitive $F$ of $f$ on $\closedint a b$ satisfies $\map F x = \map G x + c$, where $c$ is a constant.

Thus we conclude:

\(\ds \int_a^b \map f t \rd t\) \(=\) \(\ds \paren {\map G b + c} - \paren {\map G a + c}\)
\(\ds \) \(=\) \(\ds \map F b - \map F a\)

$\blacksquare$


Proof 2

As $f$ is continuous, by the first part of the theorem, it has a primitive. Call it $F$.

$\closedint a b$ can be divided into any number of closed subintervals of the form $\closedint {x_{k - 1} } {x_k}$ where:

$a = x_0 < x_1 \cdots < x_{k-1} < x_k = b$

Fix such a finite subdivision of the interval $\closedint a b$; call it $P$.


Next, we observe the following telescoping sum identity:

\(\text {(1)}: \quad\) \(\ds \sum_{i \mathop = 1}^k \paren {\map F {x_i} - \map F {x_{i - 1} } }\) \(=\) \(\ds \map F b - \map F a\)

Because $F' = f$, $F$ is differentiable.

By Differentiable Function is Continuous, $F$ is also continuous.

Therefore we can apply the Mean Value Theorem on $F$.

It follows that in every closed subinterval $I_i = \closedint {x_{i - 1} } {x_i}$ there is some $c_i$ such that:

$\map {F'} {c_i} = \dfrac {\map F {x_i} - \map F {x_{i - 1} } } {x_i - x_{i - 1} }$

It follows that:

\(\ds \map F {x_i} - \map F {x_{i - 1} }\) \(=\) \(\ds \map {F'} {c_i} \paren {x_i - x_{i - 1} }\)
\(\text {(2)}: \quad\) \(\ds \map F b - \map F z\) \(=\) \(\ds \sum_{i \mathop = 1}^k \map {F'} {c_i} \paren {x_i - x_{i - 1} }\) from $(1)$
\(\ds \) \(=\) \(\ds \sum_{i \mathop = 1}^k \map f {c_i} \paren {x_i - x_{i - 1} }\) because $F' = f$


From the definitions of supremum and infimum, we have for all $i$ (recall $I_i = \closedint {x_{i - 1} } {x_i}$):

$\displaystyle \inf_{x \mathop \in I_i} \map f x \le \map f {c_i} \le \sup_{x \mathop \in I_i} \map f x$

From the definitions of upper Darboux sum and lower Darboux sum, we conclude for any finite subdivision $P$:

$\displaystyle \map L P \le \sum_{i \mathop = 1}^k \map f {c_i} \paren {x_i - x_{i - 1} } \le \map U P$

Lastly, from the definition of a definite integral and from $(2)$, we conclude:

$\displaystyle \map F b - \map F a = \int_a^b \map f t \rd t$

$\blacksquare$


Proof 3


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Let $\closedint a b$ be the closed (real) interval.

We claim that closed (real) interval is a smooth 1-dimensional oriented manifold.

By Classification of Compact One-Manifolds, every compact connected 1-dimensional manifold is diffeomorphic to either a circle or a closed interval.

Therefore, the closed interval is a 1-dimensional manifold.

By Subset of Real Numbers is Interval iff Connected, since $\closedint a b$ is an interval of $\R$, it follows that $\closedint a b$ is connected.


This needs considerable tedious hard slog to complete it.
In particular: Prove that the closed (real) interval is a smooth and oriented manifold
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Let $\F$ be a smooth 0-form with compact support on the $\closedint a b$.

Let the boundary of $\closedint a b$ be:

$\partial \closedint a b$

Since the manifold is oriented, and has compact support, the integrals:

$\ds \int_{\partial \closedint a b} \F$

and:

$\int_{\closedint a b} \rd \F$

are well-defined.


This needs considerable tedious hard slog to complete it.
In particular: Prove that the integrals are well-defined given that the manifold is orientable and compactly supported
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Then, by General Stokes' Theorem:

$\ds \int_{\partial \closedint a b} \F = \int_{\closedint a b} \rd \F$

where $\d \F$ is the exterior derivative of 0-form:

$\F = \map f x dx$

It follows that:

\(\ds \ds \int_{\closedint a b} \map f x dx\) \(=\) \(\ds \int_{\closedint a b} \rd F\)
\(\ds \) \(=\) \(\ds \ds \int_{\partial \closedint a b } F\) by General Stokes' Theorem
\(\ds \) \(=\) \(\ds \int_{\set{a}^- \cup \set{b}^+} F\) Definition of Boundary of Manifold
\(\ds \) \(=\) \(\ds \map F b - \map F a\)

as required.

$\blacksquare$

Also presented as

The Fundamental Theorem of Calculus can also be presented in the following form:

Let $f$ be a real function which is continuous on the closed interval $\closedint a b$ expressed as a normal first order ODE:

$(1): \quad \dfrac {\d y} {\d x} = \map f x$

Let $c \in \R$ be an arbitrary real number.

Then there exists a unique solution $\map F x$ to $(1)$ on $\closedint a b$ such that $\map F a = c$, given by the definite integral:

$\ds \map F x = c + \int_a^x \map f t \rd t$


Sources

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