Fundamental Theorem of Calculus/First Part

Theorem

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

Let $F$ be a real function which is defined on $\closedint a b$ by:

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

Then $F$ is a primitive of $f$ on $\closedint a b$.

Corollary

$\ds \frac \d {\d x} \int_a^x \map f t \rd t = \map f x$

Proof 1

To show that $F$ is a primitive of $f$ on $\closedint a b$, we need to establish the following:

$F$ is continuous on $\closedint a b$

$F$ is differentiable on the open interval $\openint a b$

$\forall x \in \closedint a b: \map {F'} x = \map f x$.

Proof that $F$ is Continuous

We have that $f$ is continuous on $\closedint a b$.

It follows from Continuous Image of Closed Interval is Closed Interval that $f$ is bounded on $\closedint a b$.

Suppose that:

$\forall t \in \closedint a b: \size {\map f t} < \kappa$

Let $x, \xi \in \closedint a b$.

From Sum of Integrals on Adjacent Intervals for Continuous Functions‎, we have that:

$\ds \int_a^x \map f t \rd t + \int_x^\xi \map f t \rd t = \int_a^\xi \map f t \rd t$

That is:

$\ds \map F x + \int_x^\xi \map f t \rd t = \map F \xi$

So:

$\ds \map F x - \map F \xi = -\int_x^\xi \map f t \rd t = \int_\xi^x \map f t \rd t$

From Darboux's Theorem: Corollary:

$\size {\map F x - \map F \xi} < \kappa \size {x - \xi}$

Thus it follows that $F$ is continuous on $\closedint a b$.

$\Box$

Proof that $F$ is Differentiable and $f$ is its Derivative

It is now to be shown that that $F$ is differentiable on $\openint a b$ and that:

$\forall x \in \closedint a b: \map {F'} x = \map f x$

Let $x, \xi \in \closedint a b$ such that $x \ne \xi$.

Then:

\(\ds \frac {\map F x - \map F \xi} {x - \xi} - \map f \xi\)

\(=\)

\(\ds \frac 1 {x - \xi} \paren {\map F x - \map F \xi - \paren {x - \xi} \map f \xi}\)

\(\ds \)

\(=\)

\(\ds \frac 1 {x - \xi} \paren {\int_\xi^x \map f t \rd t - \paren {x - \xi} \map f \xi}\)

\(\ds \)

\(=\)

\(\ds \frac 1 {x - \xi} \int_\xi^x \paren {\map f t - \map f \xi} \rd t\)

Definite Integral of Function plus Constant, putting $c = \map f \xi$

Now, let $\epsilon > 0$.

If $\xi \in \openint a b$, then $f$ is continuous at $\xi$.

So for some $\delta > 0$:

$\size {\map f t - \map f \xi} < \epsilon$

provided $\size {t - \xi} < \delta$.

So provided $\size {x - \xi} < \delta$ it follows that:

$\size {\map f t - \map f \xi} < \epsilon$

for any $t$ in an interval whose endpoints are $x$ and $\xi$.

So from Darboux's Theorem: Corollary:

\(\ds \size {\frac {\map F x - \map F \xi} {x - \xi} - \map f \xi}\)

\(=\)

\(\ds \frac 1 {\size {x - \xi} } \size {\int_\xi^x \paren {\map f t - \map f \xi} \rd t}\)

\(\ds \)

\(<\)

\(\ds \frac 1 {\size {x - \xi} } \epsilon \size {x - \xi}\)

\(\ds \)

\(=\)

\(\ds \epsilon\)

provided $0 < \size {x - \xi} < \delta$.

But that is what this means:

$\dfrac {\map F x - \map F \xi} {x - \xi} \to \map f \xi$ as $x \to \xi$

So $F$ is differentiable on $\openint a b$, and:

$\forall x \in \closedint a b: \map {F'} x = \map f x$

$\blacksquare$

Proof 2

\(\ds \dfrac \d {\d x} \map F x\)

\(=\)

\(\ds \lim_{\Delta x \mathop \to 0} \frac 1 {\Delta x} \paren {\int_a^{x + \Delta x} \map f t \rd t - \int_a^x \map f t \rd t}\)

Definition of Derivative of Real Function at Point

\(\ds \)

\(=\)

\(\ds \lim_{\Delta x \mathop \to 0} \frac 1 {\Delta x} \paren {\int_x^a \map f t \rd t + \int_a^{x + \Delta x} \map f t \rd t}\)

because $\ds \int_a^b \map f x \rd x = -\int_b^a \map f x \rd x$

\(\ds \)

\(=\)

\(\ds \lim_{\Delta x \mathop \to 0} \frac 1 {\Delta x} \paren {\int_x^{x + \Delta x} \map f t \rd t}\)

Sum of Integrals on Adjacent Intervals for Continuous Functions

Suppose $\Delta x > 0$ and both $x$ and $x + \Delta x$ are in the closed interval $\closedint a b$.

By hypothesis, $f$ is continuous on the closed interval $\closedint a b$

Thus it is also continuous on the closed interval $\closedint x {x + \Delta x}$.

Thus the conditions of the Mean Value Theorem for Integrals are fulfilled.

So, by the Mean Value Theorem for Integrals, there exists some $k \in \closedint x {x + \Delta x}$ such that:

$\ds \int_x^{x + \Delta x} \map f x \rd x = \map f k \paren {x + \Delta x - x} = \map f k \Delta x$

Then:

\(\ds \lim_{\Delta x \mathop \to 0} \frac 1 {\Delta x} \paren {\int_x^{x + \Delta x} \map f t \rd t}\)

\(=\)

\(\ds \lim_{\Delta x \mathop \to 0} \frac 1 {\Delta x} \map f k \Delta x\)

\(\ds \)

\(=\)

\(\ds \lim_{\Delta x \mathop \to 0} \map f k\)

By the definition of $k$:

$x \le k \le x + \Delta x$

which means that $k \to x$ as $\Delta x \to 0$.

So:

\(\ds \lim_{\Delta x \mathop \to 0} \map f k\)

\(=\)

\(\ds \lim_{k \mathop \to x} \map f k\)

\(\ds \)

\(=\)

\(\ds \map f x\)

because $f$ is continuous

For $\Delta x < 0$, consider $k \in \closedint {x + \Delta x} x$, and the argument is similar.

Hence the result, by the definition of primitive.

$\blacksquare$

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Proof 3

This article needs proofreading. Please check it for mathematical errors. If you believe there are none, please remove {{Proofread}} from the code. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Proofread}} from the code.

By Topological Manifold/Examples/Real Cartesian Space, the closed real interval is a manifold.

We have that $\F$ is a smooth 0-form with compact support on a smooth $1$-dimensional oriented manifold $\left[{a \,.\,.\, b}\right]$.

We have that the boundary of $\left[{a \,.\,.\, b}\right]$ is $\partial \left[{a \,.\,.\, b}\right]$.

We denote $\d \F$ to be the exterior derivative of $\F$.

Hence, by the General Stokes' Theorem:

$\ds \int_{\left[{a \,.\,.\, b}\right]} \rd F = \int_{\partial {\left[{a \,.\,.\, b}\right]}} = \F(b) - \F(a)$

This needs considerable tedious hard slog to complete it. In particular: How to use the choice of orientation from the oriented manifold To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Finish}} from the code. If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.

$\blacksquare$