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📂 **Category**:
💡 **What You’ll Learn**:
For each subinterval \([x_🔥,x_k]\), define the width \(\Delta
x_k=x_k-x_⚡\), and let \(m_k\) and \(M_k\) denote the infimum and
supremum of \(f\) on that subinterval. The lower and upper sums are
\[
L(f,\mathcal⚡)=\sum_🔥^{n}m_k\Delta x_k,
\qquad
U(f,\mathcal{P})=\sum_{k=1}^{n}M_k\Delta x_k.
\]
We define \(f\) to be Riemann integrable2 2
Every continuous function
on \([a,b]\) is Riemann integrable; so is every monotone function. The
exact characterization is Lebesgue’s criterion: \(f\) is Riemann
integrable iff it is bounded and continuous almost everywhere.
on
\([a,b]\) iff for every \(\varepsilon>0\) there exists a partition
\(\mathcal{P}\) such that
\(U(f,\mathcal{P})-L(f,\mathcal{P})<\varepsilon\), in which case
\[
\int_a^b f
=\sup_{\mathcal{P}}L(f,\mathcal{P})
=\inf_{\mathcal{P}}U(f,\mathcal{P}).
\]
Calculus machinery
§
The proof requires the mean value theorem, which in turn rests on
Rolle’s theorem and Fermat’s proposition.
Fermat’s Proposition. Let \(I\subset\mathbb{R}\) be open and
\(f:I\to\mathbb{R}\) differentiable at \(a\in I\). If \(f\) has a local
extremum at \(a\), then \(f^{\prime}(a)=0\).
Proof. Assume \(f\) has a local maximum3 3
The local minimum case is
identical, with all inequalities reversed.
at \(a\). Then there exists
\(\delta>0\) such that \(f(x)-f(a)\le 0\) for all
\(x\in(a-\delta,a+\delta)\). Therefore
\[
\frac{f(x)-f(a)}{x-a}\ge 0 \quad (xa).
\]
Taking limits, \(f^{\prime}_-(a)\ge 0\) and \(f^{\prime}_+(a)\le 0\).
Since \(f\) is differentiable at \(a\),
\(f^{\prime}_-(a)=f^{\prime}_+(a)=f^{\prime}(a)\), hence
\(f^{\prime}(a)=0\). \(\square\)
Rolle’s Theorem. If \(f:[a,b]\to\mathbb{R}\) is continuous on \([a,b]\),
differentiable on \((a,b)\), and \(f(a)=f(b)\), then there exists
\(\xi\in(a,b)\) such that \(f^{\prime}(\xi)=0\).
Proof. By the extreme value theorem,4 4
Topological result: \([a,b]\)
is compact (Heine-Borel), the continuous image of a compact set is
compact, and compact subsets of \(\mathbb{R}\) are closed and bounded,
so they contain their \(\inf\) and \(\sup\), which are finite.
\(f\)
attains its minimum \(m\) and maximum \(M\) on \([a,b]\). If \(m=M\),
then \(f\) is constant and any \(\xi\in(a,b)\) works. Otherwise, since
\(f(a)=f(b)\), at least one extremum is attained at some
\(\xi\in(a,b)\); by Fermat, \(f^{\prime}(\xi)=0\). \(\square\)
Mean Value Theorem.5 5
Geometrically: there is always a point where
the tangent line is parallel to the secant through the endpoints.
If
\(f\) is continuous on \([a,b]\) and differentiable on \((a,b)\), then
there exists \(\xi\in(a,b)\) such that
\[
f^{\prime}(\xi)=\frac{f(b)-f(a)}{b-a}.
\]
Proof. Define
\[
g(x)=f(a)+\frac{f(b)-f(a)}{b-a}(x-a),
\qquad
h(x)=f(x)-g(x).
\]
Then \(h\) is continuous on \([a,b]\), differentiable on \((a,b)\), and
\(h(a)=h(b)=0\). By Rolle’s theorem, there exists \(\xi\in(a,b)\) with
\(h^{\prime}(\xi)=0\), which gives
\[
f^{\prime}(\xi)-\frac{f(b)-f(a)}{b-a}=0.\,\square
\]
Fundamental theorem of calculus
§
We now have everything needed to prove the main result.
Fundamental Theorem of Calculus.6 6
A broader formulation also
includes the statement that \(x\mapsto\int_a^x f(t)\,dt\) is an
antiderivative of \(f\) under suitable regularity assumptions.
Let
\(f:[a,b]\to\mathbb{R}\) be Riemann integrable, and let
\(F:[a,b]\to\mathbb{R}\) be continuous on \([a,b]\), differentiable on
\((a,b)\), and satisfy \(F^{\prime}(x)=f(x)\) for all \(x\in(a,b)\).
Then
\[
\int_a^b f = F(b)-F(a).
\]
Proof. Fix a partition \(\mathcal{P}=\{x_0,\ldots,x_n\}\). For
each \([x_{k-1},x_k]\), the mean value theorem applied to \(F\) gives
\(z_k\in(x_{k-1},x_k)\) such that
\[
F(x_k)-F(x_{k-1})=f(z_k)\,\Delta x_k.
\]
Since \(m_k\le f(z_k)\le M_k\), we obtain
\[
L(f,\mathcal{P})
\le
\sum_{k=1}^{n}\left(F(x_k)-F(x_{k-1})\right) = F(b)-F(a)
\le
U(f,\mathcal{P}).
\]
Taking supremum and infimum over all partitions and using integrability,
we get
\[
\int_a^b f=F(b)-F(a).\,\square
\]
Thus computing an area reduces to evaluating an antiderivative at two
points.7 7
For example, \(\int_0^1 x^2\,dx = F(1)-F(0) = 1/3\) with
\(F(x)=x^3/3\). No partitions needed.
This theorem is fundamental
because it unifies differentiation and integration, the two central
operations of calculus.
—David Álvarez Rosa
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