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Viewing Version
9
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'fundamental theorem of calculus'
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| Title of object: |
fundamental theorem of calculus |
| Canonical Name: |
FundamentalTheoremOfCalculusClassicalVersion |
| Type: |
Theorem |
| Created on: |
2004-03-01 10:55:53 |
| Modified on: |
2004-10-26 10:23:30 |
| Classification: |
msc:26A42 |
Revision comment (for changes between this and next version):
| Changes for correction #13696 ('Barrow's rule'). |
Preamble:
% this is the default PlanetMath preamble. as your knowledge
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\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{amsfonts}
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%\usepackage{psfrag}
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%\usepackage{graphicx}
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Content:
Let $f\colon[a,b]\to \mathbf R$ be a continuous function, let $c\in[a,b]$ be given
and consider the integral function $F$ defined on $[a,b]$ as
\[
F(x)= \int_c^x f(t)\, dt.
\]
Then $F$ is an antiderivative of $f$ that is,
$F$ is differentiable in $[a,b]$ and
\[
F'(x)=f(x)\qquad \forall x\in [a,b].
\]
The previous relation rewritten as
\[
\frac{d}{dx} \int_c^x f(t)\, dt = f(x)
\]
shows that the differentiation operator $\frac{d}{dx}$ is the inverse of the integration operator $\int_c^x$. This formula is sometimes called Newton-Leibniz formula.
On the other hand if $f\colon[a,b]\to \mathbf R$ is a continuous function
and $G\colon[a,b]\to \mathbf R$ is any antiderivative of $f$, i.e.\ $G'(x)=f(x)$ for all $x\in[a,b]$, then
\[
\int_a^b f(t) \, dt = G(b)-G(a).
\]
This shows that up to a constant, the integration operator is the inverse of the derivative operator:
\[
\int_a^x D G = G - G(a).
\] |
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