7.3: Condiciones de integrabilidad
- Page ID
- 108796
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Si\(a<b\) y\(f:[a, b] \rightarrow \mathbb{R}\) es monótono, entonces\(f\) es integrable en\([a, b]\).
- Prueba
-
Supongamos que\(f\) no disminuye. Dado\(\epsilon>0,\) que\(n \in Z^{+}\) sea lo suficientemente grande como para
\[\frac{(f(b)-f(a))(b-a)}{n}<\epsilon .\]
Para\(i=0,1, \ldots, n,\) dejar
\[x_{i}=a+\frac{(b-a) i}{n}.\]
Vamos\(P=\left\{x_{0}, x_{1}, \ldots, x_{n}\right\} .\) Entonces
\[\begin{aligned} U(f, P)-L(f, P) &=\sum_{i=1}^{n} f\left(x_{i}\right)\left(x_{i}-x_{i-1}\right)-\sum_{i=1}^{n} f\left(x_{i-1}\right)\left(x_{i}-x_{i-1}\right) \\ &=\sum_{i=1}^{n}\left(f\left(x_{i}\right)-f\left(x_{i-1}\right)\right) \frac{b-a}{n} \\ &=\frac{b-a}{n}\left(\left(f\left(x_{1}\right)-f\left(x_{0}\right)\right)+\left(f\left(x_{2}\right)-f\left(x_{1}\right)\right)+\cdots\right.\\ &\left.\quad+\left(f\left(x_{n-1}\right)-f\left(x_{n-2}\right)\right)+\left(f\left(x_{n}\right)-f\left(x_{n-1}\right)\right)\right) \\ &=\frac{b-a}{n}(f(b)-f(a)) \\ &<\epsilon . \end{aligned}\]
De ahí\(f\) que sea integrable en\([a, b]\). \(\quad\)Q.E.D.
\(\varphi: \mathbb{Q} \cap[0,1] \rightarrow \mathbb{Z}^{+}\)Sea una correspondencia uno a uno. Definir\(f:[0,1] \rightarrow \mathbb{R}\) por
\[f(x)=\sum_{q \in \underset{q \leq x}{\mathbb{Q} \cap [0,1]}} \frac{1}{2^{\varphi(q)}.}\]
Entonces\(f\) está aumentando\([0,1],\) y por lo tanto integrable en\([0,1]\).
Si\(a<b\) y\(f:[a, b] \rightarrow \mathbb{R}\) es continuo, entonces\(f\) es integrable en\([a, b]\).
- Prueba
-
Dado\(\epsilon>0,\) let
\[\gamma=\frac{\epsilon}{b-a}.\]
Dado que\(f\) es uniformemente continuo\([a, b],\) podemos elegir de\(\delta>0\) tal manera que
\[|f(x)-f(y)|<\gamma\]
siempre que\(|x-y|<\delta .\) Let\(P=\left\{x_{0}, x_{1}, \ldots, x_{n}\right\}\) sea una partición con
\[\sup \left\{\left|x_{i}-x_{i-1}\right|: i=1,2, \ldots, n\right\}<\delta .\]
Si, para\(i=1,2, \dots, n\),
\[m_{i}=\inf \left\{f(x): x_{i-1} \leq x \leq x_{i}\right\}\]
y
\[M_{i}=\sup \left\{f(x): x_{i-1} \leq x \leq x_{i}\right\},\]
entonces\(M_{i}-m_{i}<\gamma .\) De ahí
\[\begin{aligned} U(f, P)-L(f, P) &=\sum_{i=1}^{n} M_{i}\left(x_{i}-x_{i-1}\right)-\sum_{i=1}^{n} m_{i}\left(x_{i}-x_{i-1}\right) \\ &=\sum_{i=1}^{n}\left(M_{i}-m_{i}\right)\left(x_{i}-x_{i-1}\right) \\ &<\gamma \sum_{i=1}^{n}\left(x_{i}-x_{i-1}\right) \\ &=\gamma(b-a) \\ &=\epsilon . \end{aligned}\]
Así\(f\) es integrable en\([a, b]\). \(\quad\)Q.E.D.
Supongamos que\(a<b, f:[a, b] \rightarrow \mathbb{R}\) está acotado, y\(c \in[a, b] .\) Mostrar que si\(f\) es continuo encendido\([a, b] \backslash\{c\},\) entonces\(f\) es integrable en\([a, b]\).
Supongamos\(a<b\) y\(f\) es continuo\([a, b]\) con\(f(x) \geq 0\) para todos\(x \in[a, b] .\) Mostrar que si
\[\int_{a}^{b} f=0,\]
entonces\(f(x)=0\) para todos\(x \in[a, b]\).
Supongamos\(a<b\) y\(f\) es continuo en\([a, b] .\) Para\(i=0,1, \ldots, n\),\(n \in \mathbb{Z}^{+},\) vamos
\[x_{i}=a+\frac{(b-a) i}{n}\]
y, para\(i=1,2, \ldots, n,\) dejar\(c_{i} \in\left[x_{i-1}, x_{i}\right] .\) Mostrar que
\[\int_{a}^{b} f=\lim _{n \rightarrow \infty} \frac{b-a}{n} \sum_{i=1}^{n} f\left(c_{i}\right).\]
En la notación de Ejercicio\(7.3 .3,\) llamamos la aproximación
\[\int_{a}^{b} f \approx \frac{b-a}{n} \sum_{i=1}^{n} f\left(c_{i}\right)\]
una aproximación de regla derecha si\(c_{i}=x_{i},\) una aproximación de regla izquierda si\(c_{i}=x_{i-1},\) y una aproximación de regla de punto medio si
\[c_{i}=\frac{x_{i-1}+x_{i}}{2}.\]
Estos son ingredientes básicos para crear aproximaciones numéricas a integrales.