26 Propiedades de las integrales

Teorema VII.1.

Sea f:I→ℝf\colon I\to\mathbb{R}, con I⊆ℝI\subseteq\mathbb{R}.

Si f es continua en [a,b]⇒f es integrable en [a,b].\text{Si }f\text{ es continua en }[a,b]\Rightarrow f\text{ es integrable en }[% a,b]\mathchar 46\relax

Además, el área comprendida entre la gráfica de una función continua positiva, el eje XX y las rectas x=ax=a y x=bx=b, es igual a

∫baf.\int^{b}_{a}f\mathchar 46\relax
Demostración.

Tampoco se ha visto en clase. ∎

Ejemplo VII.1.

f(x)=cf(x)=c, con c∈ℝc\in\mathbb{R}.

Sean a,b∈ℝa,b\in\mathbb{R} y consideramos el intervalo [b,a][b,a]. Sea P={x0,x1,…,xn}⊂[a,b]P=\left\{x_{0},x_{1},\ldots,x_{n}\right\}\subset[a,b].

L(f,P)=∑i=1nmi(xi−xi−1)=∑i=1nc(xi−xi−1)=c∑i=1n(xi−xi−1)=c(xn−x0)=c(b−a){}L(f,P)=\sum_{i=1}^{n}m_{i}(x_{i}-x_{i-1})=\sum_{i=1}^{n}c(x_{i}-x_{i-1})=c% \sum_{i=1}^{n}(x_{i}-x_{i-1})=c(x_{n}-x_{0})=c(b-a)
U(f,P)=∑i=1nMi(xi−xi−1)=c∑i=1n(xi−xi−1)=c(b−a)U(f,P)=\sum_{i=1}^{n}M_{i}(x_{i}-x_{i-1})=c\sum_{i=1}^{n}(x_{i}-x_{i-1})=c(b-a)

Por tanto, U(f,P)=L(f,P)=c(b−a)∀PU(f,P)=L(f,P)=c(b-a)\;\forall P. Es decir, ff es integrable y ∫baf(x)dx=∫bacdx=c(b−a)\int^{b}_{a}f(x)dx=\int^{b}_{a}cdx=c(b-a).

f(x)=1f(x)=11133∫131dx=2\int_{1}^{3}1\,\mathrm{d}x=2
Ejemplo VII.2.

∫10xdx=?\int^{1}_{0}xdx=?. Consideramos una particion P={0,1n,2n,3n,…,nn}P=\left\{0,\frac{1}{n},\frac{2}{n},\frac{3}{n},\ldots,\frac{n}{n}\right\}. Esta particion es equiespacial ya que la diferencia entre cada elemento es siempre 1n\frac{1}{n}. Por ejemplo, si n=4n=4 la particion seria P4={0,14,24,34,44}P_{4}=\left\{0,\frac{1}{4},\frac{2}{4},\frac{3}{4},\frac{4}{4}\right\}. En este caso,

U(x,P4)=M1(x1−x0)+M2(x2−x1)+M3(x3−x2)+M4(x4−x3)==x1⋅14+x2⋅14+x3⋅14+x4⋅14=14⋅14+12⋅14+34⋅14+1⋅14{}U(x,P_{4})=M_{1}(x_{1}-x_{0})+M_{2}(x_{2}-x_{1})+M_{3}(x_{3}-x_{2})+M_{4}(x_% {4}-x_{3})=\\ {}=x_{1}\cdot\frac{1}{4}+x_{2}\cdot\frac{1}{4}+x_{3}\cdot\frac{1}{4}+x_{4}% \cdot\frac{1}{4}=\frac{1}{4}\cdot\frac{1}{4}+\frac{1}{2}\cdot\frac{1}{4}+\frac% {3}{4}\cdot\frac{1}{4}+1\cdot\frac{1}{4}
L(x,P4)=m1(x1−x0)+m2(x2−x1)+m3(x3−x2)+m4(x4−x3)==x0⋅14+x1⋅14+x2⋅14+x3⋅14=0⋅14+14⋅14+12⋅14+34⋅14{}L(x,P_{4})=m_{1}(x_{1}-x_{0})+m_{2}(x_{2}-x_{1})+m_{3}(x_{3}-x_{2})+m_{4}(x_% {4}-x_{3})=\\ {}=x_{0}\cdot\frac{1}{4}+x_{1}\cdot\frac{1}{4}+x_{2}\cdot\frac{1}{4}+x_{3}% \cdot\frac{1}{4}=0\cdot\frac{1}{4}+\frac{1}{4}\cdot\frac{1}{4}+\frac{1}{2}% \cdot\frac{1}{4}+\frac{3}{4}\cdot\frac{1}{4}

Ahora vemos el caso general. Para PnP_{n},

U(x,Pn)=∑j=1nMj⋅(xj−xj−1)=∑j=1nxj⋅1n=∑j=1njn⋅1n=1n2∑j=1njU(x,P_{n})=\sum_{j=1}^{n}M_{j}\cdot(x_{j}-x_{j-1})=\sum_{j=1}^{n}x_{j}\cdot% \frac{1}{n}=\sum_{j=1}^{n}\frac{j}{n}\cdot\frac{1}{n}=\frac{1}{n^{2}}\sum_{j=1% }^{n}j
L(x,Pn)=∑j=1nmj⋅(xj−xj−1)=∑j=1nxj−1⋅1n=∑j=1nj−1n⋅1n=1n2∑j=1n(j−1)L(x,P_{n})=\sum_{j=1}^{n}m_{j}\cdot(x_{j}-x_{j-1})=\sum_{j=1}^{n}x_{j-1}\cdot% \frac{1}{n}=\sum_{j=1}^{n}\frac{j-1}{n}\cdot\frac{1}{n}=\frac{1}{n^{2}}\sum_{j% =1}^{n}(j-1)

Por tanto, nos queda que U(x,Pn)=1n2∑j=1njU(x,P_{n})=\frac{1}{n^{2}}\sum_{j=1}^{n}j y L(x,Pn)=1n2∑j=1n(j−1)L(x,P_{n})=\frac{1}{n^{2}}\sum_{j=1}^{n}(j-1).

Sea U(f)=ı´nf(U(f,P))U(f)=\mathop{\operator@font\acute{{\imath}}nf}(U(f,P)) y L(f)=sup(L(f,P))L(f)=\sup(L(f,P)). Es obvio que L(f)≤U(f)L(f)\leq U(f). En particular,

1n2∑j=1n(j−1)=L(f,Pn)≤L(f)≤U(f)≤U(f,Pn)=1n2∑j=1nj\frac{1}{n^{2}}\sum_{j=1}^{n}(j-1)=L(f,P_{n})\leq L(f)\leq U(f)\leq U(f,P_{n})% =\frac{1}{n^{2}}\sum_{j=1}^{n}j

Ademas, ∑j=1nj=n(n+1)2\sum_{j=1}^{n}j=\frac{n(n+1)}{2} y ∑j=1n(j−1)=∑j=1nj−∑j=1n1=n(n−1)2−n\sum_{j=1}^{n}(j-1)=\sum_{j=1}^{n}j-\sum_{j=1}^{n}1=\frac{n(n-1)}{2}-n.

Por tanto, L(f,Pn)=1n2⋅n(n+1)2−n=n2+n2n2−nn2⟶n→∞12L(f,P_{n})=\frac{1}{n^{2}}\cdot\frac{n(n+1)}{2}-n=\frac{n^{2}+n}{2n^{2}}-\frac% {n}{n^{2}}\overset{n\rightarrow\infty}{\longrightarrow}\frac{1}{2} y U(f,Pn)=1n2n(n+1)2=n2+n2n2⟶n→∞12U(f,P_{n})=\frac{1}{n^{2}}\frac{n(n+1)}{2}=\frac{n^{2}+n}{2n^{2}}\overset{n% \rightarrow\infty}{\longrightarrow}\frac{1}{2}.

Entonces, L(f)=U(f)=12L(f)=U(f)=\frac{1}{2} y ff es integrable ⇒∫10xdx=12\Rightarrow\int^{1}_{0}xdx=\frac{1}{2}.

f(x)=xf(x)=x011
Proposición VII.1.

Sean ff y gg integrables en [a,b][a,b], entonces:

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    f+gf+g es integrable en [a,b][a,b] y ∫ba(f+g)=∫baf+∫bag\int^{b}_{a}(f+g)=\int^{b}_{a}f+\int^{b}_{a}g.

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    λf\lambda f es integrable en [a,b][a,b] y ∫baλf=λ∫baf\int^{b}_{a}\lambda f=\lambda\int^{b}_{a}f.

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    |f|\left|f\right| es integrable en [a,b][a,b] y ∫baf≤∫ba|f|.\int^{b}_{a}f\leq\int^{b}_{a}\left|f\right|\mathchar 46\relax

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    f⋅gf\cdot g es integrable en [a,b][a,b] pero, en general, ∫baf⋅g≠∫baf⋅∫bag\int^{b}_{a}f\cdot g\neq\int^{b}_{a}f\cdot\int^{b}_{a}g.

Proposición VII.2.

Sea f:I→ℝf\colon I\to\mathbb{R}, I⊂ℝI\subset\mathbb{R} y a,b,c∈ℝa,b,c\in\mathbb{R} con a≤b≤ca\leq b\leq c . Entonces

f es integrable en [a,c]⇔f es integrable en [a,b] y [b,c].f\text{ es integrable en }[a,c]\iff f\text{ es integrable en }[a,b]\text{ y }[% b,c]\mathchar 46\relax

Además, ∫caf=∫baf+∫cbf\int^{c}_{a}f=\int^{b}_{a}f+\int^{c}_{b}f.

Proposición VII.3.

Sean ff y gg integrables en [a,b][a,b]:

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    Si f(x)≥0∀x∈[a,b]f(x)\geq 0\quad\forall x\in[a,b], entonces ∫baf≥0\int^{b}_{a}f\geq 0.

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    Si f(x)≥g(x)∀x∈[a,b]f(x)\geq g(x)\quad\forall x\in[a,b], entonces ∫baf≥∫bag\int^{b}_{a}f\geq\int^{b}_{a}g.