26 Propiedades de las integrales

Teorema 26.1.

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

Si ⁢f⁢ es continua en ⁢[a,b]⇒f⁢ es integrable en ⁢[a,b].\displaystyle\text{Si }f\text{ es continua en }[a,b]\Rightarrow f\text{ es % integrable en }[a,b].

Además, el área comprendida entre la gráfica de una función continua positiva, el eje X\displaystyle X y las rectas x=a\displaystyle x=a y x=b\displaystyle x=b, es igual a

∫abf.\displaystyle\int^{b}_{a}f.
Demostración.

Tampoco se ha visto en clase. ∎

Ejemplo.

f⁢(x)=c\displaystyle f(x)=c, con c∈ℝ\displaystyle c\in\mathbb{R}. Sean a,b∈ℝ\displaystyle a,b\in\mathbb{R} y consideramos el intervalo [b,a]\displaystyle[b,a]. Sea P={x0,x1,…,xn}⊂[a,b]\displaystyle P=\{x_{0},x_{1},\ldots,x_{n}\}\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)\displaystyle 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)⁢∀P\displaystyle U(f,P)=L(f,P)=c(b-a)\;\forall P. Es decir, f\displaystyle f es integrable y ∫abf⁢(x)⁢𝑑x=∫abc⁢𝑑x=c⁢(b−a)\displaystyle\int^{b}_{a}f(x)dx=\int^{b}_{a}cdx=c(b-a).

Ejemplo.

∫01x⁢𝑑x=?\displaystyle\int^{1}_{0}xdx=?. Consideramos una particion P={0,1n,2n,3n,…,nn}\displaystyle P=\{0,\frac{1}{n},\frac{2}{n},\frac{3}{n},\ldots,\frac{n}{n}\}. Esta particion es equiespacial ya que la diferencia entre cada elemento es siempre 1n\displaystyle\frac{1}{n}. Por ejemplo, si n=4\displaystyle n=4 la particion seria P4={0,14,24,34,44}\displaystyle 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⋅14U(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⋅14L(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 Pn\displaystyle P_{n},

U⁢(x,Pn)=∑j=1nMj⋅(xj−xj−1)=∑j=1nxj⋅1n=∑j=1njn⋅1n=1n2⁢∑j=1nj\displaystyle U(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)\displaystyle 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=1nj\displaystyle U(x,P_{n})=\frac{1}{n^{2}}\sum_{j=1}^{n}j y L⁢(x,Pn)=1n2⁢∑j=1n(j−1)\displaystyle L(x,P_{n})=\frac{1}{n^{2}}\sum_{j=1}^{n}(j-1). Sea U⁢(f)=ı´⁢nf(U⁢(f,P))\displaystyle U(f)=\mathop{\operator@font\acute{{\imath}}nf}(U(f,P)) y L⁢(f)=sup(L⁢(f,P))\displaystyle L(f)=\sup(L(f,P)). Es obvio que L⁢(f)≤U⁢(f)\displaystyle 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\displaystyle\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\displaystyle\sum_{j=1}^{n}j=\frac{n(n+1)}{2} y ∑j=1n(j−1)=∑j=1nj−∑j=1n1=n⁢(n−1)2−n\displaystyle\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+n2⁢n2−nn2⁢⟶n→∞⁢12\displaystyle L(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)=1n2⁢n⁢(n+1)2=n2+n2⁢n2⁢⟶n→∞⁢12\displaystyle U(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)=12\displaystyle L(f)=U(f)=\frac{1}{2} y f\displaystyle f es integrable ⇒∫01x⁢𝑑x=12\displaystyle\Rightarrow\int^{1}_{0}xdx=\frac{1}{2}.

Proposición 26.1.

Sean f\displaystyle f y g\displaystyle g integrables en [a,b]\displaystyle[a,b], entonces:

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    f+g\displaystyle f+g es integrable en [a,b]\displaystyle[a,b] y ∫ab(f+g)=∫abf+∫abg\displaystyle\int^{b}_{a}(f+g)=\int^{b}_{a}f+\int^{b}_{a}g.

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    λ⁢f\displaystyle\lambda f es integrable en [a,b]\displaystyle[a,b] y ∫abλ⁢f=λ⁢∫abf\displaystyle\int^{b}_{a}\lambda f=\lambda\int^{b}_{a}f.

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    |f|\displaystyle\left|f\right| es integrable en [a,b]\displaystyle[a,b] y ∫abf≤∫ab|f|.\displaystyle\int^{b}_{a}f\leq\int^{b}_{a}\left|f\right|.

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

Proposición 26.2.

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

f⁢ es integrable en ⁢[a,c]⇔f⁢ es integrable en ⁢[a,b]⁢ y ⁢[b,c].\displaystyle f\text{ es integrable en }[a,c]\Leftrightarrow f\text{ es % integrable en }[a,b]\text{ y }[b,c].

Además, ∫acf=∫abf+∫bcf\displaystyle\int^{c}_{a}f=\int^{b}_{a}f+\int^{c}_{b}f.

Proposición 26.3.

Sean f\displaystyle f y g\displaystyle g integrables en [a,b]\displaystyle[a,b]:

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

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