27 Teoremas sobre Integrales

Teorema 27.1 (Regla de Barrow).

Sean f\displaystyle f y F\displaystyle F continuas en [a,b]\displaystyle[a,b] y F\displaystyle F derivable en (a,b)\displaystyle(a,b) tal que F′⁢(x)=f⁢(x)⁢∀x∈(a,b)\displaystyle F^{\prime}(x)=f(x)\;\forall x\in(a,b). Entonces,

∫abf⁢(x)⁢dx=[F⁢(x)]ab=F⁢(b)−F⁢(a)\displaystyle\int^{b}_{a}f(x)\;\mathrm{d}x=\left[F(x)\right]^{b}_{a}=F(b)-F(a)
Teorema 27.2 (fundamental del calculo).

Si f\displaystyle f es continua en [a,b]\displaystyle[a,b] entonces existe una funcion F\displaystyle F tal que F′⁢(x)=f⁢(x)⁢∀x∈[a,b]\displaystyle F^{\prime}(x)=f(x)\;\forall x\in[a,b].

Demostración.

Sabemos que si f\displaystyle f es continua en [a,b]\displaystyle[a,b] entonces es integrable en [a,b]\displaystyle[a,b]. Dado x∈[a,b]\displaystyle x\in[a,b], tambien se tiene que f\displaystyle f es integrable en [a,x]\displaystyle[a,x]. Definimos F⁢(x)=∫axf⁢(t)⁢𝑑t\displaystyle F(x)=\int^{x}_{a}f(t)dt. Veamos que F⁢(x)\displaystyle F(x) es derivable en c∈[a,b]\displaystyle c\in[a,b] es decir, F′⁢(c)=l⁢ı´⁢mh→0F⁢(c+h)−F⁢(c)h=f⁢(c)⇒∀ε>0⁢∃δ>0\displaystyle F^{\prime}(c)=\mathop{\operator@font l\acute{{\imath}}m}\limits_% {h\to 0}\frac{F(c+h)-F(c)}{h}=f(c)\Rightarrow\forall\varepsilon>0\;\exists% \delta>0 t.q. si h∈(−δ,δ),h≠0,\displaystyle h\in(-\delta,\delta),h\neq 0, se cumple |F⁢(c+h)−F⁢(c)h−f⁢(c)|<ε\displaystyle\left|\frac{F(c+h)-F(c)}{h}-f(c)\right|<\varepsilon.

|F⁢(c+h)−F⁢(c)h−f⁢(c)|\displaystyle\displaystyle\left|\frac{F(c+h)-F(c)}{h}-f(c)\right| =|∫ac+hf⁢(t)⁢𝑑t−∫acf⁢(t)⁢𝑑th−f⁢(c)|=|∫cc+hf⁢(t)⁢𝑑th−f⁢(c)|\displaystyle\displaystyle=\left|\frac{\int^{c+h}_{a}f(t)dt-\int^{c}_{a}f(t)dt% }{h}-f(c)\right|=\left|\frac{\int^{c+h}_{c}f(t)dt}{h}-f(c)\right|
=|∫cc+hf⁢(t)⁢𝑑th−f⁢(c)|=|∫cc+hf⁢(t)⁢𝑑th−f⁢(c)⋅1h⁢∫cc+h1⁢𝑑t⏟=1|\displaystyle\displaystyle=\left|\frac{\int^{c+h}_{c}f(t)dt}{h}-f(c)\right|=% \left|\frac{\int^{c+h}_{c}f(t)dt}{h}-f(c)\cdot\underbrace{\frac{1}{h}\int^{c+h% }_{c}1dt}_{=1}\right|
=|1h⋅[∫cc+hf⁢(t)⁢𝑑t−f⁢(c)⋅∫cc+h1⁢𝑑t]|=|1h⋅∫cc+hf⁢(t)−f⁢(c)⁢d⁢t|\displaystyle\displaystyle=\left|\frac{1}{h}\cdot\left[\int^{c+h}_{c}f(t)dt-f(% c)\cdot\int^{c+h}_{c}1dt\right]\right|=\left|\frac{1}{h}\cdot\int^{c+h}_{c}f(t% )-f(c)dt\right|
≤|1h|⁢∫cc+h|f⁢(t)−f⁢(c)|⁢𝑑t=|F⁢(c+h)−F⁢(c)h−f⁢(c)|⁢<?⁢ε\displaystyle\displaystyle\leq\left|\frac{1}{h}\right|\int^{c+h}_{c}\left|f(t)% -f(c)\right|dt=\left|\frac{F(c+h)-F(c)}{h}-f(c)\right|\overset{?}{<}\varepsilon

Sea ε>0\displaystyle\varepsilon>0. Como f\displaystyle f es continua en c\displaystyle c, entonces ∃δ>0\displaystyle\exists\delta>0 tal que ∀t∈(c−δ,c+δ)\displaystyle\forall t\in(c-\delta,c+\delta), t≠c\displaystyle t\neq c, se cumple |f⁢(t)−f⁢(c)|<ε\displaystyle\left|f(t)-f(c)\right|<\varepsilon. Luego ∀h∈(−δ,δ),h≠0\displaystyle\forall h\in(-\delta,\delta),h\neq 0, suponemos que h>0\displaystyle h>0 (similar si h<0\displaystyle h<0) y tenemos que

|F⁢(c+h)−F⁢(c)h−f⁢(c)|\displaystyle\displaystyle\left|\frac{F(c+h)-F(c)}{h}-f(c)\right| ≤1|h|⋅∫cc+h|f⁢(t)−f⁢(c)|⁢𝑑t<1h⋅∫cc+hε⁢𝑑t\displaystyle\displaystyle\leq\frac{1}{\left|h\right|}\cdot\int^{c+h}_{c}\left% |f(t)-f(c)\right|dt<\frac{1}{h}\cdot\int^{c+h}_{c}\varepsilon dt
=1h⋅ε⋅∫cc+h1⁢𝑑t=1h⋅ε⋅h=ε\displaystyle\displaystyle=\frac{1}{h}\cdot\varepsilon\cdot\int^{c+h}_{c}1dt=% \frac{1}{h}\cdot\varepsilon\cdot h=\varepsilon

*Borrador, falta reescribirlo* ∎

Integración por partes, cambios de variable.