MATEMATICA

INTEGRALES IMPROPIAS

Llamaremos integrales impropias a aquellas integrales que:

Los límites de integración crecen o decrecen sin limitación (-∞,∞)

Que las función integral sea discontinua en algún valor del intervalo de la ecuación f(x) discontinua

CASO 1:

Cuando los límites de son +∞ y -∞

∫_(-∞)^b▒f(x)dx 2. ∫_a^(+∞)▒f(x)dx

=〖lim┬(a→-∞) ∫_a^b▒〖f(X)〗〗⁡dx = 〖lim┬(b→+∞) ∫_a^b▒〖f(X)〗〗⁡dx

3. ∫_(-∞)^(+∞)▒f(x)dx

=∫_(-∞)^0▒f(x)dx + ∫_0^(+∞)▒f(x)dx

〖=lim┬(a→-∞) ∫_a^b▒〖f(X)〗〗⁡dx + 〖lim┬(a→-∞) ∫_a^b▒〖f(X)〗〗⁡dx

Si el integral impropio da un número entonces se dirá que CONVERGE a el valor encontrado. Si por el contrario el integral no existe (-∞ o +∞) entonces se dirá que DIVERGE.

Evaluar el siguiente integral:

∫_(-∞)^5▒xdx/〖〖(x〗^2+1)〗^3 = 〖=lim┬(a→-∞) ∫_a^5▒xdx/〖〖(x〗^2+1)〗^3 〗⁡

∫_(-∞)^5▒xdx/〖〖(x〗^2+1)〗^3 =∫_(-∞)^5▒〖〖〖(x〗^2+1)〗^(-3 ) dx〗

U= x2+1

Du=2xdx

(du )/2=xdx

∫▒〖u^(-3 ) du/u〗=1/2 ∫▒〖u^(-3 ) du〗

=1/2 ⌊u^(-2)/(-2)⌋ + c = (-1)/〖4u〗^2 + c = - 1/(4〖(x〗^2+1)^2 ) + c

lim┬(a→-∞)⁡〖(-1/(4〖〖(x〗^2+1)〗^2 ))^ 〗a5 = lim┬(a→-∞)⁡〖⌈(-1/(4〖(26)〗^2 ))-(-1/(4〖〖(a〗^2+1)〗^2 ))⌉^ 〗

〖=lim┬(a→-∞)〗⁡〖⌈(-1/2704)-(-1/(4〖〖(a〗^2+1)〗^2 ))⌉^ 〗

EVALUAR EL INTEGRAL

=⁡〖⌈(-1/2704)-(-1/(4〖〖(-∞〗^2+1)〗^2 ))⌉^ 〗 = -1/2704+ 1/∞

R//=-1/2704 converge

Ejercicio N° 2

∫_0^(+∞)▒dx/√(x+5)

∫_0^(+∞)▒(x+5)^(-1/2) dx =〖 lim┬(b→+∞)〗⁡〖∫_0^a▒(1+1/n) ^(-1/2) 〗 dx

∫_0^(+∞)▒(x+5)^(-1/2) dx= ∫▒u^((-1)/2) du

U=x+5 =〖2u〗^((-1)⁄2)/1

Du=dx =2〖(x+5)〗^(1⁄2) + c

(=lim)┬(b→∞)⁡〖⌈(2√(x+5))⌉^b 〗0

(=lim)┬(b→∞)⁡〖〖 ⌈(2√(b+5))⌉〗^ 〗-⌈(2√(0+5))⌉

(=lim)┬(b→∞)⁡〖〖 ⌈(2√(b+5))⌉〗^ 〗-⌈(2√5)⌉

EVALUANDO LÍMITES

=(2√(∞+5))-(2√5)

=(2√∞)-(2√5)

R// =∞ DIVERGE

3.∫_0^∞▒(〖5x〗^2 dx)/(〖(2x〗^3 〖+1)〗^5 ) = lim┬(b→∞)⁡〖∫_0^a▒(〖2x〗^3+1) ^(-5) x^2 〗 dx=∫▒u^(-5) du

u=(〖2x〗^3+1)dx =∫▒〖u^(-6)/(-6) du+c〗

du= 〖2x〗^3+dx

du=〖6x〗^2 dx

du/6= x^2 dx

4. ∫_(-∞)^(+∞)▒dx/(x^(2 )+1)= ∫_(-∞)^0▒〖dx/(x^(2 )+1)+ ∫_0^(-∞)▒〖dx/(x^(2 )+1) 〗 〗

=lim┬(a→-∞)⁡∫_a^0▒〖dx/(x^(2 )+1)+ lim┬(a→-∞)⁡∫_0^b▒dx/(x^(2 )+1) 〗

lim┬(a→-∞) arctanx|0a + lim┬(a→-∞) arctanx|b0

lim┬(a→-∞) arctan⁡(0)- arctan⁡(a)+ ⌈lim┬(a→-∞) arctan⁡(b)- arctan⁡(0) ⌉

lim┬(a→-∞) arctan⁡(a)+ ⌈lim┬(a→-∞) arctan⁡(b) ⌉

EVALUANDO LÍMITE

=-arctan(-∞) +arctan⁡(∞)

= (-π/2) + π/2

= π/2+ π/2

=π R//

∫_2^5▒xdx/√(3&x^2-25) =∫_2^n▒3/(4 ) (x^2-25)^(2⁄3) =lim┬(b→5) ̇ ⁡∫_2^n▒〖3(x^2-25)〗^(2⁄3)/4

∫_2^5▒(x^2-25)^(〖-1/〗_3 ) xdx

U=x^2-25dx

Du=2x-0dx

du/2=xdx ∫_2^5▒〖u^(-y3) du/2〗 = 1/(2 ) ∫▒〖u^(-1/3)+〗 c

x^2-25=0 1/2 [3/2 u^(2⁄3) ]+c

3/4 (u^(2⁄3) )= [3/4 (x^2-25)^(2⁄3) ]

x^2=25

x=√25 lim┬(n-5)⁡∫_2^n▒〖3(x^2-25)^(2⁄3) 〗 =[〖3(n^2-25)〗^(2⁄3)/4] + [((2^2 )-25)^(2⁄3)/4] =R/ -5.7

Ejercicios:

∫_0^3▒dx/√(3&x-1) ∫_0^3▒〖(x-1)^((-1)⁄3) dx=∫_0^1▒〖(x-1)^(1⁄3) dx+∫_1^3▒〖(x-1)^((-1)⁄3) dx〗〗〗

x-1=0 (lim)┬(n→1)⁡〖∫_0^n▒(x-1)^(1⁄3) +(lim)┬(m→1)⁡∫_m^3▒〖(x-1)^((-1)⁄3) dx〗 〗

x=0

(lim)┬(n→1)⁡〖[〖3(x-1)〗^(2⁄3)/2]^n+(lim)┬(m→1)⁡〖[〖3(x-1)〗^(2⁄3)/2]^3 〗 〗

=(lim)┬(n→1)⁡〖((3(n-1))/2)^(2⁄3)+(lim)┬(n→1)⁡〖(〖3(n-1)〗^(2⁄3)/2)+(lim)┬(m→1)⁡〖(〖3(3-1)〗^(2⁄3)/2)+((3(n-1))/2)^(2⁄3) 〗 〗

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