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APORTE TRABAJO COLABORATIVO FASE 2
CALCULO INTEGRAL
PRESENTADA POR:
TUTORA:
DELFINA REYES
UNIVERSIDAD NACIONAL ABIERTA Y A DISTANCIA UNAD
ADMINISTRACIÒN DE EMPRESAS CERES CURUMANI
MAYO DE 2015
Evaluar las siguientes integrales impropias.
1. ∫0
1
1n ( x )dx
∫0
1
ln (x)dx = xlnx−∫ 1x∗x∗dx = xlnx−∫ dx
u=lnx dv=dx =xlnx−x+c
v=1xdx v=x
∫0
1
lnx dx=xlnx−x 10
= 1 ln 1−1−0=−1
2.∫2
∞ 1
(x−1)2dx∫
2
∞1
( x−1 )2dx lim
b→∞∫
2
b1
( x−1 )2dx=u= x−1
du=dx
limb→∞
∫2
b1u2 du=lim
b→∞∫
2
b
u−2du= limb→∞
u−1
−1
¿−limb→∞
1u=− lim
b→∞
1x−1
b2
=limb→∞
1b−1
−11
limb→∞
1b−1
−1= 1∞−1
−1=0−1=−1
3. ∫−∞
∞e−5 xdx
∫−∞
∞
e−5 xdx= limA→∞B→∞
∫A
B
e−5x dx=¿¿ u=−5x
du=−5dx dx= du−5
limA→∞B→∞
−15
∫A
B
eu∗du= limA→∞B→∞
−15∫A
B
eu∗du= limA→∞B→∞
−15e−5x B
A
−15
limA→∞B→∞
e−5 x−e−5 A=−15 ( lim
A→∞B→∞
1e5B−
1e5 A )=0−∞=−∞
4. ∫2
54+x
√ x2−4dx=∫
2
54
√ x2−4dx+∫
2
5x
√x2−4dx
∫2
54
√X2−4dx X=2 sec α
X2=4 sec2α dx=2 sec α∗tanα dx
∫ 4 sec α∗tanα dx
√4 sec2−4=∫ 4
2∗sec α∗tan α dx
√ tan2α=2∫ sec α∗dx
2∫ sec α dx=2∫ sec α∗(sec α+ tanα )
sec α+ tanα∗dx=2∫ sec
2α+sec α+ tanαsec α+ tanα
∗dx
u=sec α+ tanα
du=( sec α tan α+sec2 )dx2∫ duu =2 ln (u )
2 ln (sec α+ tanα )+c
x=2 sec α sec α= x2
x2=4 sec2α
x2=4 (tan 2α−1 )=¿ 4tan2α−4
x2+44
=tan2α
tanα=√ x2+44
¿2 ln( x2 +√ x2+44 )
5¿2
a) 2ln ( 52+√ 29
4 )−2 ln (1+√2 )
b) ∫2
5x
√ x2−4 dx u=x2−4
du= 2xdx dx=du2x
∫x
u12. *du2x
=12∫ duu1/2=
12∫u1 /2du=1
2u1 /2
12
+c
(x2−4 )12 5
2=√x2−4
52
=√21 - 0
∫2
54+x
√ x2−4dx=2 ln(5
2+√ 29
4 )−2nl (1+√2 )+√21