CALCULO INTEGRAL

TRABAJO COLABORATIVO FASE 2

PRESENTA:RAUL ARTURO ECHEVERRY ARANGO CODIGO: 8.157.635ANGELA MARIA ZULUAGA RAMIREZ CODIGO 32.392.170

DIEGO MAURICIO ATEHORTUAMICHAEL PEREZ

CLAUDIA MARCELA ECHEVERRI

GRUPO: 100411_333

TUTOR:

JACKSON ARIEL URRUTIA

TECNOLOGÍA EN AUTOMATIZACIÓN ELECTRÓNICA

CEAD: MEDELLIN

2015

Evaluar las siguientes integrales impropias

1.

∫1

∞

(1−x )e− xdx

limb→∞

∫1

b

(1−x ) e−x dx

U=(1−x )dv=e− xdx

du=−dx v=−e−x

limb→∞ (−(1−x ) e−x|b1−∫1

b

e−x dx)limb→∞ (−(1−x ) e−x|b1+e− x|b1 )limb→∞

{ [−(1−b )e−b+(1−1 )e−1 ]+[e−b−e−1 ] }

[0+0+0−e−1 ]

¿−e≈2.71 (Laintegral converge )

2.

∫−∞

∞e x

1+e2xdx

∫−∞

0e x

1+e2xdx+∫

0

∞ex

1+e2 xdx

limA→−∞

∫A

0ex

1+(ex )2dx+ lim

B→∞∫0

Bex

1+(ex )2dx

U=ex du=exdx

limA→−∞

∫A

0du1+u2

+ limB→∞

∫0

Bdu1+u2

tan−1 (u )|10+ tan−1 (u )|∞1[ π2−0]+[ π2− π4 ]¿ π2

(Laintegral es convergente )

3.

∫0

1dx3√ x

lima→0

∫a

1

x−13 dx

lima→0 [ x

23

23

]|1alima→0 [ 32 x

23]|1a

¿ [ 32 (1 )−32

(0 )]¿ 32

(La integral esconvergente )

4.

∫0

π2

cos ( x )√1−sen ( x )

dx

limb→ π

2

∫0

bcos ( x )

√1−sen ( x )dx

U=1−senx du=−cosxdx

limb→ π

2

∫0

b−du√u

− limb→ π

2

∫0

b

u−12 du

− limb→ π

2 [ u12

12

]|b0− limb→ π

2

[2u12 ]|b0−2 lim

b→π2

[√1−senx ]|b0−2 lim

b→π2

[0−(1 ) ]

¿2 (Laintegral es convergente )

5.

∫ x3 (x4+3 )2dx

∫ x3 (x4+3 )2dx

U=x4+3du=4 x3dx

∫u2 du414∫ u

2du

14 ( u33 )+c112u3

112

(x4+3 )3

6.

∫0

13

(4+√ x )dx

U=4+√x du= 1

2√ xdx

√ x=U=4

∫4

52 (u−4 )du

u

2∫4

5

(1−4u )du

2 (u−4 ln|u|)|542 [ (5−4 ln|5|)−(4−4 ln|4|) ]

2 [5−4 ln|5|−4+4 ln|4|]

2[1+ln( 45 )4]

2+ ln( 45 )8

¿0.21

7.

∫ dx

x2√4+x2

x=2 tanθ d x=2 sec 2θdθ

∫ 2 sec2θdθ

(2tanθ )2√4+4 tan2θ

28∫ sec2θdθ

tan 2θ√1+ tan2θ

14∫ sec 2θ

tan2θ√sec2θdθ

14∫ sec2θtan2θ

dθ

14∫

1cosθ

sen2θcos2θ

dθ

14∫ cos2θcosθsen2θ

dθ

14∫

cosθ

sen2θdθ

U=senθ du=cosθdθ

14∫

du

u2

14

(−u−1)+c

−14u

+c

−14 senθ

+c

−1

4 ( x

√4+x2 )+c

−√4+x24 x

+c

8.

∫ x2

√ x2−4dx

x=2 secθ dx=2 secθ tanθ dθ

∫ (2 secθ)22 secθtanθ

√4 sec2θ−4dθ

82∫ sec

3θ tanθ

√ tan2θdθ

4∫ sec3θdθ

4∫ sec2θ secθ dθ

u=secθ dv=sec2θdθ

du=secθ tanθ v=tanθ

¿4 ¿

∫ sec3θdθ=4 secθtanθ−4∫ sec3θdθ+∫ secθdθ

5∫ sec3θdθ=4 secθ tanθ+ln|secθ+ tanθ|+c

∫ sec3θ=45 ( x2 √ x2−42 )+ln|x2 + √x2−4

2 |+c¿ 15x √x2−4+ ln|x+√ x−4

2 |+c

9.

∫ x2 sen ( x )dx

U=x2dv=senxdx

du=2 xdx v=−cosx

¿−x2cosx+2∫ xcosx dx

U=xdv=cosxdx

du=dx v=senx

¿−x2cosx+2 ( xsenx−∫ senx dx)

¿−x2cosx+2 ( xsenx+cosx+c )

¿−x2cosx+2 xsenx+2cosx+c

10.

∫ (3 x+5 )x3−x2−x+1

dx

∫ 3x+5( x−1 ) (x2−1 )

dx

∫ 3x+5( x−1 ) ( x−1 ) (x+1 )

dx

3 x+5( x−1 ) ( x−1 ) ( x+1 )

= A(x−1 )

+ B

( x−1 )2+ C

( x+1 )

3 x+5=A ( x−1 ) ( x+1 )+B ( x+1 )+C ( x−1 )2

si x=1 B=4

si x=−1C=12

si x=0 A=−109

−109 ∫ 1

x−1dx+4∫ ( x−1 )−2dx+ 1

2∫1x+1

dx

−109ln|x−1|+4 (−( x−1 )−1 )+ 1

2ln|x+1|+c

−109ln|x−1|− 4

( x−1 )+ 12ln|x+1|+c

11.

∫0

π4

sen3 (2 x )cos4 (2x )dx

∫0

π4

sen (2 x ) sen2 (2x ) cos4 (2x )dx

∫0

π4

sen (2 x ) (1−cos2 (2 x ))cos4 (2 x )dx

∫0

π4

sen (2 x ) (cos4 (2 x )−cos6 (2 x ) )dx

U=cos (2 x )du=−2 sen (2 x )dx

x=0→U=1

x=π4→U=0

−12∫1

0

(u4−u6)du

12∫1

0

(u4−u6 )du

12 ( u

5

5−u

7

7 )|1012 [(15−17 )−(0 )]135

12.

∫ (ex cosh ( x )+ln ( x ) )dx

∫ excosh ( x )dx+∫ ln ( x )dx

Identidad Trigonometrica

cosh ( x )= ex+e−x

2

u=ln xdv=dx

du=1xdx v=x

∫ ex( ex+e−x2 )dx+( x ln x−∫ dx )

12∫ (e2 x+1 )dx+( x lnx−x )+c

12 ( 12 e2 x+x )+ ( xlnx−x )+c

14e2x+ x

2+ xlnx−x+c