Compilacion de Aportes

7/25/2019 Compilacion de Aportes

1/37

Hallar la solucin de las siguientes integrales paso a paso, teniendo en cuentalas propiedades de las integrales indefnidas, las cuales son consecuencia delas aplicadas en la dierenciacin.

EJERCICIO 1

SILVIA

x

3+x2x2

dx

ddx [f1 X

3+X2

X2

dx ]

X

F.1( 3+X2)X

2 dx

ddx

XF( 3+X2)

X2

dx

ddx

XF( 3+X2)

X2

d1

x

ddx

XF( 3+X2)

X2

dx

ddx


2/37

XF( 3+X2)d

X2

x1

d

dx

XF( 3+X2)dx

X2

ddx

X

x ( f( 3+X2)d)xx

ddx

ANGELA

X3+X2

x2 dx=x

3

x2dx+ x

x2dx2 1

x2dx

xdx+ 1x

dx2x2dx

x

2

2+ ln|x|2x

1

(1 )

x

2

2+

2

x+ ln|x|+c

JHON


3/37

XF.1(3+X2)

X2 dx

XF( 3+X2)

X2 dx

x3+x2

x2 dx

ddx[f1 X

3+X2

X2 dx ] ddx

XF( 3+X2)

X2

d1

x

XF( 3+X2)

X2

dx

XF( 3+X2)d

X2

x1

d

dx

XF(3+X2)dx

X2

Xx ( f( 3+X2)d)

xx

ddx

OSCAR

1. x3+x2

x2 dx = (x

3

x2+

x

x22x2 )dx


4/37

= x3

x2dx+ x

x2dx 2

x2dx

= x dx+1

x

dx2 1

x

2dx

=(x2

2+lnx+

2

x )+C

LEONARDO

1 x

3+x2

x2

dx = (x3

x2+

x

x22x2 )dx

= x3

x2dx 2

x2dx+ x

x2dx

= x dx2 1

x2dx+ 1

xdx

=x

2

2+

2

x+lnx+C

EJERCICIO 2

SILVIA

sec

2(x )

tan(x )dx

ddx

[f

(

d1

sec2 (x )

tan

(x )

. x

)]

ddx [ f( d sec

2 (x )

tan(x).x1 )]

o


5/37

ddx [ f( d sec

2 (x )x

tan (x ))]o

ddx

[f1

(dsec2 (x ))x

tan (x )]o

ddx [fd sec

2 (x )x

tan (x)]o

fd1

ddx [ sec

2 (x )x

tan (x)12]o

fd ddx [sec

2 (x )x

tan(x)12]

fd ddx[tan (x )

1

2 ddx [sec2 (x )xsec2 (x )xddx ] [ tan (x )

1

2 ](tan (x )

1

2) ]

fd ddx[tan (x )

1

2 ddx

[sec2 (x )x ]sec2 (x )x ddx

[ tan (x )1

2 ]( tan(x )) ]

fd ddx[tan (x )

1

2 ddx(sec2 (x )+x ddx[sec2(x )])sec2 (x)xddx[ tan (x )

1

2 ]tan(x) ]

fd ddx[

tan (x )

1

2 ddx(sec

2

(x )+2xs+sec (x )+xddx[sec(x )])sec

2

(x )xddx[ tan (x )

1

2

]tan(x ) ]

fd ddx[tan (x )

1

2 ddx

(sec2 (x )+2xsec (x ) tan(x))sec2 (x )x ddx

[ tan (x )1

2 ]tan(x ) ]


6/37

fd ddx[tan (x )

1

2 ddx

(sec2 (x )+2xsec (x ) tan(x))sec2 (x )x ddx

[ tan (x )1

2 ]tan(x ) ]

fd ddx[tan (x )

12 d

dx(sec2 (x )+2xsec (x ) tan(x))sec2 (x )x d

dx[ tan (x )12 ]

tan(x ) ] fd

ddx[tan (x )

1

2 ddx

(sec2 (x )+2xsec (x )2 (x) tan(x))sec2 (x )x ddx

[ tan(x)1

2 ]tan(x) ]

fd ddx[

tan (x )12 d

dx(sec2 (x )+2xsec (x )2 (x ) tan(x))sec2 (x )x d

dx [ tan(x)12+1222 d

dx] [ tan(x) ]tan (x ) ]

fd ddx[tan (x )

1

2 ddx

(sec2 (x )+2xsec (x )2 (x) tan(x))sec2 (x )x ddx [ tan(x)

1

2+12

2 ddx ] [ tan(x) ]

tan(x) ]

fd ddx[

tan (x )

1

2 (sec2

(x )+2xsec (x )2

(x ) tan(x))1

2

2 sec

4

(x )x

d

dx[tan

(x )]1

2x2

tan(x) ]JHON

sec2(x)

tan (x )dx

1 u

du 1

u=u

12 u

12 du=

u12 +1

12+1

=tan

12 +

1

(x )12

+1=2tan (x )=2tan (x )+c

sec2(x )

tan (x )dx=2tan (x )+c


7/37

CLAUDIA

sec2(x)

tan (x )

dx

1u

du

1

u=u

12

u12 du

u

12 +1

12 +1

tan

12

+1(x)

12

+1

2tan (x )

2tan (x )+c

sec2

(x ) tan (x ) dx=2tan (x )+c

OSCAR

sec2(x )

tan(x)dx = Integral por sustitucin

= tan (x)=t ;dt=sec2(x )dx

= dt sec2(x)dx; se despejadx

=dx=

dt

sec2(x )


8/37

Por identidadcos

2 (x )= 1

sec2(x )

= dx=cos2 (x ) dt

Entonces

sec2 (x )cos2(x )

t dt

1

cos2(x )

cos2(x )

t dt

1

cos2(x )

cos2(x )

t dt

1t

dt

t1 /2 dt

t

12+1

12 +1

=t1 /2

1

2

=2t ; reemplazo t=tan(x )

sec2(x )

tan(x)dx=2tan (x )+C

LEONARDO

sec2(x )

tan(x)dx

Integral por sustitucin


9/37

= tan (x)=t ;dt=sec2(x )dx

= dt sec2(x)dx; se despejadx

=dx= dt

sec2(x )

Por identidad cos2 (x )=

1

sec2(x )

= dx=cos2 (x ) dt

Entonces :

sec2 (x )cos2(x )

t dt

1

cos2(x )

cos2(x )

t dt

1

cos2(x )

cos2

(x )

t dt

1t

dt

t1 /2 dt

t

12+1

12 +1

=t12

1

2

=2t ;

reemplazot=tan(x)


10/37

sec2(x )

tan(x)dx=2tan (x )+C

EJERCICIO 3

SILVIA

(1+3)2

3 x dx

1+32=(9x2+6x+1)

9x

2+6x+13

xdx

9x

2

xdx+ 6x

xdx+ 13

xdx

9x2

3xdx=

27x8

3

8

6xx

dx=18x

5

3

5

1x

dx=3x

2

3

2

27x

8

3

8 +

18x5

3

5 +

3x2

3

2

1+32

ANGELA

(1+3x3x)2

dx= 9x2+6x+1

3x

dx


11/37

9x

2

3

xdx+ 6x3

xdx+ 13

xdx

9x2 . x13 dx+ 6x . x

13 dx+x

13 dx

9x

8

3

8

3

+6x

5

3

5

3

+x

2

3

2

3

+c

27

8 x

8

3+18

5 x

5

3+3

2x

2

3

+ c

JHON1+32

1+32=(9x2+6x+1 )= 9x2+6x+1

3xdx= 9x

2

xdx+ 6x

xdx+ 13x

dx 9x2

3xdx=

27x8

3

8 6x

xdx=

(1+3)2

3 x dx

CLAUDIA

(1+3x)2

3 x dx

1+32=(9x2+6x+1)

9x2+6x+1

3

xdx

9x2

x

dx+ 6x

x

dx+ 13x

dx

9x2

3xdx=

27x8

3

8

6xx

dx=18x

5

3

5


12/37

1x

dx=3x

2

3

2

27x

8

3

8

+18x

5

3

5

+3x

2

3

2

+c

OSCAR

LEONARDO

(1+3x )2

3xdx= 1+3x+dx

2

3xdx

13x

dx+ 3x3x

dx+ 9x2

3

xdx

x13 dx3x

2

3 dx+9x5

3 dx

3x2

3

2 +

9x5

3

5 +

27x8

3

8 +c

EJERCICIO 4


13/37

JHON

tan3 (x ) dx= tan (x) sec 2 (x) dx tan (x) dx.

CLAUDIA

tan3(x)dx

tanx=sinxcosx

sen2x=1cos2x

Reemplazando

tan3

dx=sen3x

cos3x dx

Descomponiendo la potencia del numerador

sen2x . senx

cos3x

dx

Usandox

1cos2 .senx

cos

3x

Por el mtodo de sustitucin de variablesu=cosxdu=sen xdx

dx= dusenx

Reemplazando en trminos du

(1u)2 senx

u3 senxdu

eparando los integrales


15/37

"EO#$R%O

Sustitucion

El co&unto de todas las antirediadas de '()* se llaa integral inde'inida de ' respecto a ), -

se denota por el si!olo f(x ) dx=F(x )+C . Resoler las siguientes integrales

inde'inidas:

EJERCICIO /

SILVIA 2+9 3x

3x2dx

2+93x

x2

3

dx

39u+2du

39u+2du


16/37

3v9

dv

31

9 v dv

31

9

v1

2+1

1

2+1

31

9

(93x+2)

1

2+1

1

2+1

2

9(9

3x+2)3

2

=2

9(9

3x+2)3

2+C

JHON

2+93x

3x2dx

2+93

x

x2

3

dx 39u+2du39u+2du 3

v9

dv

31

9 v dv 3

1

9

v1

2+1

1

2+1

31

9

(93x+2)

1

2+1

1

2+1

2

9(9

3x+2)3

2

=

2

9(9

3x+2)3

2+C

CLAUDIA


17/37

2+93x

3

x2dx

3x2 = x2

3 , se asue 0ue x0

2+93x

x2

3

dx

e aplica integracin por sustitucin:

39u+2du

e saca la constante:

39u+2du

#ueaente se aplica la integracin por sustitucin:

3 v9

dv


31

9 v dv

"uego se aplica la regla de la potencia:

31

9

v1

2+1

1

2+1

e sustitu-e en la ecuacin:

31

9

(93x+2)

1

2+1

1

2+1

e sipli'ica:


18/37

2

9(9

3x+2)3

2

Por ltio se agrega una constante a la solucin:

2

9(9

3x+2)32+C

OSCAR

2+93x

3

x2dx

2+93x

3

x2

dx=

(2+9x1

2 )1/2

x2

/3

aceos: u=2+9x1 /3=du=

913

x2/3

dx

du=3x2 /3 dx

du=3dx

x2/3

du3=dx

x2 /3

1

3 u1/2 du=

1

21

3 u

3

2+C=2

9(2+9x

1

3)3

2

+C

x

2+93

2

9

"EO#$R%O


19/37

2+93x

3

x2dx

2+93x

3

x2dx

(2+9x

1

2)1/2

x2 /3

Integral por sustitucin:

u=2+9x1 /3=du=91

3 x2/3dx

du=3x2 /3 dx

du=3dx

x2/3

du

3

=dx

x2 /3

1

3u

1

2 du=

1

21

3u

3

2+C

2

9(2+9x

1

3)3

2

+C

x2+9 3

2

9


20/37

EJERCICIO

SILVIA

x

3x4 dx

123V2 d

1

2 1

3v2dv

1

2 3cos (v )

33 sen2(v) dv

1

23 cos (v )

33 sen2(v )

1

23 cos(v )

3 sen2 ( v )+1dv

123 cos(v )cos2(v)3 dv

1

23 cos(v)

cos(v)3dv

1

23 1

3dv

12

3 13 v

1

23

1

3arcSen

1

3x

2


21/37

1

2arcSen(x

2

3 )JHON

x3x4

dxIntega! "n#ed"ata

xa2x2

=1

2arcSen (xa )+c=12 arcSen (x2 )+c

ANGELA

x

3x

4dx Integral inediata

xa2x2

=1

2arc Sen(xa )+c

1

2arc Sen (x2 )+c

CLAUDIA

x3x4

dx

Esdela forma dxa2x2

x=a sen

dx=acosd

x

2

=3

sen

2xdx=3cosd

xdx=3cosd 2


22/37

x22

(3)2x

3

2(3sen)

2

3cosd

3cosd2(33 sen2)

32 cos

3(1sen2 )d

32 cos

3cos2d

323

coscos

d

1

2 1d

1

2+c

Enc ontramos


23/37

x2=3Sen

x2

3=sen

=arcsen (x

2

3)

ue!o x3x 4

dx=arcsen(x

2

3)+c

OSCAR

=

=

LEONARDO

32

1 2

1 xSen

=


24/37

EJERCICIO 5

JHON

sen (4x ) cos (3x )dx =!"#$%&' sen"()& * sen")& + d) = "#$%& !sen"()& d)*"#$%& ! sen")& d) = "#$#-&co)"()& "#$%& cos")& *

CLAUDIA

sen(4x)cos (3x)dx

e usa la identidad: cos ( t) sen (s )=sen (s+ t)+sen(st)

2

sen (4x+3x )+sen4x3x2

dx


1

2 sen (4x+3x )+sen (4x3x )dx

e aplica la regla de la sua:

1

2( sen (4x+3x )dx+ sen (4x3x )dx)

sen (4x+3x ) dx=17 cos (7x )

sen (4x3x ) dx=cos (x )

12 (

17 cos (7x )cos (x ))

e agrega una constante a la solucin:

1

2 (17 cos (7x )cos (x ))+ C


25/37

OSCAR

sen (4x ) cos (3x )dx

dx=7dxdx=dx

7

senx2

x1

1dx

senx14

dxcosx14

cos (7x )14

cosx2

cos (7x )14

cosx2

+C

LEONARDO

sen (4x ) cos (3x )dx

INTEGRAL POR SSTIT!ION

dx=7dx dx=dx7

senx2

x1

1dx

senx14

dx=cosx14

cos (7x )

14


26/37

cosx2

cosx2 +C

EJERCICIO 6

ANGELA

cos

3 (t)+1

cos2 (t)

dt= Integral inediata 1

cos2x

dx=tan! x

cos3 (t)

cos2 (t)

dt+ 1cos

2 (t)dt=

cos (t) dt+ 1cos

2 ( t)dt=

Sen (t)+ tan ( t)+c

JHON

.

cos (t) dt+ 1cos

2 (t)dt==Sen (t)+ tan (t)+c

cos

3 (t)+1

cos2 (t)

dt=cos

3 (t)

cos2 (t)

dt+ 1cos

2 (t)dt=

OSCAR


27/37

LENARDO

( )

=

+=

dtt

dtt

t

2

2

seccos

cos

1cos

7n teorea generalente posee un nero de preisas 0ue de!en ser enueradas o

aclaradas de anteano. "uego e)iste una conclusin, una a'iracin lgica o ate8tica,

la cual es erdadera !a&o las condiciones dadas. El contenido in'oratio del teorea es la

relacin 0ue e)iste entre las 9iptesis - la tesis o conclusin.

EJERCICIO

ANGELA

". f(x )=x x2+16 [0,3 ] ;9 alor edio f

" (c )= f(# )f( a )#a

f(x )=2x2+16

x2+16 f(# )=39+16

2x2+16

x2+16=

15030

f( # )=3.5

f( a )=002+16

2x2+16=5x2+16

f( a )=0


28/37

x1=1,68992

Valores ediosx2=1,68992

JHON

f(x )=x x2+16 [0,3 ] /0 valor medio f" (c )=

f( # )f(a )#a

f(x )=2x2+16

x2+16f( # )=39+16

2x2+16

x2+16=

15030

f(# )=3.5 f( a )=002+16

2x2+16=5x2+16 f( a )=0

OSCAR

F(X)=XX2+16

fpromen [0,3]

x2x (+16)dx

fprom=0

3

fprom=1

3[1+x

3

3+16]

9+16+30

fprom

=1

3

R


29/37

a

#

f(x )= (#a )F(C)

0

3

x x2+16dx=(30 ) f(c)

0

3

x x2+16dx

x x2+16dx

$=x2+16

d$=2x dx

$d$2

1

2 $

12 d$=

1

2

$3

2

3 /2+c

x2

+163

+c

$3

3+c=

6

x2+13

0

3

x x2+16dx=


30/37

(3)2+163(o)

+162

15.6253

40963

125

3

64

3=

125643

=61

3

$9ora

61

3= (30 ) . f(c)

61

9=f(c)

c c2+16=619

c

(c2+16 )9

61

c2(c2+16)=

3721

81


31/37

c4+16c2

3721

81 =0

x=

16%

16

24 (1 )3721

81

2(1)

x=16%256+1488481

2

x=16%3562081

2

16+20.972

=2,48

LEONARDO

F(X)=xx2+16

fpromen [0 "3]

x2x (+16)dx

fprom=0

3

fprom=13[1+

x

3

3+16]


32/37

9+16+30

fprom=1

3

EJERCICIO 1>

i se supone 0ue la po!lacin undial actual es de 5 il illones - 0ue la po!lacin dentro

de t a?os est8 dada por la le- del creciiento e)ponencial p (t)=e0.023 t

. Encuentre, la

po!lacin proedio de la tierra en los pr)ios 3> a?os.

CLAUDIA

p&=7mil millones ta'os=p (t)=e0,023t

p promen30a'os

valor promedios= 1

300 0

30

7e0,023 t

dt 1

30 [ 70,023 e0.023 t]

1

30

[ 7

0,023 e

0.023 t.30

7

0,023

] 1

30.

7

0,023[e0.691 ] = 8,883045 miles de millones

OSCAR

p=7mll

p=30a'os

7 (30 )=(210)0 .023=3 "772milmillones

R


33/37

LEONARDO

%atos:

p&=7milmillones

p=30a'os

7 (30 )=(210)0 .023=3 "772mil millones de personas

EJERCICIO 11

Sip (x )=

1

x3

cos (t) dt. Deter#inar

dp

dx=

d

dx1

x3

cos (t)dt.

ANGELA

+c

dpdx1

x3

cos (t) dt=dpdx

Sen (x)+1

x3

dp

dx

(Sen (x3 )Sen (1 ) )

3x2cosx3

OSCAR

Deter#inar

LEONARDO

Deter#inar


34/37

( ) ( ) 3233

33

cos311

cos

xxsensenxdx

dsensenx

senttdt xX

=++

=

EJERCICIO 12

CLAUDIA

$plicar el segundo teorea 'undaental del c8lculo para resoler:

0

(4

sen3 (2x ) cos (2x ) dx

a

#

f(x ) dx= f(# )f( a ) f) (x )=f(x)

0

(4

sen3 (2x ) cos (2x ) dx

$=2x

d$2=dx

0

(2

sen2 ($ ) sen ( $ ) cos ( $ ) d $

2

120

2(

2 sen2($)sen ( $ )(cos$)2

d$

1

20

(2

(1cos2 $) (sen$ )(cos$)d$


35/37

0

(

2

sen$.cos$d$+0

(

2

cos2$ . sen$. cos$d$

1

2

v=sen($)

dv=cos ($ ) . d$

t=cos ($)

d t=sen ( $ ) d$

0

(2

vdv+0

(2

cos3$ sen$.d$

1

2

+0

(2

t3dt

v2

2 0

(2

1

2

(2

2

20

t4

40

(2

1

2

=

(2

8

(4

64

1

2 [(4

2

2

(16

4

4]=12


36/37

(

2

16

(4

128

OSCAR

0

(/4

sen3 (2x ) cos (2x )dx=por sustituciont=2x ;dt=2dx

1

20

(/ 4

sen3 (t) cos ( t) dt

sustitu*endode nuevou=sen (t); du=cos (t) dt

1

20

(/ 4

u3du

1

2 [u3+1

3+1 ] {(4

0

=1

2 [u4

4] {(4

0

sen3 (2x )cos (2x ) dx=1

2 [sen4(t)4 ]{

(4

0

=1

2 [sen4 (2x )4 ]{

(4

0

0

(/4

1

2[sen4(2

(4)

4 ]1

2 [sen4 (20 )4 ]

1

2 [14 ]12[0 ]=18

LEONARDO

sen3 (2x )cos (2x ) dx=

0

(/4


37/37

Por el @todo de sustitucin t=2)A dt=2 d)

1

20

(/ 4

sen3 (t) cos ( t) dt

sustitu*endode nuevou=sen (t); du=cos (t) dt

1

20

(/ 4

u3du

1

2 [u3+1

3+1 ] {(4

0

=1

2 [u4

4] {(4

0

sen3 (2x )cos (2x ) dx=1

2 [sen4(t)4 ]{

(4

0

=1

2 [sen4 (2x )4 ]{

(4

0

0

(/4

1

2[sen4(2

(4)

4 ]1

2 [sen4 (20 )4 ]

1

2 [14 ]12[0 ]=18

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