PRIMERA PRACtica de Analisis3

7/21/2019 PRIMERA PRACtica de Analisis3

1/60

I. SUMATORIAS:Encontrar una frmula para cada una de las siguientessumatorias

ai=1

n

i5 (resuelto por RAMOS PALOMINO Jhordy Maycol)

SOLUCI!:

Usando la identidad: i6( i1 )6=i6(i1)6

"esarrollando el segundo miem#ro

i6( i1 )6=i6[ i66 i5+15 i420 i3+15 i26 i+1 ]

i6( i1 )6=+6 i515 i4+20 i315 i2+6 i1

Aplicando sumatorias a am#os miem#ros

i=1

n

[ i6 (i1 )6 ]=i=1

n

[6 i515 i4+20 i315 i2+6 i1 ]

Recordando: i=1

n

[ f(i)f(i1)]= f( i )f(0) $aciendo%

f( i )=i6 f(n )=n6 ; f(0 )=0

i=1

n

[ i6 (i1 )6 ]=n6

&untando tenemos:

n6=6i=1

n

i515i=1

n

i4+20i=1

n

i315i=1

n

i2+6i=1

n

ii=1

n

1

6i=1

n

i5=n6+15i=1

n

i420i=1

n

i3+15i=1

n

i26i=1

n

i+i=1

n

1

6i=1

n

i5=n6+15(n)(n+1)(6 n3+9n2+n1)

30 20

( n2 )(n+1)2

4 +15

(n ) (n+1 ) (2n+1 )6

6(n ) (n+1 )

2 +n


2/60

6i=1

n

i5=n6+(6n5+15n4+10n3n)

2 (5n4+10n3+5n2)+

(10n3+15n2+5n )2

(3n2+3n )+n

6

i=1

n

i5=2n6+6n5+15n4+10n3n10n420n310n2+10n3+15n2+5n6n26n+2n

2

6i=1

n

i5=2n6+6n5+5n4n2

2

i=1

n

i5=n2(2n4+6n3+5n21)

12

i=1

n

i5

=n2 (n+1 )2(2n2+2n1)

12 RPTA

bi=1

n

(i6) .. (resuelto porCA!A"O #!ISP$ % Jose Rolando )

SOLUCIO!:

Sabiendo que:

i=1

n

( f( i)f( i1 ))=f(n )f(0 ) 1rare&latelesc'pica

Sea : f( i)=(i+1 )7

i=1

n

i6

=[(i+1 )7

( i+11 )7

]=(n+1 )7

1

i=1

n

i5=[ ( i7+7 i6+21 i5+35 i4+35 i3+21i2+7 i+1 )( i )7 ]=(n+1 )71

7i=1

n

i6+21i=1

n

i5+35i=1

n

i4+35i=1

n

i3+21i=1

n

i2+7i=1

n

i+i=1

n

1=(n+1 )71


3/60

7i=1

n

i6+21[ 2n6+6n5+5n4n2

12 ]+35 [ 6n5+15n4+10n3n

30 ]+35 [ n4+2n3+n2

4 ]+21[ 2n3+3n2+n

6 ]+7 [n2+n2 ]

7

i=1

n

i6+

[210n6+1050n5+2100n4+2170n3+1260n2+410n

60

]=

[(n7+7n6+21n5+35n4+35n3+21 n2+7n+1)

7i=1

n

i6=[( n7+7n6+21n5+35n4+35n3+21n2+7n+1 )](1[ 210n6+1050n5+2100n4+2170 n3+1260n2+41

60

7i=1

n

i6=60n7

60 +

420n6

60 +

1260n5

60 +

2100n4

60 +

2100n3

60 +

1260n2

60 +

420n60

210n6

60

1050n5

60

2100n4

60

2170

60

i=1

n

i6=[ 60n7+210n6+210n570n3+10n

420 ]

i=1

n

i6=[ 6n7+21n6+21n57n3+n

42 ]

i=1

n

i6=

[

n (6n6+21n5+21n47n2+1)42

]i=1

n

i6=[n (n+1 )(6n5+15n4+6n36n2n+1)

42 ]

c i=1

n

(i7 ) ..(resuelto por !AMANI"ARAN)A %Obed)

'ara $allar su frmula usaremos una identidad:

i8( i1 )8=i8 (i1)8

'ara el primer miem#ro usaremos la primera reglatelescpica:

i=1

n

[ f(i ) f( i1 )]= f(n ) f(0 )

f(n )=n8 f(0 )=0


4/60

[ f( i ) f(i1 )]=n80

i=1

n

i=1

n

[ i8( i1 )8

]=i=1

n

[ i8( i1 )8

]

n8=i=1

n

[i8(i88 i7+28 i656 i5+70 i456 i3+28 i28 i+1) ]

n8=i=1

n

[8 i728 i6+56 i570 i4+56 i328 i2+8 i1 ]

Como $emos notado de#emos de sa#er las frmulas de

i6% i5 %i 4 % i3 % i2% i

(a conocemos las siguientes formulas:

i=1

n

[ i ]=n (n+1 )

2

i=1

n

[ i2 ]=n (n+1 ) (2n+1 )6

i=1

n [ i3 ]= [n (n+1 )]2

4

i=1

n

[ i4 ]=n (n+1 )(6n3+9n2+n1 )30

i=1

n

[ i5 ]=n2 (n+1 )2(2n2+2n+1)

12

i=1

n

[ i6 ]=n (n+1 )(6n5+15n4+6n36 n2n+1)

42

Entonces a$ora reempla)amos en nuestro e*ercicio:

n8=i=1

n

[8 i728 i6+56 i570 i4+56 i328 i2+8 i1 ]

n8=8i=1

n

i728i=1

n

i6+56i=1

n

i570i=1

n

i 4+56i=1

n

i328i=1

n

i2+8i=1

n

ii=1

n

1

n8=8i=1

n

i728[n (n+1)(6n5+15n4+6n36n2n+1)

42 ]+56[ n2 (n+1 )2(2n2+2n+1)

12 ]70 [n (n+1 )(6n3+30


5/60

n8=8i=1

n

i72 [n (n+1 )(6n5+15n4+6n36n2n+1) ]

3 +

14n2 (n+1 )2(2n2+2n+1)3

7n (n+1 )(6n3+9n2+

3

n8=8i=1

n

i7[12n7+42n6+42n514 n3+2n ]

3 +

(28n6+84n5+98n4+56n3+14 n2)3

42n5+105n4+70n3

3

n8+12n7+14n68n42n3+n2

3=8

i=1

n

i7

3n8+12n7+14 n68n42n3+n2

24 =

i=1

n

i7

i=1

n

i7=n2 (3n6+12n5+14n48n22n+1)

24

d i=1

n

(i8 )*(Resuelto por RO+RI)O R$)INAL+O % Cristhian A *)

SOLUCI!:

Usando identidad:i9( i1 )9=i9(i1 )9

Aplicando sumatorias:

[i9( i1 )9 ]=

i=1

n

[i9( i1 )9 ]

i=1

n

Aplicando la +eraRegla Telescpica en el +, miem#ro:

i=1

n

[ f(i ) f( i1 )]= f(n ) f(0 )

Sea: f( i )=i9 f(n )=n9; f(0 )=0

Rempla)ando:

f(n )f(0 )=i=1

n

[i9(i99 i8+36 i784 i6+126 i5126 i4+84 i336 i2+9i1)]

n9=i=1

n

[9 i836 i7+84 i6126 i5+126i484 i3+36 i29 i+1 ]

n9=9i=1

n

(i8 )36i=1

n

(i7 )+84i=1

n

(i6 )126i=1

n

( i5 )+126i=1

n

( i4 )84i=1

n

(i3 )+36i=1

n

(i2 )9i=1

n

( i )+n


6/60

n9n=9i=1

n

(i8 )36n2 (3n6+12n5+14 n47n2+2 )2

24 +

84 n (6n6+21n5+21n47n2+1 )2

42

126n2 (n+1 )2(12

n9n=9i=1

n

(i8 )32

(3n8+12n7+14 n67n4+2n2 )+2 (6n7+21n6+21n57n3+n )212 (2n6+6n5+5n4

n9n=9i=1

n

(i8 )(9n8+36n7+42n621n4+6n2 )

2 + (12n7+42n6+42n514n3+2n)(42n

6+126n5+102

n9n=9i=1

n

(i8 )+(45n8180n7210n6+105n430n2 )

10 +

(120n7+420n6+420n5140n3+20n )10

+(

n9n=9i=1

n

(i8 )+(45n860n7+42n520n37 n )

10

10n310n10 =9i=1

n

(i8

)(45n8+60n742n5+20n3+7n )

10

10 n910n+45n8+60n742n5+20n3+7n10

=9i=1

n

(i8 )

10n9+45n8+60n742n5+20n33n90

=i=1

n

(i8 )

i=1

n

(i8 )=n(10n8+45n7+60n642n4+20n23)

90

fi=1

n

(i10)..(resuelto por ,$N+$-! LOP$- %Luisa)

SOLUCI!:

'or la primera regla telescpica:

i=1

n

( f( i)f( i1))=f(n)f(0)

Sea: f( i )=(i+1 )11

i=1

n

((i+1 )11i11 )=(n+1)111


7/60

i=1

n

(i11+11i10+55i9+165 i8+330i7+462i6+462 i5+330i4+165 i3+55 i2+11i1+1i0i11)=n11+11n10+55

i=1

n

(11i10+55i9+165 i8+330i7+462i6+462 i5+330i4+165 i3+55 i2+11i1+1)=n11+11n10+55n9+165n8

11i=1

n

i10+55i=1

n

i9+165i=1

n

i8+330i=1

n

i7+462i=1

n

i6+462i=1

n

i5+330i=1

n

i4+165i=1

n

i3+55i=1

n

i2+11i=1

n

i1+i

11i=1

n

i10+55(2n10+10n9+15n812n64 n5+10n43n2

20 )+165(10n9+45n8+40n730n652n5+20

90

11i=1

n

i10

=n11

+11n10

+55n9

+165n8

+330n7

+462n6

+462n5

+330n4

+165n3

+55n2

+11nn

112

(n2+n )556

(2n3+3n2+n )1654

(n4+2n3+n2 )11(6n5+15n4+10n3n )772

(2n6+6n5+5n4n2)

11i=1

n

i10=n11+n10(112)+n9(552 553)+n8(1654 1652 1654)+n7(2203 16566)+n6(33+55

11i=1n

i10=n11112

n102756

n9165n89133

n78252

n6+ 1213

n58912

n41543

n31654

n2556

n

i10=12n1166n10550n81980n83652n74950n6+484n5+5346n4616n3495n2110n

132

i=1

n

hi=.

n

i2..(resuelto por R$"$S #!ISP$% In/s 0i1ena)

SOLUCIO!:

Aplicandola12re&la telesc'pica &enerali3ada :


8/60

( ) ( ) ( ) ( )1 1n

i k

f i f i f n f k=

=

( ) ( )3

: 1donde f i i= +

( ) ( )3 33 3

1 1

n

i k

i i n k =

+ = +

3 2 3 3 2 33 3 1 3 3 1

n

i k

i i i i n n n k =

+ + + = + + +

2 3 2 33 3 1 3 3 1

n

i k

i i n n n k =

+ + = + + +

2 3 2 33 3 1 3 3 1

n n n

i k i k i k

i i n n n k = = =

+ + = + + +

( )2 2

2 3 2 33 3 1 3 3 1

2

n

i k

n k n k i n k n n n k

=

+ ++ + + = + + +

( )2 2 2 3 2 33 3 3 33 1 3 3 12 2 2 2

n

i k

i n k n k n k n n n k =

+ + + + + = + + +

2 3 2 3 2 23 3 3 33 3 3 1 1

2 2 2 2

n

i k

i n n n k n k n k n k =

= + + + + +

3 2 3 2

2

3 1 3 1

2 2 2 2

3

n

i k

n n n k k k

i=

+ + + =

( ) ( ) 3 22

1 2 1

6 3 2 6

n

i k

n n n k k ki

=

+ += +

ii=.

n

(i3 ) *(Resuelto por RO+RI)O R$)INAL+O%Cristhian A*)

SOLUCI!


9/60

Usando identidad:i4( i1 )4=i4(i1)4


[i4 (i1 )4

]=

i=.

n

[i4( i1 )4

]i=.

n

Aplicando la +eraRegla Telescpica -eneral en el +,miem#ro:

i=.

n

[ f(i )f( i1 )]=f(n )f(.1 )

Sea: f( i )=i4

f(n )=n

4

; f( .1 )= (.1 )

4

Rempla)ando:

f(n )f(.1 )=i=.

n

[ i4(i44 i3+6 i24 i+1)]

n4(.1 )4=

i=.

n

[4 i36 i2+4 i1 ]

n4(.1 )4=4

i=.

n (i3 )6i=.

n (i2 )+4i=.

n( i)

i=.

n(1 )

n4(.1 )4=4i=.

n

(i3 )6i=.

n

(i2 )+4 [n(n+1 ).(.1 )2 ](n.+1 )

n4(.1 )4=4i=.

n

(i3 )6[n (n+1 ) (2n+1 ).(.1 ) (2.1 )6 ]+4 [ n (n+1 ).(.1 )2 ](n.+1 )

n4(.1 )4=4i=.

n

(i3 )n (n+1 ) (2n+1 )+.(.1 ) (2.1 )+2n (n+1 )2.( .1 ) (n.+1 )

n4(.44.3+6.24.+1)=4i=.

n

(i3 )2n33n2n+2.33.2+.+2n2+2n2.2+2.n+.1

n4.4+4 .36 .2+4 .1+n.+1=4i=.

n

( i3 )2n3n2+n+2.35 .2+3.


10/60

n4.4+4 .36 .2+4 .+n.+2n3+n2n2.3+5.23.=4i=.

n

(i3 )

n4+2n3+n2.4+2.3.2

4

=

i=.

n

( i3 )

i=.

n

(i3 )=n2 (n2+2n+1 ).2 (.22.+1 )

4

i=.

n

(i3 )=n2 (n+1 )2.2 (.1 )2

4

r i=1

n

(ei

) .. (resuelto porCA!A"O#!ISP$ % Jose Rolando )

SOLUCI!

Aplicandola12re&latelesc'pica:

( ) ( ) ( ) ( )1

1 0

n

i

f i f i f n f=

=

( ) 1: idonde f i e +=

1 1

1

ni i n

i

e e e e+ +

=

=

( ) 11

1

ni n

i

e e e e+

=

=

( ) 11

1

ni n

i

e e e e+

=

=

( )( )

1

1 1

nni

i

e ee

e

+

=

=

ui=1

n

(sen (i )) ..(resuelto por R$"$S #!ISP$ % In/s 0i1ena)

SOLUCI!:


11/60

!sare1oslaidentidad:sen (a ) *sen (b )=1

2[cos (ab )cos (a+b )]

Sea :a=i ;b=1

2

Aplica1ossu1atoriasa a1bost/r1inos

(sen (i) * sin( 12 ))=12i=1n

[cos(i12 )cos(i+ 12 )]i=1

n

( sen ( i))=12i=1

n

[cos( i+12 )cos(i

12 )]

sen ( 12 )i=1n

( sen ( i))=i=1

n

[cos(i+12 )cos(i12 )]2 sen( 12 )i=1

n

Resol4e1osel se&undo 1ie1brode la ecuaci'n%usando :

i=1

n

( f( i)f( i1 ))=f(n )f(0 ) 1rare&latelesc'pica

Sea:f( i)=

cos

( i+1

2

) ; f( i1

)=cos

(i1

2

)i=1

n

[cos(i+ 12 )(cos(i12 ))]=cos(n+ 12 )cos( 12 )Por lo tanto :


12/60

( sen ( i))=i=1

n

[cos(i+ 12 )cos(i12 )]2sin ( 12 )

i=1

n

( sen ( i))=cos(n+ 12 )cos( 12 )

2sin (12 )i=1

n

4 i=1

n

senh (i ) *(Resuelto por RO+RI)O R$)INAL+O %Cristhian A *)

SOLUCI!:

Sa#emos:senh ( i )=

e iei

2

Rempla)ando

i=1

n

senh ( i )=i=1

n

( eiei

2 )

i=1

n

senh ( i )=i=1

n

(e i

2

ei

2

)i=1

n

senh ( i )=12i=1

n

(ei )12i=1

n

(ei )

'rimero $allamos

i=1

n

(ei )

Usando identidad:


13/60

e iei1=eiei1


[eie i1 ]=i=1

n

[e ie i1 ]

i=1

n

Aplicando la +eraRegla Telescpica -eneral en el +,miem#ro:

i=1

n

[ f(i )f( i1 )]=f(n )f(0 )

Sea: f( i )=ei f(n )=en ; f(0 )=1

Rempla)ando:

f(n )f(0 )=i=1

n

[e ie i1 ]

en1=i=1

n

(eie ie1 )

en1=i=1

n

(e i)1ei=1

n

(e i )

en

1=e1

e i=1n

(ei

)

i=1

n

( ei )=e(en1)e1

Segundo $allamos

i=1

n

(ei )

Usando identidad:eiei+1=eiei+1


[eiei+1 ]=i=1

n

[eiei+1 ]

i=1

n

Aplicando la +eraRegla Telescpica -eneral en el +,

miem#ro:


14/60

i=1

n

[ f(i )f( i1 )]=f(n )f(0 )

Sea: f( i )=ei f(n )=en ; f(0 )=1

Rempla)ando:f(n )f(0 )=

i=1

n

[eiei+1 ]

en1=i=1

n

( eiei e )

en1=i=1

n

( ei )ei=1

n

( ei )

en1=(1e )i=1

n

(ei )

i=1

n

( ei )=en11e

'or ultimo rempla)amos en:

i=1

n

senh ( i )=e (en1)2 (e1 )

en12 (1e )

i=1

n

senh ( i )= e (en

1)+( en

1 )2 (e1 )

II. REAS USA!"O SUMATORIAS:

1. Calcular el rea de la figura limitada por y=64552

4 y el eje de las abscisas.

(Resuelto por Rodrigo Reginaldo, Cristhian A.)

SOLUCIN:


15/60

alla!do los pu!tos de

i!tersecci"!:

0=64552

4

0=256552

4

0=5 (5256 )

Donde: 5=05=256

Sabemos #ue:

6 5=ban 6 5=

256

n

Ci=a+i 6 5Ci=256 i

n

f(Ci )=64 ( 256 in )( 256 in )

2

4 f(Ci)=

47642 in

4 7642i2

n2

Luego:

8rea=l91n {i=1

n

[ f(Ci ) ] 7 6 5}8rea=l91

n {i=1n

[47642i

n

47642i2

n2 ]( 256n)}8rea=l91

n: {( 256n) [47642

n i=1

n

( i)47642

n2

i=1

n

(i2 )]}


16/60

8rea=l91n{( 256n)(4764

2

n )(n (n+1 )2 )( 256n)(47642

n2 )(n (n+1) (2n+1 )

6 )}8rea=l91

n

{(256727642 )

(

n+1

n)

(

256727642

3

)(

(n+1 ) (2n+1 )

n2

)}8rea=l91

n: {(237643 )(1+ 1n )(23 7643

3 )(1+ 1n )(2+ 1n )}8rea=23 7643+l91

n:( 2

37643

n )( 237643

3 )[(1+ l91n( 1n ))(2+l91n:( 1n ))]8rea=23 7643+0(2

37643

3 )[(1+0) (2+0 )]

8rea=23 76432( 23 7643

3 )8rea=

3723764324 7643

3

8rea=2097152

3u2

$. allar el rea compre!dida e!tre y=152 % y el eje

5 desde [3 ;3 ]

(Resuelto por Rodrigo Reginaldo, Cristhian A.)

SOLUCIN:El rea de esta funcin no se puede determinar ya ue se !a hasta el infinito en el

e"e y y en el e"e 5 % el rango dado no se interseca, por lo tanto su rea es

indeterminada


17/60

&.Calcular el rea formada por y=senh (5 ); y el eje ' desde [5 %5 ]())))*+SU+L,O -O* *++S /UIS-+% I!0s ime!a2

6 5=(ba ) /n

6 5=5(5)

n =

10

n

Ci=a+(6 5 ) i

Ci=5+(10

n) i

f(Ci )=senh (5+10 in)8rea=l 9 1

n :[i=1n

(senh(5+ 10 in))( 10n)]

8rea=l 9 1n :(

10

n)[i=1n

(senh(5+ 10 in))]Si :senh (5+y )=senh (5 )*cosh (y )+senh (y ) *cosh (5 )

8rea=l 9 1n :

(10

n)[i=1

n

(senh (5 ) *cosh

(10 i

n)+senh

(10 i

n)*cosh (5 )

)]


18/60

8rea=l 9 1n :(

10

n)[i=1n

(senh (5) *cosh ( 10 in))+i=1n

(senh( 10 in) *cosh (5))]8rea=l 9 1

n :

(

10

n)[senh (5 )

i=1

n

cosh

(

10i

n)+cosh (5 )

i=1

n

senh

(

10 i

n)]#allando:i=1

n

senh( 10 in)+ado :

cosh

(i+1

)cosh

(i1

)=cosh

(i+1

)cosh

(i1

) 1

$or la identidad.

cosh (A+, )=cosh (A )cosh ( , )+senh (A ) senh (, )2

cosh (A, )=cosh (A) cosh ( , )senh (A ) senh (, )3

Restando % y & se tiene.

cosh (A+, )cosh (A, )=2 senh (A ) senh ( , )

i=1

n

[cosh( 10 in + 10n)cosh ( 10in 10n)]=i=1n

2 senh( 10 in) *senh( 10n)

i=1

n

[cosh( 10 in + 10n)cosh ( 10in 10n)]=2 senh( 10n)i=1n

senh( 10 in)

Anali'ando el primer miemro.

f(i+1 )=cosh ( 10in + 10n)% f(i1)=cosh( 10 in 10n)

Entonces aplicamos la segunda regle telescpica.

f( n+1)=cosh(10n (n+1 ))+ f( n)=cosh (10) % f(1 )=cosh ( 10n ) % f(0)=1


19/60

i=1

n

[cosh( 10 in + 10n)cosh ( 10in 10n)]=2 senh( 10n)i=1n

senh( 10 in)

cosh

(10

n(n+1)

)+cosh (10)cosh

(10

n)1=2 senh

(10

n)i=1

n

senh

(10 i

n)

i=1

n

senh( 10 in)=cosh( 10n (n+1))+cosh (10)cosh ( 10n )1

2 senh(10

n)

#allando:

i=1n

cosh

(10 in)

senh ( i+1)senh ( i1 )=senh (i+1 )senh ( i1 ) .1

$or la identidad.

senh (A+, )=senh (A )cosh (, )+cosh (A ) senh (, ) ..2

senh (A, )=senh (A )cosh

( , )cosh

(A) senh (, ) .3


senh (A+, )senh (A, )=2cosh (A ) senh ( , )

i=1

n

senh( 10 in + 10n)senh (10 in 10n)=i=1n

2cosh( 10 in) *senh( 10n)

i=1

n

[senh(

10 in +10n)senh (

10 in 10n)]=

2 senh(10n)i=1

n

cosh

(10 in)


f(i+1 )=senh (10 in + 10n) % f(i1)=senh( 10 in 10n)

Entonces aplicamos la segunda regla telescpica.


20/60

f( n+1)=senh( 10n (n+1 ))+ f( n)=senh (10) % f(1 )=senh ( 10n ) % f(0)=0

i=1

n

[senh

(10 i

n

+10

n)senh

(10 i

n

10

n)]=2 senh

(10

n)i=1

n

cosh

(10 i

n)senh ( 10n (n+1))+senh (10 )senh( 10n)0=2 senh( 10n)i=1

n

cosh ( 10in)

i=1

n

cosh ( 10 in)=senh ( 10n (n+1))+senh (10 )senh( 10n)0

2 senh (10

n )

8rea=l 9 1n :(

10

n)[senh (5 )i=1n

cosh ( 10in)+cosh (5 )i=1n

senh(10 in)]#allando el corchete:

8rea=l 9 1n :

(

10

n)[senh (5 )

(senh ( 10n (n+1))+senh (10 )senh( 10n)

2senh (10n )

)+cosh (5 )

(cosh (10n (n+1))+cos

2 sen


21/60

3.+!co!trar el rea formada por y=sen (25 ) ; y el eje ' desde

[3 = %2 =] . ())))*+SU+L,O -O* *++S /UIS-+ % I!0s ime!a 2

6 5=(ba ) /n

6 5=2=(3=)

n =

5=n

Ci=a+(6 5 ) i

Ci=3 =+( 5 =n) i

f(Ci )=sen (2(3=+ 5 =in))

8rea=l 9 1n :[i=1

n

sen (2(3=+ 5=in))( 5=n)]8rea=l 9 1

n :(5 =n)[i=1

n

sen(2(3 =+ 5=in))]


22/60

8rea=l 9 1n :(

5 =n)[i=1

n

sen(6 =+10=i

n )]

Si :sen (5+y )=sen (5 ) *cos (y )+cos (5) *sen(y)

8rea=l 9 1n :(

5 =n)[i=1

n

sen (6=)*cos( 10 =in )+cos (6=) *sen ( 10=in )]8rea=l 9 1

n :(5 =n)[ sen (6=)i=1

n

cos( 10=in )+cos (6 =)i=1n

sen( 10 =in )]

#allando:

i=1

n

sen( 10=in )+ado:

cos ( i+1 )cos (i1 )=cos ( i+1 )cos ( i1 )1

$or la identidad.cos (A+, )=cos (A ) cos (, )sen (A ) sen ( , ) 2

cos (A, )=cos (A )cos ( , )+sen (A ) sen ( , ) 3


cos (A+, )cos (A, )=2 sen (A ) senh ( , )

i=1

n

[cosh(10 =i

n +

10=n)cosh( 10=in 10=n)]=i=1

n

2 senh( 10 =in )* senh( 10=n)

i=1

n

[cosh( 10 =in + 10=n)cosh( 10=in 10=n)]=2 senh( 10 =n)i=1n

senh( 10=in )


f(i+1 )=cosh (10=i

n +

10=

n) % f(i1)=cosh (

10 =i

n

10 =

n)


23/60

Entonces aplicamos la segunda regle telescpica.

f( n+1)=cosh(10 =n (n+1 ))+ f(n )=cosh (10 =) % f(1 )=cosh ( 10=n ) % f(0 )=1

i=1

n

cosh( 10 =in + 10=n)cosh( 10=in 10=n)=2 senh( 10 =n)i=1n

senh( 10=in )

cosh ( 10=n (n+1))+cosh (10 =)cosh ( 10=n)1=2 senh ( 10=n)i=1n

senh( 10=in )

i=1

n

senh

(

10=i

n )=

cosh( 10=n (n+1 ))cosh (10=)+cosh ( 10 =n)+12 senh (

10 =n )

#allando:

i=1

n

cosh ( 10 =in )senh ( i+1)senh ( i1 )=senh (i+1 )senh ( i1 ) .1

$or la identidad.

senh (A+, )=senh (A )cosh (, )+cosh (A ) senh (, ) ..2

senh (A, )=senh (A ) cosh ( , )cosh (A) senh (, ) .3


senh (A+, )senh (A, )=2cosh (A ) senh ( , )

i=1

n

[senh( 10 =in + 10=n)senh ( 10=in 10 =n)]=i=1n

2cosh ( 10=in ) *senh( 10 =n)

i=1

n

[senh( 10 =in + 10=n)senh ( 10=in 10 =n)]=2 senh (10 =n )i=1n

cosh ( 10=in )



24/60

f(i+1 )=( 10=in +10 =n) % f( i1 )=( 10 =in 10=n)

Entonces aplicamos la segunda regla telescpica.

f( n+1)=senh( 10 =n (n+1 ))+ f(n )=senh (10 =) % f(1 )=senh ( 10=n ) % f(0 )=0

i=1

n

[senh( 10 =in + 10=n)senh ( 10=in 10 =n)]=2 senh (10 =n )i=1n

cosh ( 10=in )

senh ( 10=n (n+1 ))+senh (10=)senh ( 10=n )=2 senh (10 =n)i=1n

cosh ( 10=in )

i=1

n

cosh ( 10 =in )=senh( 10 =n (n+1 ))+senh (10 =)senh (10 =n)

2 senh( 10 =n )

8rea=l 9 1n :(

5 =n)[ sen (6=)i=1

n

cos( 10=in )+cos (6 =)i=1n

sen( 10 =in )]#allando el corchete:

8rea=l 9 1n :

(5 =n)[ sen (6 =) * senh(10=

n (n+1 ))+senh (10 =)senh( 10=n)

2 senh( 10=n) +cos (6 =) *

cosh ( 10=n (n


25/60

4. Calcular el rea compre!dida e!tre y=538 5 y eje desde [4,5 ]

(Resuelto por Chahuayo uispe, *os+ R.)

-/C012:

De: 0=538

0=(5+2 )(5245+4)

5=2


26/60

#allando A1 [4 ;2 ]

6 5=ba

n

=2

n

ci=a+i 6 5=4+2 i

n

f(ci)=

[(4+

2i

n)

3

+8

]A

1=l 9 1n :

i=1

n

f(c i) 6 5

A1=l 9 1n:

i=1

n

{[(4+ 2in)3

+8]2n }

A1=l 9 1n :

i=1

n

{[(2 i4 n

n )3

+8]2n}

A1=l91

n :i=1

n

[(8 i364n348n i2+96n2 i+8n3

n3 )2

n ]A1=l91

n :i=1

n

(16i3112n396n i2+192n2i

n4 )


27/60

A1=l91n :

i=1

n

(16i3

n4

112

n 96

i2

n3+192

i

n2 )

A1=

[l91n :

16

n4 i=1n

i

3

l91n :112

n i=1n

1 l91n :96

n3 i=1n

i

2

+ l91n :192

n2i=1n

i

]A1=[16l91n : n

2 (n+1 )2

(4 ) n4 112 l91

n:

nn96l91

n :

n (n+1 ) (2n+1 )

(6 )n3 192 l91

n :

n (n+1 )

(2 )n2 ]

A1=[4 l91n: (n2+2n+1)

n2 112l91

n :116 l91

n :

(n+1 ) (2n+1)

n2 +96 l91

n :

(n+1 )n ]

A1=[4 l91n:(1+ 2n + 1n2 )112 l91n :116l91n :(1+ 1n )(2+ 1n )+96 l91n :(1+ 1n )]A1=[4 (1 )112 (1 )16 (2 )+96 ]

A1=[44 ]

A1=44u2

#allando A2 [2 ;0 ]

6 5=ba

n =

2

n ci=a+i 6 5=2+

2 in f(ci)=[(2+ 2in)

3

+8]

A2= l91n :i=1

n

f( c i) 6 5

A2= l91n :

i=1

n

{[(2+ 2 in)3

+8]2n}

A2= l91n :

i=1

n

{[( 2i2nn )3

+8 ]2n }


28/60

A2=l91

n :i=1

n

[(8 i38n324n i2+24 n2i+8n3

n3 )2

n ]

A2= l91n :i=1

n

(16i348n i2+48n2i

n4 )A2= l91

n :i=1

n

(16i3

n4

48 i2

n3 +

48 i

n2 )

A2=[ l91n : 16n4 i=1n

i3 l91n :

48

n3i=1

n

i2+ l91n :

48

n2i=1

n

i ]

A2=[16l91n : n2 (n+1 )2

(4 ) n4 48l91

n :

n (n+1 ) (2n+1 )

(6 )n3 +48 l91

n :

n (n+1 )

(2 )n2 ]A2=[4 l91n :(n

2+2n+1n2 )8l91n :

(n+1 ) (2n+1 )

n2 +24l91

n :

(n+1)n ]

A2=

[4 l91

n :(1+2

n+1

n2 )8 l91n :(1+1

n )(2+1

n )+24l91n :(1+1

n )

]A2=[4 (1 )8 (2 )+24 (1 )]A

2=[12 ]

A2=12u2

#allando A3 [0 ;5 ]

6 5=ba

n =

5

n ci=a+i 6 5=0+

5 in f(c i)=[( 5in)

3

+8 ]

A3= l91

n :i=1

n

f( c i) 6 5


29/60

A3= l91n :

i=1

n

{[( 5in)3

+8]5n }

A3= l91n :i=1n

[(125 i3+8n3

n3 )5

n

]A

3= l91n :

i=1

n

(625i3+40n3

n4 )

A3=[ l91n : 625n4i=1n

i3+ l91n :

40

n

i=1

n

1]

A3=[625l91n : n2 (n+1 )2

(4 ) n4 +40 l91

n:

nn ]

A3=[6254 l91n :(n

2+2n+1n2 )+40 l91n :

nn ]

A3=

[

625

4 l91

n :

(1+

2

n+1

n

2

)+40 l91

n:1

]A3=[6254 (1 )+40 (1 )]

A3=[7854]

A3=785

4

u2

Entonces el rea total es:

AT=A1+A2+A3

AT=44+12+785

4

AT=1009

4 u2


30/60

6. Calcular el rea compre!dida e!tre y=52+16 ; y=52+9*

(Resuelto por Chahuayo uispe, *os+ R.)

SOLUCIN:

#allando los puntos de interseccin en 0=25225 :

0=(255 ) (25+5 )

5= 5

2; 5=

5

2

#allando A1

f(5 )=252+25 [0 ; 52 ]

6 5=ba

n

= 5

2n

c i=a+i 6 5= 5i

2n

f(ci)=

(

5 i

2n )

2

+25


31/60

A1= l91

n :i=1

n

f( c i) 6 5

A1

= l91n :i=1

n

{[

( 5 i

2n )

2

+25

] 5 i

2n }A1= l91

n :i=1

n

{[25 i2

2n2 +25] 5 i2n }

A1= l91

n :i=1

n

[125 i2+125n2

22n2 ]

A1=125

22l91n :i=1

n

(n2i2

n2 )

A1=125

22 [ l91n:i=1n

n2

n2 l91

n:i=1

ni2

n2 ]

A1=125

22 [ l91n:i=1n

1 l91n :

i=1

n n ( n+1 ) (2n+1 )

(6 ) n2 ]A1=

125

22 [11

3 ]

A1=125

32u2

Entonces el rea de la figura es:1=A2

A

AT=A1+A2

AT=250

3 2u2

III. AREAS USA!"O I!TE-RALES SIM'LES:+/ Todo 0II/


32/60

1.1 Calcular el rea de la fgura limitada por

y=64552

4 y el eje de las abscisas.

SOLUCI!:

64552

4=0

256552

4 =0

5 (2565)4

=0

5=2565=0

-r12ca:

Integracin Simple:

(64552

4)

0

256

640

256

5140

256

52

6452

2|0

256

512

3|0

256

32(2562 )(256

3

12 )


33/60

12 (2097152 )1677721612

=2097152

3u2

1.2 Hallar el rea comprendida entre y=1

52 , y el eje 5

desde [3 ;3 ] . (Resuelto por C$a$ua3o 4uispe5 &os6

RolandoSOLUCI!:

'ara: y=1

52; 52R

!o cual indica "ue 5=0 en una as#ntota entonces no e$iste

intersecci%n con el eje entonces el dominio est dado por[3 ;0>


34/60

A1=

5

0

sinh5

A1=[ cosh5 ]50

A1=cosh (0 )cosh (5)A1=cosh (0 )cosh (5)

coshu=eu+eu

2

A1=e0+e0

2

e5+e5

2

A1=1

e5(e10+1)2

A1=1(e10+1)

2e5

Reempla)ando el 7alor de: e=2.718

A1=1(2.71810+1)

2 (2.718 )5

A1=73.171u2

'ero como el 1rea no puede ser negati7o:A1=73.171u

2

A2=0

5

sinh5

A1=[ cosh5 ]05

A1=cosh (5 )cosh (0)

coshu= eu+eu

2

A1=e5+e5

2

e0+e0

2

A1=e5+e5

2 1

A1=

e5(e10+1)2

1


35/60

A1=(e10+1)

2e5 1

Reempla)ando el 7alor de: e=2.718

A1=(2.71810+1)

2 (2.718)5 1

A1=73.171u2

AT=A1+A2

AT=73.171+73.171

AT=146.342u2

1.) &ncontrar el rea 'ormada por y=sin (25 ) , y el eje 5

desde [3 =;2 =]

SOLUCI!:

sin(25)d5

4=25

d4=2d5

sin (25 )=12 sin (25 )(2d5)

sin (25 )=1

2 sin (4 ) d4


36/60

sin (25 )=12cos (4 )+c

sin (25 )=12cos (25 )+c

A1=

3=

5=/2

sin(25)

A1=[12 cos(25 )]3=

5=/2

A1=1

2

cos2

(5 =

2

)+1

2

cos2 (3=)

A1=1

2cos (6 =)

1

2cos (5 =)

A1=1

2+1

2=1

A2=5=/2

2=

sin(25) A2=[

1

2 cos

(25 )]

5=/2

2=

A2=12 cos2 (2 =)+

1

2cos2 (

5=2 )

A2=1

2cos (5 =)

1

2cos (4 =)

A2=

1

2

1

2=1=1

A3=

2=

3=/2

sin(25) A3=[12 cos (25 )]2=3=/2

A3=12 cos2(3=2 )+ 12 cos2 (2=)

A3=12 cos (4 =)12 cos (3=)


37/60

A3=1

2+1

2=1

A4=3=/2

=

sin (25 )

A4=[12 cos(25)]3=/2=

A4=

12

cos2 (=)+1

2cos2 (

3=2

)

A4=

1

2

cos (3=)1

2

cos (2=)

A4=12

1

2=1=1

A5=

=

=/2

sin (25) A5=[12 cos(25 )]==/2

A5=

12 cos2

(=2

)+1

2cos2 (=)

A5=1

2cos (=)

1

2cos( =2)

A5=1

2+1

2=1

A6= =/2

0

sin (25) A6=[12 cos (25)]0

=/2

A6=12 cos2 (0 )+

1

2cos2 (

=2 )

A6=

1

2cos (=)

1

2cos (0 )

A6=12 12=1=1


38/60

A7=0

=/2

sin (25) A7=[12 cos (25)]0=/2

A7

=1

2 cos2 (0 )+

1

2cos2 (

=

2)

A7=1

2cos (0)

1

2cos ( =)

A7=12

1

2=1=1

A8

=

=/2

=

sin (25)

A8=[12 cos (25)]=/2=

A8=

12

cos2 (=)+1

2cos2 (

=2)

A8=

1

2 cos (=)

1

2 cos (2 =)

A8=12

1

2=1=1

A9= =

3=/2

sin (25)

A9=[12 cos(25)]=

3=/2

A9=12 cos2(3 =2)+ 12 cos2 (=)

A9=1

2cos (2=)

1

2cos (3 =)

A9=1

2+1

2=1


39/60

A10= 3=/2

2=

sin(25)

A10

=

[1

2 cos(25 )

]3=/2

2=

A10=12 cos2 (2=)+ 1

2cos2 (

3 =2 )

A10=

1

2cos (3 =)

1

2cos (4 =)

A10=1

2

1

2

=1=1

AT=10 (1 )

AT=10u2

1.* Hallar el rea comprendida entre y=538 , y el eje 5

desde [4 ;5 ] ...(Resuelto por Chahuayo Quispe, Jos

Rolando )SOLUCI!

"e: f(5 )=538

5=2con 5 desde [4 ;5 ]

5 [4 ;2 ] [2 ;0 ] [0;5 ]


40/60

8allando: A1

A1=4

2

(538)d5

A1=54

4| 2

485|24

A1=

(2 )4(4 )4

4 8 (2+4 )

A1=240

4 16

A1=44u2

8allando: A2

A2=2

0

(538 )d5

A2=54

4| 0285| 02A2=

(0 )4(2)4

4 8 (0+2 )

A2=164 16

A2=12

A2=12u2

8allando: A3

A3=0

2

(538 )d5

A3=54

4| 5085|50A3=

(5 )4(0 )4

4 8 (50 )

A3=625

4 40

A3=785

4


41/60

A3=785

4 u2

Entonces el 1rea total es:

AT=A1+A2+A3 AT=44+12+785

4

AT=1009

4 u2

1.+ Hallar el rea comprendida entre y=arctan (5 ) , y el eje 5 desde

[5 ;5 ] ..(Resuelto por Rodrigo Reginaldo, Cristhian A.)

SOLUCIO!:

reaT=A1+A2

8allando A1

A1=5

0

arctan (5 ) * d5

'or partes:u=arctan (5 )d5=d4

du=dd5

arctan (5 )d5= d4

du= 1

1+52d5 5=4

arctan (5 ) * d5=5arctan (5 )5

(

1

1+5

2

)d5


42/60

arctan (5 ) * d5=5arctan (5 )( 51+52 )d5

'or sustitucin:

u1=1+52

du1=25d5

d5=du125

arctan (5 ) * d5=5arctan (5 )

(5

u1)

du1

25

arctan (5 ) * d5=5arctan (5 )12(1u1)du1

arctan (5 ) * d5=5arctan (5 )12ln (u1 )

arctan

(5 ) * d5=5arctan (5 )

1

2ln

(1+5

2

)

"nde :

5

0

arctan (5 ) * d5=[5arctan (5 )12ln (1+52 )]50

A1=[5arctan (5 )12 ln (1+52 )]50

u2

A1=[0+5arctan

(5

)+

1

2ln

(1+ (

5)

2

) ]5

0

u

2

A1=[06.867+1.629] u2

A1=1062

125u2

9inalmente:

8reaT=2A1


43/60

8reaT=2124

125u2

1.. Hallar el rea comprendida entre y=52

9 ,y=52+16 , y el eje 0 desde [3,6 ]

(. -esuelto por H/0/ 3/-/4/ 5bedSOLUCI!

'ara el gr12co:

6gualamos ambas 'unciones para saber cules son lospuntos "ue tienen en com7n. &ntonces:

529=52+16

25225=0

5=>3,535

&l primer punto en com7n ser: (3,535 ;3,496 )

&l segundo punto en com7n ser: (3,535 ;3,496 )

'ara $allar el 1rea integraremos restando las

funciones:

3,535

3,535

[ (52+16)(529)]d5

3,535

3,535

(252+25) d5

[253

3+255 ]

3,535

3,535

u2

[58,92(58,92 )]u2


44/60

117,84u2

1.18 Hallar el rea comprendida entre y=59,y=e5

, y el

eje 5 desde [3 ;6 ] .(Resuelto por Chahuayo Quispe,

Jos Rolando )SOLUCI!

8allando: A1

A1=3

0

(e5 )d5

A1=e5| 0

3

A1=(e0e3 )

A1=1+1

e3

A1=11

e3u2

8allando: A2

A2=0

6

(e55+9 )d5

A2=e5| 6

053 /2|6

0+95|6

0

A2=(e6e0 )(63 /203 /2 )+9 (60 )

A2=1

e6+166+54

A2=1

e6+166+54u2

&ntonces el rea total es: AT=A1+A2

AT=11

e3

1

e6+166+54u2

AT=(56 2e666)u2

2 Hallar el rea de la regi%n 'ormada por

y2=165254 ;2552+16y2=400


45/60

SOLUCI!:

8allando los puntos de: y2=165254

y=165254

y=5 1652

5=01652?0

5=0(54 ) (5+4 )?0

5=0 (54 ) (5+4 )@0

5=05 [4 ; 4 ]

8allando los puntos de: 2552+16y2=400

y=

400255

2

16

4002552

16?0

(5520 ) (55+20 )?0

(5520 ) (55+20 ) @0

5[45 ;45] 8allando los puntos de interseccin:

y2=165254 ;2552+16y2=400

2552+16 (165254 )=400

281521654=400

165428152+400=0

(5216) (165225 )=0(45+5 ) (54 ) (455 ) (5+4 )=0

5=45=45=54 5=

5

4


46/60

8allando: A1

f(5 )=165254[0 ;54 ]

6 5=ban = 5

4 n ci=a+i 6 5= 5 i

4nf(c i)=5 in

16(5 in)

2

A1= l91

n :i=1

n

f( c i) 6 5

A1= l91n :

i=1

n {[ 5 in16( 5 i4n )2] 54 n }

A1= l91n :

25

4

i=1

n

[ in2 16( 5 i4n )2]

8allando: A2

f(5 )=51652[ 54;4 ]

6 5=

ba

n =

11

4 n

ci=a+i 6 5=

5

4 +

11i4 n

f(c i)=5

16(

5

4 +

11i4n)

2


47/60

A2= l91

n :i=1

n

f( c i) 6 5

A2= l91n :

i=

1

n

{[5 in

16( 54 + 11

i4n)

2

]11

4 n

}A2= l91n :

55

4

i=1

n

[ in2 16( 54 + 11i4n)2]

"onde el 1rea total es:AT=4 (A1+A2 )

AT=4 ( l91n : 254i=1n [ in2 16( 5 i4n )

2]+ l91n: 554i=1n [ in2 16( 54 + 11i4n)

2])AT=25 l91

n:i=1

n

[ in2 16( 5 i4 n )2

]+55l91n :i=1n

[ in2 16(54 +11i4 n )2

]9 Hallar el rea de la regi%n 'ormada por y

2=542552 ; 52+y2=36

(resuelto por &D&; !5


48/60

8allando los puntos de interseccin5 igualamos las dosecuaciones:

y2=542552 ; 52+y2=36

542552=3652

54245236=0

5=b >b

24ac2a

5=(24 )>(24)

24(1)(36)

2(1)

5=24>125

2

51=12+65 ; 52=1265

Entonces:

5 52

25[3652]d5

A1=

5

6

5 52253652

[ ]d55

6

A1=5

6

Resol7emos las integrales por sustitucintrigonom6trica:

55225

[]d5

5

6


49/60

55=sec

5=5 sec

d5=5 sec*tand

Reempla)ando:

5sec 25 sec225[(5 sectand)]

5

6

sec[sec25( 21) (sectand )]

255

6

sectan2

[ (sectand )]

25(5)5

6

sectan[ (sectand )]

25(5)5

6

25(5)5

6

sec2 tan2d

25(5)5

6

sec2 (sec 21)d

sec 4d25(5)5

6

sec2d

25(5)5

6


50/60

25 (5 )5

6

sec2sec2d25(5)5

6

sec2d

25

(5

)5

6

sec2 (tan2 +1)d25(5

)tan

25 (5 )5

6

sec2 tan2d+25 (5 )5

6

sec2 d25 (5 )tan

25 (5 )5

6

sec2 tan2d+25 (5 ) t&25 (5 ) tan

25 (5 )5

6

tan2d (t&)

[125tan3 ]56

Rempla)ando la tan3 :

[125

3

((5225)

3

53

)]5

6

(6225)3

3

(25225 )

3

3

11113

3652

[]d5

5

6

cos=3652

6

sin=56

5=6 sin


51/60

=acrsin(56 )

d5=6 cosd

Rempla)ando:

3652

[ ]d5=5

6

6cos .6cos d

5

6

5

6

36cos2 d

5

6

36cos2 d

cos2=

1+cos22

5

6

36.1+cos2

2 d

185

6

cos2 d

18[5

6

d+5

6

cos2 d ]18[+ 12 sin2 ]5

6

sin 2=2 sincos

18[+ 12 .2 sincos]56


52/60

18 [+sincos ]56

Rempla)ando los 7alores 3a o#tenidos:

18[arcsin (56 )+56*

36526 ]5

6

[18arcsin(56 )+5

36522 ]5

6

18arcsin( 66 )+ 63662

2 [18arcsin( 56 )+ 5365

2

2 ]1620 [1024.26 ]=595.74

A1=11113

+595.74

A1=607.9

AT=2(607.9)u2

Hallar el rea de la regi%n 'ormada por y= 16

52+1 , y el eje0

desde [6,10 ] (.resuelto por H/0/6 3/-/4/ 5bedSOLUCI!:

'ara el gr12co:

!a 7nica restricci%n de la 'unci%n ser#a:52+1B0

52B1

3 pues siendo una 'unci%n cuadrada nunca llegar#a a ser

negati=a


53/60

'ara $allar el 1rea integraremos la funcin:

6

0

( 1652+1 )d5+010

( 1652+1 )d5&mpecemos por:

6

0

( 1652+1 )d5trica:

?eniendo de la 'orma : u2+a2 , usaremos: u=a tan

&n ese caso tenemos:5= tan

d5=( sec )2d

cos= 1

52+1

(cos )2= 1

52+1

&ntonces aplicamos:

166

10

( 1

52+1 )d516 (cos )2 (sec )2 d

16 d

16 ( )=16arctan (5 )

16 [arctan (5 )]60

16 [arctan (0 )arctan (6 )]


54/60

16[0(80,53o 7 =180o )]16

[0

(80,53o 7

=

180o

)]16 [1,405 ] u26

10

( 1652+1 )d5=22,48u2!uego @allamos:

0

10

( 1652+1 )d5Como ya sabemos el resultado de la integral:

16

52+1d5=16 arctan (5 )

&ntonces aplicamos a la integral:

0

10

( 1652+1 )d516 [arctan (5 )]0

10

16 [arctan (10)arctan (0 )] u2

16 [84,289o0 ]u2

16[84,289o 7 =180o ]u216 [1,471 ] u2

0

10

( 1652+1 )d5=23,536 u2

Luego sumamos am#as integrales:

6

0

( 1652+1 )d5+010

( 1652+1 )d5

(22,48+23,536 ) u2

46,016u2


55/60

) Hallar el rea de la regi%n 'ormada por

y=|5265+10|; 5=10 ; 5=16 ; y=0 .(Resuelto por Ramos

Palomino, Jhordy Maycol)SOLUCI!:

|5265+10|={5265+10 ; si 5265+10?0

52+6510 ;s i5265+10


56/60

A1=2=

=

sen (5 ) * d5

A1=[cos (5 )]2==

u2

A1=[cos (=)cos (2=)]u2

A1=[11 ] u2

A1=2u2

7i!alme!te:8reaT=4A1

8reaT=8u2

3) #allar el rea de la regin formada por y=cos (5 ) ; 5=3= % 5=5= % y=0

SOLUCIN: I!tegraci"! Simple:

169=2

5=

( cos5 )

16sen 5|9=2

5=

16 [sen (5 =)sen(9 =2 )]

16 [0(1)]=16u2

44)#allar el rea de la regin formada por 52

3+y2

327=0

0;;;..


57/60

A1=0

a

(a2

352

3)3

2

d5

acemos la sustituci"! trigo!om0trica:

5=a t3

d5=3at2 dt

Los lmites de i!tegraci"! so!:

5=0 % t=0

5=0 % t=1

+!to!ces se tie!e:

0

1

(a2

352

3)3

2

(3at2 dt)=3a20

1

(1t2)3

2 dt

acemos u!a 8e; ms la sustituci"! trigo!om0trica:

t=sin

dt=cos

Sus lmites de i!tegraci"! ser!:

t=0 % =u

t=1 %==2

Se tie!e:

(1sin2 )3

2 sincosd=3a20

=2

cos3 sin2cosd

3a20

=2


58/60

(1sin2 )3

2 sincosd=3a20

=2

cos4 sin2 d

3a2

0

=2

(1sin2 )3

2 sincosd=3a2

8

0

=2

(1cos2)2 (1+cos2 ) d

3a20

=2

(1sin2 )32 sincosd=

3a2

8

0

=

2

(1cos2cos22+cos3 ) d

3a20

=2

d0

=2

cos21

20

=2

(1+cos2 )+0

=2

cos2(1sin22)d

0

=

2

3a20

=2

(1sin2 )3

2 sincosd=3a2

8

d1

20

=2

cos2d (2)1

20

=2

d1

40

=2

cos2d (2 )+1

20

=2

cos2d (2)1

20

=2

sin2d (sen2)

0

=2

3a20

=2

(1sin2 )3

2 sincosd=3a2

8

3a20

=2

(1sin2 )3

2 sincosd=3a2

8[12 sen2+ 12 14 sen 2+ 12 sen213 sin3 ]0=/2


59/60

3a20

=2

(1sin2 )3

2 sincosd=3a2 =32

A1=3

a

2

=32

AT=4 (3a2 =32

)

AT=3 a2 =

8

*eempa;a!do el 8alor de a:

52

3+y2

3=a2

3

+l 8alor apro'imado de ! la grfica es de 1&?:

52

3+y2

3=1402

3

AT=

3 (140)2 =

8

AT=7350= u2

4%.Calcular el rea descrita por (524 )2 (y4 )2=100

()). resuelto por C@U@O /UIS-+% ABose2

SOLUCIN

9espeja!do < y =% te!emos:

(524 )2 (y4 )2=100

(y4 )2=(524 )2100

y4=>(524)2

100

y=>(524 )2100+4


60/60

alla!do el domi!io:

(524 )2100?0

(52

4 )2

?100

524?10 ; 524 ?10

52?14 ; 52?6

5@145 ?14 ; 52 ?6

;14+> 52R

14 ;:

5

14 ;:

5

PRIMERA PRACtica de Analisis3

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