x= y2

4−

ln ( y )2

∴ dxdy

= y2− 1

2 y=1

2( y− y−1 )

L=∫1

e

√1+( dxdy )

2

dy=∫1

e

√1+ 14

( y2−2+ y−2 ) dy

L=∫1

e

√ 44+ y2−2+ y−2

4dy=∫

1

e

√ y2+2+ y−2

4dy

∫1

e √ y2+2+ 1

y2

4dy=∫

1

e √ y4+2 y2+1

y2

4dy=∫

1

e

√ y4+2 y2+14 y2 dy

L=∫1

e √( y2+12 y )

2

dy=∫1

ey2+12 y

dy=12∫1

e

ydy+12∫1

edyy

L=[ y2

4+

ln ( y )2 ] ¿1

e = e2+14

f ( x )=1x+

f ' ( x )1!

+f ' ' ( x )2!

+f ' ' ' ( x )

3 !+

f ' ' ' ' ( x )4 !

+…

f n ( x )=−1

x2+ 2

x3− 6

x 4+ 24

x5−120

x6+…

f n ( x )= (−1 ) 1!

x2+(−1 )2 2!

x3+(−1 )3 3 !

x4+(−1 ) 4 4 !

x5+ (−1 )5 5 !

x6+…

f (x)=1x+

(−1 )x2 +

(−1 )2

x3 +(−1 )3

x4 +(−1 )4

x5 +(−1 )5

x5 +…

∑n=1

+∞ (−1 )n+1

xn =1x+

(−1 )x2 ( x−2 )+ (−1 )2

x3 ( x−2 )2+ (−1 )3

x4 ( x−2 )3+…

∑n=0

+∞ (−1 )n

2n+1 ( x−2 )n=12+

(−1 )22 ( x−2 )+ (−1 )2

23 ( x−2 )2+ (−1 )3

24 ( x−2 )3+…

2x3 =∑

n=2

+∞ (−1 )n

2n+1 n (n−1 ) ( x−2 )n−2; evaluando x=3

227

=∑n=2

+∞ (−1 )n

2n+1 n (n−1 )

∑n=1

+∞ (−1 )n−1

(2 n−1 ) (2n )= 1

1∗2− 1

3∗4+ 1

5∗6− 1

7∗8+…

L=2+∫π3

π2

√[2−2cos (θ ) ]2+ [2 sen (θ ) ]2 dθ+∫π3

π2

√ [2 cos (θ ) ]2+[−2 sen (θ ) ]2dθ

L=2+∫π3

π2

√4−8cos (θ )+4cos2 (θ )+4 sen2 (θ ) dθ+∫

π3

π2

√4cos2 (θ )+4 sen2 (θ ) dθ

L=2+∫π3

π2

√8−8 cos (θ )dθ+∫π3

π2

√4dθ=2+2√2∫π3

π2

√1−cos (θ ) dθ+2∫π3

π2

dθ

L=2+2√2∫π3

π2

√2 sen2( θ2 )dθ+2∫

π3

π2

dθ=2+4∫π3

π2

|sen ( θ2 )|dθ+2∫

π3

π2

dθ

L=2+4∫π3

π2

sen( θ2 )dθ+2∫

π3

π2

dθ=2+[−8cos( θ2 )+2θ ] ¿π

3

π2

L=2+[−4 √2+4√3+ π3 ]=2+4√3−4√2+ π

3

1−cos (θ )2

=sen2(θ2 )→ 1−cos (θ )=2 sen2( θ

2 )

f ( x )=ln ( x )+ f ' ( x )+ f ' ' ( x )+ f ' ' ' ( x )+ f ' ' ' ' (x )+…

f n ( x )=1x− 1

x2+ 2

x3− 6

x4+ 24

x5−120

x6+…

f n ( x )=0 !x

+(−1 ) 1 !

x2+(−1 )2 2 !

x3+(−1 )3 3 !

x4+(−1 )4 4 !

x5+(−1 )5 5 !

x5+…

f ( x )=ln ( x )+ 0 !x

+(−1 )1 !

x2 +(−1 )22 !

x3 +(−1 )3 3 !

x4 +(−1 ) 4 4 !

x5 +…

ln ( x )+∑n=0

+∞ (−1 )n

xn+1 n!=1x+

(−1 )x2 ( x−1 )+ (−1 )2

x3 (x−1 )2+ (−1 )3

x4 ( x−1 )3+…

∑n=0

+∞ (−1 )n

2n+1 ( x−2 )n=12+

(−1 )22 ( x−2 )+ (−1 )2

23 ( x−2 )2+ (−1 )3

24 ( x−2 )3+…

∑n=2

+∞

an ( x−3 )n ;converge en x=−1.1 , entonces convege en x=7

limn → ∞|an+1 ( x−3 )n+1

an ( x−3 )n |<1=limn → ∞|an+1 ( x−3 )

an|<1=|x−3|< an

an+1

Si tomamos an=1

2n→2−3<|x−3|<2+3 →−1<|x−3|<5

−1<|−1.1−3|<5 → 1<|−4.1|<5Verdadero

−1<|7−3|<5→−1<|4|<5 Verdadero

La serie de Taylor es:

f ( x )=f (a )+ f ' (a )1!

( x−a )+ f ' ' (a )2!

( x−a )2+ f ' ' ' (a )3 !

(x−a )3+…

La serie de Maclaurin es:

f ( x )=cos ( x )

f ( x )=f (0 )+ f ' (0 )1 !

x+f ' ' (0 )

2 !x2+

f ' ' ' (0 )3 !

x3+…

f n ( x )=cos ( x )− sen ( x )1 !

x−cos ( x )

2!x2+

sen ( x )3 !

x3+cos ( x )

4 !x4−

sen ( x )5!

x5−…

f ( x )=cos (0 )− sen (0 )1!

x−cos (0 )

2!x2+

sen (0 )3 !

x3+cos (0 )

4 !x4−

sen (0 )5 !

x5−…

cos (x )=1−0− 12 !

x2+0+ 14 !

x4−0− 16 !

x6+0+ 18 !

x8−0− 110!

x10+…

cos (x )−1=−12 !

x2+ 14 !

x4− 16 !

x6+ 18!

x8− 110 !

x10+…

1−cos ( x )= 12!

x2− 14 !

x4+ 16 !

x6− 18 !

x8+ 110 !

x10− 112!

x12+…

1−cos ( x )x2

= 12!

− 14 !

x2+ 16 !

x4− 18 !

x6+ 110!

x8− 112!

x10+…

1−cos ( x )x2 =∑

n=0

+∞ (−1 )n x2 n

(2n+2 ) !

h ( x )=∫0

x1−cos (t )

t2 dt=∫0

x

(∑n=0

+∞ (−1 )nt 2n

(2n+2 ) ! )dt

h ( x )=∫0

x

( 12 !

− 14 !

t 2+ 16 !

t 4− 18 !

t6+ 110 !

t 8− 112!

t 10+…)dt

h ( x )=( 12 !

t− 13 ∙ 4 !

t 3+ 15 ∙6 !

t 5− 17 ∙8 !

t7+ 19 ∙ 10!

t 9− 111 ∙12 !

t 11+…) ¿0x

h ( x )= 12!

x− 13 ∙ 4 !

x3+ 15 ∙6 !

x5− 17 ∙ 8 !

x7+ 19 ∙ 10!

x9− 111 ∙12 !

x11+…

h ( x )=∑n=0

+∞ (−1 )n x2 n+1

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

limn → ∞| (−1 )n+1

x2n+3

(2n+3 ) (2n+4 ) !(−1 )n x2 n+1

(2n+1 ) (2n+2 )!|=lim

n → ∞| (−1 )n+1x2 n+3 (2n+1 ) (2n+2 )!

(−1 )n x2n+1 (2n+3 ) (2n+4 ) (2n+3 ) (2n+2 ) !|

limn→ ∞| (−1 ) x2 (2n+1 )

(2n+3 ) (2n+4 ) (2n+3 )|<1=¿|x2|limn→ ∞

(2n+1 )(2n+3 ) (2n+4 ) (2n+3 )

<1¿

Converge para todos los reales (−∞ ,+∞)

f ( x )=ex

f ( x )=1+ 11 !

( x )+ 12!

x2+ 13 !

x3+ 14 !

x4+…

f ( x )=∑n=0

+∞1

n !xn≡ 1∑

n=0

+∞xn

n!;Si tomamos x=1

V C=V R0 (1+8∗10−6℃−1 (20℃ ) )

V R=100 cm 3 (1.00016 )=100.016 cm 3

Volumen inicial del recipienteV R0=1 cm 2 S

Volumen inicial del liquidoV L0=100 cm 3

V L=V L 0 (1+BL∗∆ T ) → V L=100 cm 3∗(1+12∗10−4℃−1 (20℃ ))

V L=100 cm3∗(1.024 )=102.4 cm 3

∆ V =|102.4 cm 3−100.016 cm3|=2.384 cm 3 respecto a lamarca

mcu∗C cu ∆ T=−mAl∗C Al∆ T →mcu∗C cu ( T E−0 )=−mAl∗CAl (T E−100℃ )

mcu∗C cuT E=mAl∗C Al (100℃−T E ) → mAl=mcu∗Ccu T E

C Al (100℃−T E )

Dcu=D0 cu (1+α cu T E ) DAl=D0 Al (1+α Al (T E−100℃ )) D cu=DAl

D0 Al (1+α Al (T E−100℃ ))=D0cu (1+αcu T E )

D0 Al+D0 Al α Al T E−D0 Al α Al100℃=D 0 cu+D0 cu α cu T E

D0 Al α Al T E−D 0 cu αcu T E=D0 Al α Al 100℃+D0 cu−D0 Al

T E ( D0 Al α Al−D0 cu α cu )=D 0Al α Al100℃+D0 cu−D0 Al

T E=D 0 Al α Al100℃+D0cu−D0 Al

D0 Al α Al−D0cu αcu

=2.54533∗α Al∗100℃−0.00533

2.54533∗α Al−2.54000¿αcu

T E=2.54533∗23∗10−6∗100℃−0.00533

2.54533∗23∗10−6−2.54000∗17∗10−6 =5.854259∗10−3−5.33∗10−3

5.854259∗10−5−4.318∗10−5

T E=5.24259∗10−4

1.536259∗10−5 =34.13℃

mAl=21.6g∗387∗34.13℃

900∗(100℃−34.13℃ )=285299.5 g℃

59283℃=4.81 g

Sea f una función con derivada superior en el punto x = 1 de la cual se conoce que:

f (n)=−2 f (n−1) (1 )

n, n∈N ; f (0 ) (1 )= f (1 )=1

f ( x )=ln (x+1)

f (n) (0 )=ln (0+1 )+ 10+1

− 1

(0+1 )2+ 2

(0+1 )3− 6

(0+1 )4+…

f (0 )= ln (0+1 )0 !

+0 ! x1!

−1! x2

2 !+ 2 ! x3

3!−3 ! x4

4 !+ 4 ! x5

5 !−5 ! x6

6 !+…

f (0 )=∑n=0

+∞ (−1 )n xn+1

(n+1), luego∑

n=1

+∞ (∫0

x (−1 )nt n+1

n+1dt)=∑

n=1

+∞ ( (−1 )n

n+1∫0

x

t n+1 dt)∑n=0

+∞ ( (−1 )n

n+1 [ tn+2

n+2 ] x0)=¿∑

n=0

+∞ (−1 )n xn+2

(n+1 ) (n+2 )=∫

0

x

ln (t +1)dt ¿

∑n=0

+∞ [ (−1 )n xn+2

n+1−

(−1 )n xn+2

n+2 ]<1=limn→∞ |(−1 )n x (n+1 )

(n+3 ) |<1

∑n=0

+∞ (−1 )n

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

xn+2∫0

x

ln (t+1)dt

∑n=0

+∞ (−1 )n

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

xn+2 [( x+1 ) ln ( x+1 )−x−12

x2]