Tarea 2 Ejercicos C

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Ejercicio 1.c Integración por sustitución ∫

π‘₯ 𝑑π‘₯ (6π‘₯ 2 + 5)3

∫

1 1 βˆ™ 𝑑𝑒 3 (𝑒) 12

∫

1 𝑑𝑒 12(𝑒)3

1 1 ∫ 𝑑𝑒 12 (𝑒)3 1 ∫ π‘’βˆ’3 𝑑𝑒 12 1 1 (βˆ’ π‘’βˆ’2 + 𝐢) 12 2 βˆ’ βˆ’

1 +𝐢 24𝑒2

1 +𝐢 24(6π‘₯ 2 + 5)2

Ejercicio 2.c IntegraciΓ³n por partes ∫ sinβˆ’1(π‘₯)𝑑π‘₯ sinβˆ’1(π‘₯)π‘₯ βˆ’ ∫ π‘₯ sinβˆ’1 (π‘₯)π‘₯ βˆ’ ∫ sinβˆ’1(π‘₯)π‘₯ βˆ’ ∫

π‘₯ √1 βˆ’ π‘₯ 2 π‘₯

√1 βˆ’ π‘₯ 2 1

𝑑π‘₯

𝑑π‘₯

1 𝑑𝑒 βˆšπ‘’ βˆ’2

sinβˆ’1(π‘₯)π‘₯ βˆ’ ∫ βˆ’ sinβˆ’1(π‘₯)π‘₯ βˆ’ βˆ’ ∫

βˆ™

1 2βˆšπ‘’ 1 2βˆšπ‘’

𝑑𝑒 𝑑𝑒

1 1 sinβˆ’1 (π‘₯)π‘₯ + ∫ 𝑑𝑒 2 βˆšπ‘’

1 1 sinβˆ’1(π‘₯)π‘₯ + ∫ π‘’βˆ’2 𝑑𝑒 2 1 1 sinβˆ’1(π‘₯)π‘₯ + (2𝑒2 + 𝐢) 2 1

sinβˆ’1 (π‘₯)π‘₯ + 𝑒2 + 𝐢 1

sinβˆ’1(π‘₯)π‘₯ + (1 βˆ’ π‘₯ 2 )2 + 𝐢 1

x sinβˆ’1(π‘₯) + (1 βˆ’ π‘₯ 2 )2 + 𝐢

Ejercicio 3.c Sustitución Trigonométrica y Fracciones parciales. ∫

π’™πŸ βˆšπ’™πŸ βˆ’ πŸ’

𝒅𝒙

Aplicamos sustituciΓ³n trigonomΓ©trica 𝒙 = πŸπ’”π’†π’„πœ½ ∫ πŸ’π’”π’†π’„πŸ‘ 𝜽 π’…πœ½ πŸ’ βˆ™ ∫ π’”π’†π’„πŸ‘ 𝜽 π’…πœ½ Aplicamos reducciΓ³n = πŸ’(

= πŸ’(

π’”π’†π’„πŸ 𝜽 βˆ™ π’”π’Šπ’πœ½ 𝟏 + βˆ™ ∫ π’”π’†π’„πœ½π’…πœ½) 𝟐 𝟐

π’”π’†π’„πŸ 𝜽 βˆ™ π’”π’Šπ’πœ½ 𝟏 + 𝑰𝒏|π’•π’‚π’πœ½ + π’”π’†π’„πœ½|) 𝟐 𝟐 𝟏

Remplazamos 𝜽 por(𝒂𝒓𝒄𝒔𝒆𝒄 (𝟐 𝒙)) = πŸ’(

𝟏 𝟏 π’”π’†π’„πŸ (𝒂𝒓𝒄𝒔𝒆𝒄 (𝟐 𝒙)) βˆ™ π’”π’Šπ’(𝒂𝒓𝒄𝒔𝒆𝒄 (𝟐 𝒙)) 𝟐 𝟏 𝟏 𝟏 + 𝑰𝒏 |𝒕𝒂𝒏(𝒂𝒓𝒄𝒔𝒆𝒄 ( 𝒙)) + 𝒔𝒆𝒄(𝒂𝒓𝒄𝒔𝒆𝒄 ( 𝒙))|) 𝟐 𝟐 𝟐 Simplificamos 𝟏

𝟏

= πŸ’ (πŸ– π’™βˆšπ’™πŸ βˆ’ πŸ’ + 𝟐 𝑰𝒏 |

βˆšπ’™πŸ βˆ’πŸ’+𝒙 𝟐

|)+C

Ejercicios 4.c Integral Impropias 0

∫ βˆ’βˆž 0

∫ βˆ’βˆž 0

4∫ βˆ’βˆž 0

4∫ βˆ’βˆž

4 𝑑π‘₯ 4 + π‘₯2 1 𝑑π‘₯ 4 + π‘₯2 1 𝑑π‘₯ 4 + π‘₯2 1 𝑑π‘₯ 22 + π‘₯ 2

1 π‘₯ 4 ( arctan( )]) 2 2 βˆ’βˆž arctan(0) arctan( 2 ) 4( βˆ’ ) 2 2 βˆ’2 arctan(

βˆ’βˆž ) 2

Tabla links videos explicativos. Nombre Estudiante Ejemplo: Adriana Granados

Ejercicios Link video explicativo sustentados Ejercicio Link. asignado por el tutor

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