Ejercicio 2.docx

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Ejercicio c. ∫ π’”π’†π’βˆ’πŸ (𝒙)𝒅𝒙 Re-expresando la integral ∫ 𝒂𝒓𝒄𝒔𝒆𝒏(𝒙)𝒅𝒙 Aplicando integraciΓ³n por partes: 𝑒 = π‘Žπ‘Ÿπ‘π‘ π‘’π‘›(π‘₯) 𝑣=1 Aplicando la fΓ³rmula de integraciΓ³n por partes 𝑓(π‘₯)𝑔(π‘₯) βˆ’ ∫ π‘“κž‹(π‘₯)𝑔(π‘₯)𝑑π‘₯ π‘₯π‘Žπ‘Ÿπ‘π‘ π‘’π‘›(π‘₯) βˆ’ ∫ π‘₯π‘Žπ‘Ÿπ‘π‘ π‘’π‘›(π‘₯) βˆ’ ∫

1 √1 βˆ’ π‘₯ 2 π‘₯ √1 βˆ’ π‘₯ 2

βˆ— π‘₯𝑑π‘₯ 𝑑π‘₯

Vamos a resolver la integral aplicando integraciΓ³n por sustituciΓ³n 𝑒 = 1 βˆ’ π‘₯2 𝑑𝑒 = βˆ’2π‘₯𝑑π‘₯ Reemplazando π‘₯π‘Žπ‘Ÿπ‘π‘ π‘’π‘›(π‘₯) βˆ’ ∫ βˆ’

1 2βˆšπ‘’

𝑑𝑒

Sacamos la constante de la integral segΓΊn: ∫ π‘Ž βˆ— 𝑓(π‘₯)𝑑π‘₯ = π‘Ž ∫ 𝑓(π‘₯)𝑑π‘₯ 1 1 π‘₯π‘Žπ‘Ÿπ‘π‘ π‘’π‘›(π‘₯) + ∫ 𝑑𝑒 2 βˆšπ‘’ 1 π‘₯π‘Žπ‘Ÿπ‘π‘ π‘’π‘›(π‘₯) + ∫ π‘’βˆ’1/2 2 Resolviendo la integral

1

1 π‘’βˆ’2+1 π‘₯π‘Žπ‘Ÿπ‘π‘ π‘’π‘›(π‘₯) + βˆ— 2 βˆ’1 + 1 2 Reemplazando el valor de 𝑒 1

1 (1 βˆ’ π‘₯ 2 )βˆ’2+1 π‘₯π‘Žπ‘Ÿπ‘π‘ π‘’π‘›(π‘₯) + βˆ— 1 2 βˆ’2 + 1 1

Resolviendo la operaciΓ³n βˆ’ 2 + 1 =

βˆ’1+2 2

1

=2

Por lo tanto, 1

1 (1 βˆ’ π‘₯ 2 )2 π‘₯π‘Žπ‘Ÿπ‘π‘ π‘’π‘›(π‘₯) + βˆ— 1 2 2 1 √1 βˆ’ π‘₯ 2 π‘₯π‘Žπ‘Ÿπ‘π‘ π‘’π‘›(π‘₯) + βˆ— 1 2 2 Aplicando las propiedades de fracciones, multiplicaciΓ³n en cruz. 1 2√1 βˆ’ π‘₯ 2 π‘₯π‘Žπ‘Ÿπ‘π‘ π‘’π‘›(π‘₯) + βˆ— 2 1 Eliminamos los tΓ©rminos comunes 2 π‘₯π‘Žπ‘Ÿπ‘π‘ π‘’π‘›(π‘₯) + √1 βˆ’ π‘₯ 2 Agregamos la constante π‘₯π‘Žπ‘Ÿπ‘π‘ π‘’π‘›(π‘₯) + √1 βˆ’ π‘₯ 2 + 𝑐

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