integral ejercicios
DESCRIPTION
integral resueltaTRANSCRIPT
UNIVERSIDAD INTERAMERICANA PARA EL DESARROLLO
ALUMNA: FÁTIMA MARTÍNEZ CHAVEZ
CARRERA: ADMINISTRACIÓN DE EMPRESAS
ÁLGEBRA
RESUELVE LAS SIGUIENTES INTEGRALES
1.-
Forma en excel
y= 4/5 x^ 5 + 2
integral= 0.133333 x^ 6 + 2 x + C
2.-
𝑦=4/5 𝑥^5+2∫1▒〖𝑓 ( )"𝑥 " 〗 dx=∫1▒〖 (4/5 ^5+2"𝑥 " )𝑑𝑥〗=4/5 ∫1▒ ^5𝑥 𝑑𝑥+2∫1▒𝑑𝑥=4/5 (𝑥^(5+1)/(5+1))+2𝑥=(2𝑥^6)/15+2𝑥+𝑐
𝑦=16𝑥^3+2𝑥+3∫1▒〖𝑓 ( )"𝑥 " 〗 dx=∫1▒〖 (16 ^3+2 +3"𝑥 𝑥 " )𝑑𝑥〗=16∫1▒ ^𝑥 3 𝑑𝑥+2∫1▒𝑥𝑑𝑥+3∫1▒𝑑𝑥=16(𝑥^(3+1)/(3+1))+2( ^(𝑥 1+1)/(1+1))+3𝑥=4 ^𝑥 4+ ^𝑥 2+3𝑥+𝑐
forma excel
y= 16 x^ 3 + 2 x^ 1 + 3
integral= 4 x^ 4 + 1 x^ 2 + 3 x + C
3.-
y= 3 x^ 1/3 + 4
integral= 2.25 x^ 1 1/3 + 4 x + c
Evalua las siguientes integrales4.-
Evalua de 2 a 3 [2,3]
y= 4 x^ 2 + 3
integral: 1.33 x^ 3 + 3 x
limite sup: 3limite inf: 2
Exaluando :en 3: 45en 2: 16.66667
resultado de la integral: 28.333
5.- Evalua de 1 a 2 [1,2]
∫1▒〖𝑓 ( )"𝑥 " 〗 dx=∫1▒〖 (16 ^3+2 +3"𝑥 𝑥 " )𝑑𝑥〗=16∫1▒ ^𝑥 3 𝑑𝑥+2∫1▒𝑥𝑑𝑥+3∫1▒𝑑𝑥=16(𝑥^(3+1)/(3+1))+2( ^(𝑥 1+1)/(1+1))+3𝑥=4 ^𝑥 4+ ^𝑥 2+3𝑥+𝑐
𝑦=√(3&3𝑥)+4∫1▒〖𝑓 ( )"𝑥 " 〗 dx=∫1▒〖 ( 3𝑥+4"𝑥 " )𝑑𝑥〗=∫1▒〖〖 3𝑥〗^(1/3) 𝑑𝑥+4∫1▒𝑑𝑥〗=9/4 〖 〗𝑥 ^(4/3)+4𝑥+𝑐
𝑦=4𝑥^2+3
∫24_𝑎^𝑏▒〖𝑓 (𝑥)[𝑎,𝑏]𝑑𝑥=∫24_2^3▒〖 (4 ^2 "+3𝑥 " ) 𝑑𝑥〗〗 =4∫24_2^3▒〖𝑥 ^2 𝑑𝑥+3∫24_2^3▒𝑑𝑥=〗 4/3 ^3+3𝑥 𝑥]■8(3@2)=4/3 〖 (3)〗 ^3+3(3)-[4/3 〖 (2)〗 ^3+3(2)]=28.33
∫1▒𝑓(𝑥)𝑑𝑥
𝑦=5𝑥^3+6x+2
∫24_𝑎^𝑏▒〖𝑓 (𝑥)[𝑎,𝑏]𝑑𝑥=∫24_1^2▒〖 (5 ^3 "+6x+2𝑥 " ) 𝑑𝑥〗〗 =5∫24_1^2▒〖𝑥 ^3 𝑑𝑥+6∫_1^2▒𝑥𝑑𝑥+3∫24_1^2▒𝑑𝑥=〗 5/4 ^𝑥 4+3𝑥^2+2𝑥]■8(2@1)=29.75
y= 5 x^ 3 + 6 x^ 1 + 2
integral= 1.25 x^ 4 + 3 x^ 2 + 2 x
evaluandolimite sup: 2limite inf: 1
Exaluando :en 2: 36en 1: 6.25
resultado de la integral: 29.75
∫24_𝑎^𝑏▒〖𝑓 (𝑥)[𝑎,𝑏]𝑑𝑥=∫24_1^2▒〖 (5 ^3 "+6x+2𝑥 " ) 𝑑𝑥〗〗 =5∫24_1^2▒〖𝑥 ^3 𝑑𝑥+6∫_1^2▒𝑥𝑑𝑥+3∫24_1^2▒𝑑𝑥=〗 5/4 ^𝑥 4+3𝑥^2+2𝑥]■8(2@1)=29.75
∫1▒𝑓(𝑥)𝑑𝑥
UNIVERSIDAD INTERAMERICANA PARA EL DESARROLLO
ALUMNA: FÁTIMA MARTÍNEZ CHAVEZ
CARRERA: ADMINISTRACIÓN DE EMPRESAS
ÁLGEBRA
∫1▒〖𝑓 ( )"𝑥 " 〗 dx=∫1▒〖 (4/5 ^5+2"𝑥 " )𝑑𝑥〗=4/5 ∫1▒ ^5𝑥 𝑑𝑥+2∫1▒𝑑𝑥=4/5 (𝑥^(5+1)/(5+1))+2𝑥=(2𝑥^6)/15+2𝑥+𝑐
∫1▒〖𝑓 ( )"𝑥 " 〗 dx=∫1▒〖 (16 ^3+2 +3"𝑥 𝑥 " )𝑑𝑥〗=16∫1▒ ^𝑥 3 𝑑𝑥+2∫1▒𝑥𝑑𝑥+3∫1▒𝑑𝑥=16(𝑥^(3+1)/(3+1))+2( ^(𝑥 1+1)/(1+1))+3𝑥=4 ^𝑥 4+ ^𝑥 2+3𝑥+𝑐
Evalua de 2 a 3 [2,3]
Evalua de 1 a 2 [1,2]
∫1▒〖𝑓 ( )"𝑥 " 〗 dx=∫1▒〖 (16 ^3+2 +3"𝑥 𝑥 " )𝑑𝑥〗=16∫1▒ ^𝑥 3 𝑑𝑥+2∫1▒𝑥𝑑𝑥+3∫1▒𝑑𝑥=16(𝑥^(3+1)/(3+1))+2( ^(𝑥 1+1)/(1+1))+3𝑥=4 ^𝑥 4+ ^𝑥 2+3𝑥+𝑐
∫1▒〖𝑓 ( )"𝑥 " 〗 dx=∫1▒〖 ( 3 +4"∛ 𝑥 " )𝑑𝑥〗=∫1▒〖〖 3𝑥〗^(1/3) 𝑑𝑥+4∫1▒𝑑𝑥〗=9/4 〖 〗𝑥 ^(4/3)+4𝑥+𝑐
∫24_𝑎^𝑏▒〖𝑓 (𝑥)[𝑎,𝑏]𝑑𝑥=∫24_2^3▒〖 (4 ^2 "+3𝑥 " ) 𝑑𝑥〗〗 =4∫24_2^3▒〖𝑥 ^2 𝑑𝑥+3∫24_2^3▒𝑑𝑥=〗 4/3 ^3+3𝑥 𝑥]■8(3@2)=4/3 〖 (3)〗 ^3+3(3)-[4/3 〖 (2)〗 ^3+3(2)]=28.33
∫24_𝑎^𝑏▒〖𝑓 (𝑥)[𝑎,𝑏]𝑑𝑥=∫24_1^2▒〖 (5 ^3 "+6x+2𝑥 " ) 𝑑𝑥〗〗 =5∫24_1^2▒〖𝑥 ^3 𝑑𝑥+6∫_1^2▒𝑥𝑑𝑥+3∫24_1^2▒𝑑𝑥=〗 5/4 ^𝑥 4+3𝑥^2+2𝑥]■8(2@1)=29.75
∫24_𝑎^𝑏▒〖𝑓 (𝑥)[𝑎,𝑏]𝑑𝑥=∫24_1^2▒〖 (5 ^3 "+6x+2𝑥 " ) 𝑑𝑥〗〗 =5∫24_1^2▒〖𝑥 ^3 𝑑𝑥+6∫_1^2▒𝑥𝑑𝑥+3∫24_1^2▒𝑑𝑥=〗 5/4 ^𝑥 4+3𝑥^2+2𝑥]■8(2@1)=29.75