Miercoles 15 De Julio Calculo Integral
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MIERCOLES 15 DE JULIO HORAS ) GRUPO 361 RESOLVER LOS SIGUIENTES LIMITES
∫ (2x + 11 ) dx = 4
∫( 5 – 3x ) dx = 2
∫ ( 8x
2
∫
+ 5 ) 5xdx =
( 7x3 + 5 ) 5 x2dx =
∫ ( 2x + 6 )
4
dx =
∫ ln ( 2x + 10 ) dx = ∫ ln ( 5x + 6 ) dx = ∫ 5/x dx = ∫x/( 2x
2
+ 5 ) dx =
∫ dx/ ( x + 4 ) = ∫e
4x
∫e
dx =
2X + 6
∫
dx =
e8x+1 dx =
CALCULO INTEGRAL ( 2
∫ (2x + 11 ) dx = 4
∫( 5 – 3x ) dx = 2
∫ ( 8x
2
∫
+ 5 ) 5xdx =
( 7x3 + 5 ) 5 x2dx =
∫ ( 2x + 6 )
4
dx =
∫ ln ( 2x + 10 ) dx = ∫ ln ( 5x + 6 ) dx = ∫ 5/x dx = ∫x/( 2x
2
+ 5 ) dx =
∫ dx/ ( x + 4 ) = ∫e
4x
∫e
dx =
2X + 6
∫
dx =
e8x+1 dx =
CALCULO INTEGRAL ( 2
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