Question

Calculo diferencial integral de limite superior 2 limite inferior 0 e elevado ala x que elevado ala 2

Answer 1

Respuesta:

n = 6

= ∑ (1/ n !) (2^(2n+1)) /(2n+1)

n = 0

realizar la sumatoria

=(1/ 0 !) (2^(2*0+1)) /(2*0+1) + (1/ 1 !) (2^(2*1+1)) /(2*1+1) +...+ (1/ 6 !) (2^(2*6+1)) /(2*6+1)

Explicación:

Hola posible solución de ∫ e^x^2

∞

e^x= ∑ (x^n / n !)

n =0

∞

e^x^2 = ∑ (x^2n / n !)

n =0

  ∞

∫ ∑ (x^2n / n !)

n =0

 ∞

∑ (1/ n !) (x^(2n+1)) /(2n+1) I 0 2

n = 0

reemplazas x con 2, y 0

 ∞

= ∑ (1/ n !) (2^(2n+1)) /(2n+1) - ∑ (1/ n !) (0^(2n+1)) /(2n+1)

n = 0

se considera n=6

 n= 6

=∑ (1/ n !) (2^(2n+1)) /(2n+1)

n = 0

realizar la sumatoria

= (1/ 0 !) (2^(2*0+1)) /(2*0+1) + (1/ 1 !) (2^(2*1+1)) /(2*1+1) +...+ (1/ 6 !) (2^(2*6+1)) /(2*6+1)