Question

Determine a integral indefinida de f left parenthesis x right parenthesis equals x squared square root of e to the power of x end root.

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Answer 1

Resposta:

Se for  f(x) = x²*√eˣ =x² * e^(x/2)

∫ x² * e^(x/2) dx

Fazendo por Partes

u=x² ==> du=2x dx

dv=e^(x/2) dx   ==>∫  dv=∫ e^(x/2) dx  ==>v =2* e^(x/2)

∫ x² * e^(x/2) dx  = 2*x²* e^(x/2) -  ∫ 2* e^(x/2)  *  2x dx

∫ x² * e^(x/2) dx  = 2*x²* e^(x/2) - 4* ∫x* e^(x/2) dx   (1)

calculando ∫x* e^(x/2) dx

Fazendo por Partes

u=x ==>du=dx

dv=e^(x/2) dx   ==>∫  dv=∫ e^(x/2) dx  ==>v =2* e^(x/2)

∫x* e^(x/2) dx =2 * x *  e^(x/2)  - ∫ 2* e^(x/2)  dx

∫x* e^(x/2) dx =2 * x *  e^(x/2)  - 2 *∫  e^(x/2)  dx

∫x* e^(x/2) dx =2 * x *  e^(x/2)  - 4 e^(x/2)   (2)

(2)  em (1) ficamos com

∫ x² * e^(x/2) dx  = 2*x²* e^(x/2) - 4* [2 * x *  e^(x/2)  - 4 e^(x/2)] + c

∫ x² * e^(x/2) dx  =   [2x²-8x +16] * e^(x/2)+ c

∫ x² * e^(x/2) dx  =   [2x²-8x +16] * √eˣ+ c

Letra C