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Faça u'(x) = √(x+4) e v(x)=x. Então ∫vdu = vu - ∫udv é o mesmo que:
∫x√(x+4) dx = (2/3)x(x+4)^(3/2) - ∫(x+4)^(3/2) dx
O último termo é uma integral simples, cujo resultado é (4/15)(x+4)^(5/2), logo
∫x√(x+4) dx = (2/3)x(x+4)^(3/2) - (4/15)(x+4)^(5/2) + C
Note que (4/15)(x+4)^(5/2) = (4/15)(x+4)(x+4)^(3/2), então dá para simplificar a expressão para:
∫x√(x+4) dx = [(2/3)x- (4/15)(x+4)](x+4)^(3/2) + C
∫x√(x+4) dx =(2/15)(3x-8)(x+4)^(3/2) + C

Como é uma integral definida,
∫x√(x+4) dx = (2/15)(3x-8)(x+4)^(3/2) de 0 a 32
Em 0, -(2/15)64 = -128/15
Em 32, (2/15)(3*32-8)216 = 38016/15
Logo, 38016/15+128/15 = 38144/15

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