∫(上限2,下限0)x根号下(2x-x^2)dx求大神解答. 弄不懂了.希望详细一点谢谢了

3个回答

  • ∫(0,2) x√(2x - x²) dx

    = ∫(0,2) x√[- (x² - 2x + 1) + 1] dx

    = ∫(0,2) x√[1 - (x - 1)²] dx

    令x - 1 = sinθ,dx = cosθ dθ

    x = 0 --> θ = - π/2

    x = 2 --> θ = π/2

    = ∫(- π/2,π/2) (1 + sinθ)|cosθ| * cosθ dθ

    = ∫(- π/2,π/2) (1 + sinθ)cos²θ dθ

    = ∫(- π/2,π/2) cos²θ dθ + ∫(- π/2,π/2) sinθcos²θ dθ

    = 2∫(0,π/2) (1 + cos2θ)/2 dθ + ∫(- π/2,π/2) cos²θ d(- cosθ)

    = [θ + (1/2)sin2θ] |(0,π/2) - (1/3)[cos³θ] |(- π/2,π/2)

    = π/2