Question

解一下这题

Answer 1

解：
∫(0,1) x√(2x - x²) dx
= ∫(0,1) x√[- (x² - 2x + 1) + 1] dx
= ∫(0,1) x√[1 - (x - 1)²] dx
令x - 1 = sinθ,dx = cosθ dθ
x = 0 --> θ = - π/2
x = 1 --> θ = 0
= ∫(- π/2,0) (1 + sinθ)|cosθ| * cosθ dθ
= ∫(- π/2,0) (1 + sinθ)cos²θ dθ
= ∫(- π/2,0) cos²θ dθ + ∫(- π/2,0) sinθcos²θ dθ
= ∫(-π/2,0) (1 + cos2θ)/2 dθ + ∫(- π/2,0) cos²θ d(- cosθ)
= [θ/2 + (1/4)sin2θ] |(-π/2,0) - (1/3)[cos³θ] |(- π/2,0)
= -π/4