因为,4x-x^2-3=1-(x-2)^2 设 x-2=cosθ,θ∈【0,π】,
则dx=-sinθdθ,x=0不行,最小取1,θ=π,x=2,θ=0
∫[根号下4x-x^2-3] dx= ∫sinθ(-sinθ)dθ=∫0.5[cos2θ-1]dθ=【0.25sin2θ-0.5θ】|(θ从π到0)=π/2
因为,4x-x^2-3=1-(x-2)^2 设 x-2=cosθ,θ∈【0,π】,
则dx=-sinθdθ,x=0不行,最小取1,θ=π,x=2,θ=0
∫[根号下4x-x^2-3] dx= ∫sinθ(-sinθ)dθ=∫0.5[cos2θ-1]dθ=【0.25sin2θ-0.5θ】|(θ从π到0)=π/2