Question

∫(上限1下限0)√(4-x²)dx

请问结果是否为(2Ω+3√3)〔希望能用x=2sint证明〕

Answer 1

let
x=2sinu
dx=2cosu du
x=0, u=0
x=1, u=π/6
∫(0->1) √(4-x^2) dx
=4∫(0->π/6) (cosu)^2 du
=2∫(0->π/6) (1+cos2u) du
=2[ u +(1/2)sin2u]|(0->π/6)
=2[π/6 +(1/2)(√3/2)]
=π/3 + √3/2