高数定积分求面积的题目求曲线y=√(2x-x^2)与y=x/√3所围成的平面图形的面积 答案是π/3-√3/4 我之后还会加分

解{ y = √(2x - x^2)
{ y = x/√3
得交点：(0,0)、(3/2,√3/2)
在0 ≤ x ≤ 3/2上、y = √(2x - x^2) > y = x/√3
所以面积 = ∫[0→3/2] [√(2x - x^2) - x/√3] dx
= ∫[0→3/2] √[1 - (1 - x)^2] dx - (1/√3)∫[0→3/2] x dx
= [(- 1/2)arcsin(1 - x) - (1/2)(1 - x)√(2x - x^2) - (1/√3)(x^2/2)] |[0→3/2]
= [π/12 - √3/4] - [- π/4]
= π/3 - √3/4
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∫ √[1 - (1 - x)^2] dx、1 - x = sinz、- dx = cosz dz
= ∫ √[1 - sin^2(z)][- cosz dz]
= - ∫ cos^2(z) dz
= - ∫ (1 + cos2z)/2 dz
= (- 1/2)[z + (1/2)sin2z] + C
= (- 1/2)z - (1/2)sinzcosz + C
= (- 1/2)arcsin(1 - x) - (1/2)(1 - x)√(2x - x^2) + C