Question

定积分 求做法

Answer 1

∫(0→2) x√(2x - x²) dx
= ∫(0→2) x√[1 - (x - 1)²] dx
令x - 1 = sinz、dx = cosz dz
x = 0 ==> sinz = - 1 ==> z = - π/2
x = 2 ==> sinz = 1 ==> z = π/2
原式 = ∫(- π/2→π/2) (1 + sinz)(cosz)² dz
= ∫(- π/2→π/2) (cos²z + sinzcos²z) dz
= z/2 + (1/4)sin2z - (1/3)cos³z |(- π/2→π/2)
= (π/4 + 0 - 0) - (- π/4 + 0 - 0)
= π/2

差点没把下限看成θ了- . -
难怪最后做出的那个答案竟然这么长呢。。。

Answer 2

根号看成2分1次方。导函数就好了。普通微积分做法啊