第1个回答 2022-12-18
根据对称性求解
(x-1)根号(2x-x^2)关于(1,0)中心对称,在(0,2)上积分为0
所以原来积分=
∫(x+4)根号(2x-x^2)dx = ∫(x-1)根号(2x-x^2)dx + ∫5根号(2x-x^2)dx
=∫5根号(2x-x^2)dx , 取x-1=sint, t取从-pi/2到pi/2, dx = d(x-1) = cost dt
带入得到
=∫5根号(2x-x^2)dx = ∫5根号(1-(sint)^2) costdt
=∫5(cost)^2dt
=5/4 ∫(1+cos2t)d2t
=5/4 (t +sin2t) |-pi/2, pi/2
=5pi/4