求解∫（下限为0,上限为2）（-y·根号下（2y-y^2))dy

Question

求解∫（下限为0,上限为2）（-y·根号下（2y-y^2))dy

Answer 1

∵原式=∫(0,2)[-y√(1-(y-1)²)]dy
∴设y-1=sinx,则y=1+sinx,dy=cosxdx
故原式=-∫(-π/2,π/2)(1+sinx)cos²xdx
=-∫(-π/2,π/2)cos²xdx-∫(-π/2,π/2)sinxcos²xdx
=-1/2∫(-π/2,π/2)(1+cos(2x))dx+∫(-π/2,π/2)cos²xd(cosx)
=-1/2[x+sin(2x)/2]│(-π/2,π/2)+[cos³x/3]│(-π/2,π/2)
=-1/2(π/2+π/2)+(0+0)
=-π/2.