求解∫(下限为0,上限为2)(-y·根号下(2y-y^2))dy
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求解∫(下限为0,上限为2)(-y·根号下(2y-y^2))dy
![求解∫(下限为0,上限为2)(-y·根号下(2y-y^2))dy](/uploads/image/z/5903942-14-2.jpg?t=%E6%B1%82%E8%A7%A3%E2%88%AB%EF%BC%88%E4%B8%8B%E9%99%90%E4%B8%BA0%2C%E4%B8%8A%E9%99%90%E4%B8%BA2%EF%BC%89%EF%BC%88-y%C2%B7%E6%A0%B9%E5%8F%B7%E4%B8%8B%EF%BC%882y-y%5E2%29%29dy)
∵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.
∴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.
求解∫(下限为0,上限为2)(-y·根号下(2y-y^2))dy
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