设函数f(x)在[0,t]上连续,记F(t)=∫0tdz∫0zdy∫0y(y-z)2f(x)dx.求F’(t).

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问题 设函数f(x)在[0,t]上连续,记F(t)=∫0tdz∫0zdy∫0y(y-z)2f(x)dx.求F’(t).

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答案F(t)对应的积分区域为 Ω(t)={(x,y,z)|0≤x≤y,0≤y≤z,0≤z≤t} 先交换累次积分∫0zdy∫0y(y-z)2f(x)dx的次序,因为 σxy={(x,y)|0≤x≤y,0≤y≤z}={(x,y)|0≤x≤z,x≤y≤z} 则交换积分次序,得 F(t)=∫0tdz∫0zf(x)dx∫xz(y-z)2dy=[*]∫0tdz∫0z(z-x)3f(x)dx 再对上式交换积分次序,因σxz={(x,z)|0≤x≤z,0≤z≤t}={(x,z)|x≤z≤t,0≤x≤t},故 F(t)=[*]∫0tdz∫0z(z-x)3f(x)dx=[*]∫0tf(x)dx∫xt(z-x)3dz=[*]∫0t(t-x)4f(x)dx =[*][t40tf(x)dx-4t30txf(x)dx+6t20tx2f(x)dx-4t∫0tx3f(x)dx+∫0tx4f(x)dx] 由变限积分求导公式,得 F’(t)=[*]∫0t(t-x)3f(x)dx

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