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二重积分的分部积分法

2023-07-02 13:29 作者:编程会一点建模不太懂  | 我要投稿

题目选自2011年考研数学一

已知函数f(x%2Cy)具有二阶连续偏导数,

f(1%2Cy)%3D0%2Cf(x%2C1)%3D0%5Ciint%5Climits_D%7Bf%5Cleft(%20x%2Cy%20%5Cright)%20dxdy%7D%3Da

D%3D%5Cleft%5C%7B%20%5Cleft(%20x%2Cy%20%5Cright)%20%7C0%5Cle%20x%5Cle%201%2C0%5Cle%20y%5Cle%201%20%5Cright%5C%7D%20%0A

计算二重积分

I%3D%5Ciint%5Climits_D%7Bxyf_%7Bxy%7D%5E%7B''%7D%5Cleft(%20x%2Cy%20%5Cright)%20dxdy%7D%0A

本题中给出的函数是抽象函数,常见的极坐标这类二重积分计算方法失效,只能尝试直角坐标法计算,同时考虑到被积函数中存在函数f(x%2Cy)偏导数,可以联想到一元函数中的分部积分法。

I%3D%5Ciint%5Climits_D%7Bxyf_%7Bxy%7D%5E%7B''%7D%5Cleft(%20x%2Cy%20%5Cright)%20dxdy%7D%0A

首先化为累次积分

I%3D%5Cint_0%5E1%7Bxdx%7D%5Cint_0%5E1%7Byf_%7Bxy%7D%5E%7B''%7D%5Cleft(%20x%2Cy%20%5Cright)%20dy%7D%0A

其中对y积分就行凑微分

%5Cint_0%5E1%7Byf_%7Bxy%7D%5E%7B''%7D%5Cleft(%20x%2Cy%20%5Cright)%20dy%7D%3D%5Cint_0%5E1%7Bydf_%7Bx%7D%5E%7B'%7D%5Cleft(%20x%2Cy%20%5Cright)%7D%0A

%3Dyf_%7Bx%7D%5E%7B'%7D%5Cleft(%20x%2Cy%20%5Cright)%20%5Cmid_%7By%3D0%7D%5E%7By%3D1%7D-%5Cint_0%5E1%7Bf_%7Bx%7D%5E%7B'%7D%5Cleft(%20x%2Cy%20%5Cright)%20dy%7D%0A

%3Df_%7Bx%7D%5E%7B'%7D%5Cleft(%20x%2C1%20%5Cright)%20-%5Cint_0%5E1%7Bf_%7Bx%7D%5E%7B'%7D%5Cleft(%20x%2Cy%20%5Cright)%20dy%7D%0A

带入原积分

I%3D%5Cint_0%5E1%7Bxdx%7D%5Cint_0%5E1%7Byf_%7Bxy%7D%5E%7B''%7D%5Cleft(%20x%2Cy%20%5Cright)%20dy%7D%0A

%3D%5Cint_0%5E1%7Bx%5Cleft(%20f_%7Bx%7D%5E%7B'%7D%5Cleft(%20x%2C1%20%5Cright)%20-%5Cint_0%5E1%7Bf_%7Bx%7D%5E%7B'%7D%5Cleft(%20x%2Cy%20%5Cright)%20dy%7D%20%5Cright)%20dx%7D%0A

%3D%5Cint_0%5E1%7Bxf_%7Bx%7D%5E%7B'%7D%5Cleft(%20x%2C1%20%5Cright)%20dx%7D-%5Cint_0%5E1%7Bx%5Cint_0%5E1%7Bf_%7Bx%7D%5E%7B'%7D%5Cleft(%20x%2Cy%20%5Cright)%20dy%7Ddx%7D%0A

其中第一个积分继续采用分部积分

%5Cint_0%5E1%7Bxf_%7Bx%7D%5E%7B'%7D%5Cleft(%20x%2C1%20%5Cright)%20dx%7D%3D%5Cint_0%5E1%7Bxdf%5Cleft(%20x%2C1%20%5Cright)%7D%0A

%3Dxf%5Cleft(%20x%2C1%20%5Cright)%20%5Cmid_%7Bx%3D0%7D%5E%7Bx%3D1%7D-%5Cint_0%5E1%7Bf%5Cleft(%20x%2C1%20%5Cright)%20dx%7D%0A

%3Df%5Cleft(%201%2C1%20%5Cright)%20-%5Cint_0%5E1%7Bf%5Cleft(%20x%2C1%20%5Cright)%20dx%7D%0A%3D0

由于题设中,f(x%2C1)%3D0,所以积分结果为0

第二个积分首先进行积分换序

%5Cint_0%5E1%7Bx%5Cint_0%5E1%7Bf_%7Bx%7D%5E%7B'%7D%5Cleft(%20x%2Cy%20%5Cright)%20dy%7Ddx%7D%3D%5Cint_0%5E1%7Bdy%7D%5Cint_0%5E1%7Bxf_%7Bx%7D%5E%7B'%7D%5Cleft(%20x%2Cy%20%5Cright)%20dx%7D%0A

单独计算对x的积分

%5Cint_0%5E1%7Bxf_%7Bx%7D%5E%7B'%7D%5Cleft(%20x%2Cy%20%5Cright)%20dx%7D%3D%5Cint_0%5E1%7Bxdf%5Cleft(%20x%2Cy%20%5Cright)%7D%0A

%3Dxf%5Cleft(%20x%2Cy%20%5Cright)%20%5Cmid_%7Bx%3D0%7D%5E%7Bx%3D1%7D-%5Cint_0%5E1%7Bf%5Cleft(%20x%2Cy%20%5Cright)%20dx%7D%0A

%3Df%5Cleft(%201%2Cy%20%5Cright)%20-%5Cint_0%5E1%7Bf%5Cleft(%20x%2Cy%20%5Cright)%20dx%7D%0A

%3D-%5Cint_0%5E1%7Bf%5Cleft(%20x%2Cy%20%5Cright)%20dx%7D%0A

由于题设中,f(1%2Cy)%3D0

所以

%5Cint_0%5E1%7Bx%5Cint_0%5E1%7Bf_%7Bx%7D%5E%7B'%7D%5Cleft(%20x%2Cy%20%5Cright)%20dy%7Ddx%7D%3D-%5Cint_0%5E1%7Bdy%7D%5Cint_0%5E1%7Bf%5Cleft(%20x%2Cy%20%5Cright)%20dx%7D%0A

%3D-%5Ciint%5Climits_D%7Bf%5Cleft(%20x%2Cy%20%5Cright)%20dxdy%7D%3D-a%0A

最终

I%3D%5Cint_0%5E1%7Bxf_%7Bx%7D%5E%7B'%7D%5Cleft(%20x%2C1%20%5Cright)%20dx%7D-%5Cint_0%5E1%7Bx%5Cint_0%5E1%7Bf_%7Bx%7D%5E%7B'%7D%5Cleft(%20x%2Cy%20%5Cright)%20dy%7Ddx%7D%3Da

题目分析:二重积分的分部积分法在当年属于第一次在考研数学中出现的考题,具有创新性,计算过程比较冗长,要分清楚对x积分还是对y积分。虽然,从未出现过,但是,题目中被积函数带有f(x,y)的导数确实起到一定的提示作用。总体来说,具有创新性和部分提示,要求计算过程思路清晰。

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