第1个回答 2019-12-25
哪一个不会？不可能让别人都给你做出来吧？先给你写一下第5题第一个
第2个回答 2019-12-05
详细过程如图，希望能帮到你解决你心中的问题
希望过程清楚明白
本回答被提问者采纳第3个回答 2019-12-05
设对于任意光滑有向闭曲面S ,都有∮∮x f ( y ) dy dz + y f ( x ) dz dx - z [ b+ f ( x + y ) ] dx dy = 0,其中函数f ( x ) 在(- ∋ ,+ ∋ ) 内连续,且f ( 1) = a( a,b 都是常数) ,求f ( 2010) .
第4个回答 2019-12-05
I(m,n)=1/(n+1)*∫(0,1) (lnx)^m d[x^(n+1)]
=1/(n+1)*(lnx)^m*x^(n+1)|(0,1)-m/(n+1)*∫(0,1) x^n*(lnx)^(m-1) dx
=-1/(n+1)*lim(x->0+) (lnx)^m*x^(n+1)-m/(n+1)*I(m-1,n)
=-m/(n+1)*I(m-1,n)
=[-m/(n+1)]*[-(m-1)/(n+1)]*I(m-2,n)
......
=m!*[-1/(n+1)]^m*I(0,n)
=m!*[-1/(n+1)]^m*∫(0,1) x^n dx
=m!*[-1/(n+1)]^m*[1/(n+1)]*x^(n+1)|(0,1)
=(-1)^m*m!/(n+1)^(m+1)
=1/(n+1)*(lnx)^m*x^(n+1)|(0,1)-m/(n+1)*∫(0,1) x^n*(lnx)^(m-1) dx
=-1/(n+1)*lim(x->0+) (lnx)^m*x^(n+1)-m/(n+1)*I(m-1,n)
=-m/(n+1)*I(m-1,n)
=[-m/(n+1)]*[-(m-1)/(n+1)]*I(m-2,n)
......
=m!*[-1/(n+1)]^m*I(0,n)
=m!*[-1/(n+1)]^m*∫(0,1) x^n dx
=m!*[-1/(n+1)]^m*[1/(n+1)]*x^(n+1)|(0,1)
=(-1)^m*m!/(n+1)^(m+1)
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