f(x)在[0,1]连续,f(x)=3x-√(1-x^2)[∫f^2(x)]dx,求f(x)
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/05/12 09:21:17
f(x)在[0,1]连续,f(x)=3x-√(1-x^2)[∫f^2(x)]dx,求f(x)
f^2(x)是f(x)的平方还是二阶导数?
如果是平方:
令k=∫[f(x)]^2dx
则f(x)=3x-k√(1-x^2)
[f(x)]^2=k^2+(9-k^2)x^2-6kx√(1-x^2)
k=∫[f(x)]^2dx
=∫[k^2+(9-k^2)x^2-6kx√(1-x^2)]dx
=k^2+(9-k^2)∫x^2dx-6k∫x√(1-x^2)dx
=k^2+(9-k^2)/3-2k
整理2k^2-9k+9=0
k=3或3/2
然后就可以得到f(x)
如果是平方:
令k=∫[f(x)]^2dx
则f(x)=3x-k√(1-x^2)
[f(x)]^2=k^2+(9-k^2)x^2-6kx√(1-x^2)
k=∫[f(x)]^2dx
=∫[k^2+(9-k^2)x^2-6kx√(1-x^2)]dx
=k^2+(9-k^2)∫x^2dx-6k∫x√(1-x^2)dx
=k^2+(9-k^2)/3-2k
整理2k^2-9k+9=0
k=3或3/2
然后就可以得到f(x)
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