已知函数f(x)在[-1,1]上连续且满足f(x)=3x-√（1-x^2)∫(0,1)f^2(t)dt,求f(x)

设∫(0,1)f^2(t)dt=k ，显然k为常数
所以f(x)=3x-√（1-x^2)∫(0,1)f^2(t)dt=3x-k√（1-x^2) ......（1）
同时又f(x)=3x-√（1-x^2)∫(0,1)f^2(t)dt=3x-√（1-x^2)∫(0,1)[3x-√（1-x^2)∫(0,1)f^2(t)dt]^2dt=3x-√（1-x^2)∫(0,1)[3x-k√（1-x^2)]^2dt .....(2)
令x=sinu , -π/2 由（1）式得f(sinu)=3sinu-kcosu
由（2）式得f(sinu)=3sinu-cosu∫(0,π/2)[3sinu-kcosu]^2dsinu=3sinu-cosu( 3-2k+2k^2/3)
比较上面两式得
k=3-2k+2k^2/3 即 2k^2-9k+9=0,解得k=3或k=3/2
当k=3/2时 函数为f(x)=3x-3/2*√（1-x^2)
当k=3时 ，函数为f(x)=3x-3√（1-x^2)

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
f(x)=3x-3√(1-x^2)或f(x)=3x-1.5√(1-x^2)

因为f(x)在[-1,1]上连续,则∫（0,1）f^2(t)dt存在,令A=∫（0,1）f^2(t)dt,于是f(x)=3x-A√(1-X^2)=>f^2(x)=9x^2-6Ax√（1-x^2)+A^2(1-X^2)又 A=∫(0,1)f^2(t)dt=∫(0,1)f^2(x)dx=∫(0,1)[9-A^2)x^2-6Ax√（1-x^2)+...

f(x)=3x-√（1-x^2)∫(0,1)f^2(t)dt
令∫(0,1)f^2(t)dt=A
f(x)=3x-A√（1-x^2)
df(x)=[3+Ax/√（1-x^2)]dx
x=0
f(0)=-A
x=1
f(1)=3
3x-f(x)=A√（1-x^2)
平方得
[3x-f(x)]^2=[A√（1-x^2)]^2
9x^2-6xf(x)+f^2(x)=A-Ax^2
f^2(x)=A-Ax^2-9x^2+6xf(x)
两边在区间[0,1]积分得
∫[0,1]f^2(x)dx=∫[0,1] [A-Ax^2-9x^2+6xf(x)]dx=A
[Ax-Ax^3/3-3x^3][0,1]+∫[0,1][6xf(x)]dx=A
A-A/3-3+∫[0,1][6xf(x)]dx=A
∫[0,1][6xf(x)]dx=A/3+3
=3∫[0,1]f(x)dx^2
=3x^2f(x)[0,1]-3∫[0,1]x^2df(x)
=9-3∫[0,1]x^2*[3+Ax/√（1-x^2)]dx
=9-3x^3[0,1]-3A∫[0,1]x^3/√（1-x^2)]dx
=6-3A/2∫[0,1]x^2/√（1-x^2)]dx^2
=6-3A/2∫[0,1] (x^2-1+1)/√（1-x^2)]dx^2
=6-3A/2∫[0,1] [-√(1-x^2)+1/√（1-x^2)]dx^2
=6-3A/2[-2/3(1-x^2)^(3/2)-2√（1-x^2)][0,1]
=6-3A/2(2/3+2)
=6-4A
所以
A/3+4A=6-3
13A/3=3
A=9/13
f(x)=3x-9/13√（1-x^2)