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)

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)+...