设f(x)=ax+b,且∫-1到1f^2(x)dx=1,求f(a)的取值范围

1=∫-1到1f^2(x)dx
=∫-1到1 (ax+b)^2dx
=∫-1到1 (a^2x^2+2abx+b^2)dx
=2∫0到1 (a^2x^2+b^2)dx
=2(a^2/3+b^2)
得a^2=3(1/2-b^2)≥0,-√2/2≤b≤√2/2
则f(a)=a^2+b
=3(1/2-b^2)+b
=-3(b^2-b/3)+3/2
=-3(b-1/6)^2+19/12
故-3(-√2/2-1/6)^2+19/12=-3(1/2+√2/6+1/36)+19/12=-√2/2≤f(a)≤19/12
也即f(a)的取值范围为[-√2/2,19/12]