设函数f(x)连续,且f(x)=x^2+2∫[1,0]f(x)dx,则f(x)=?

Question

设函数f(x)连续,且f(x)=x^2+2∫[1,0]f(x)dx,则f(x)=?

Answer 1

∫[1,0]f(x)dx,是定积分,是一个常数
设C=∫[1,0]f(x)dx
则
f(x)=x^2+2C
∫[1,0]f(x)dx
=∫(0→1)f(x)dx
=∫(0→1)(x^2+2C)dx
=(1/3)x^3+2Cx (0→1)
=2C + 1/3 =C
C=-1/3
则
f(x)=x^2-2/3