设 f(x)=x^2-x∫f(x)dx [1,2]+2∫f(x)dx [0,1],求f(x)?
∫f(x)dx [1,2]和∫f(x)dx [0,1]是常数,f(x)是2次函数,可用待定系数法.
设:f(x)=ax^2+bx+c
记:g(x)=∫f(x)dx=ax^3/3+bx^2/2+cx+d
则:g(2)=8a/3+4b/2+2c+d;g(1)=a/3+b/2+c+d;g(0)=d;
g(2)-g(1)=7a/3+3b/2+c
g(1)-g(0)=a/3+b/2+c
f(x)=x^2-x∫f(x)dx [1,2]+2∫f(x)dx [0,1]
等价:ax^2+bx+c=x^2-(7a/3+3b/2+c)x+2(a/3+b/2+c)
对比系数得:
a=1
b=-(7a/3+3b/2+c)
c=2(a/3+b/2+c)
解得:a=1,b=-10/9,c=4/9
所以:f(x)=x^2-10x/9+4/9