由密度函数及期望、方差的性质可以知道,
∫(0到1)f(x)dx=1
E(X)=∫(0到1)x*f(x)dx=0.5
D(X)=E(X^2)-[E(X)]^2=∫(0到1)x^2*f(x)dx-0.5^2=0.15
所以
∫(0到1)f(x)dx
=∫(0到1)ax^2+bx+cdx
=(ax^3/3+bx^2/2+cx)代入上下限1和0
=a/3+b/2+c=1
而
EX=∫(0到1)x*f(x)dx
=∫(0到1)ax^3+bx^2+cxdx
=(ax^4/4+bx^3/3+cx^2/2)代入上下限1和0
=a/4+b/3+c/2=0.5
DX=∫(0到1)x^2*f(x)dx-0.5^2
=∫(0到1)ax^4+bx^3+cx^2dx-0.25
=(ax^5/5+bx^4/4+cx^3/3)-0.25代入上下限1和0
=a/5+b/4+c/3-0.25=0.15
于是
a/3+b/2+c=1
a/4+b/3+c/2=0.5
a/5+b/4+c/3=0.4
解得
a=12,b=-12,c=3