问题
解答题
已知∫01(3ax+1)(x+b)dx=0,a,b∈R,试求ab的取值范围.
答案
∫01(3ax+1)(x+b)dx
=∫01[3ax2+(3ab+1)x+b]dx
=[ax3+
(3ab+1)x2+bx]1 2 | 10
=a+
(3ab+1)+b=01 2
即3ab+2(a+b)+1=0
设ab=t∴a+b=-3t+1 2
则a,b为方程x2+
x+t=0两根3t+1 2
△=
-4t≥0∴t≤(3t+1)2 4
或t≥11 9
∴a•b∈(-∞,
]∪[1,+∞)1 9