Question

已知二次函数y=fx满足f(-1)=2,f'(0)=0,∫0-1 f(x)dx=-2,求∫1-2 f(x)/x dx的值

Answer 1

设f(x)=ax^2+bx+c

f(-1)=a-b+c=2

f'(x)=2ax+b

f'(0)=b=0

∫0-1 f(x)dx=(a/3)x^3+(b/2)x^2+cx|0-1=a/3+b/2+c= -2

所以 a=6 b=0 c=-4

∫1-2 f(x)/x dx=∫1-2 (6x - 4/x) dx = 3x^2 - 4*lnx|1-2

=(3*2*2-3*1*1)-4*(ln2-ln1)

=9-4ln2