Question

求一下这道题的详细过程，谢谢

Answer 1

设f(t)=kt+b，则：
∫(a,x)(kt+b)dt
=(k/2)t^2+bt|(a,x)
=(k/2)x^2+bx-(k/2)a^2-ab
=2x^2-3x+1
所以：
k/2=2,即k=4.
b=-3.
此时，f(x)=4x-3.
则：
-2a^2+3a=1
2a^2-3a+1=0
(2a-1)(a-1)=0
所以a=1或者a=1/2.

Answer 2

∫(a->x) f(t) dt = 2x^2-3x+1
let
f(t) = ct+d
∫(a->x) f(t) dt
=∫(a->x) (ct+d) dt
=[ (1/2)ct^2 +dt] |(a->x)
=(1/2)cx^2 +dx - [(1/2)ca^2+da]
∫(a->x) f(t) dt = 2x^2-3x+1
=>
coef.of x^2
(1/2)c = 2
c=4
coef. of x
d= -3
coef. of constant
-[(1/2)ca^2+da] = 1
(1/2)ca^2+da =-1
2a^2 -3a +1 =0
(2a-1)(a-1)=0
a=1 or 1/2