f(x)=x+a,f(x)=x+∫f(t)dt(上限2下限0),a=

题目

f(x)=x+a,f(x)=x+∫f(t)dt(上限2下限0),a=

答案

f(x)=x+a
∫(0->2)f(t)dt
=∫(0->2)(t+a)dt
=∫(0->2)tdt+∫(0->2)adt
=t^2/2|(0->2)+at|(0->2)
=2+2a
f(x)=x+∫f(t)dt(上限2下限0)
=x+2+2a=x+a
2+2a=a
a=-2