Question

f（x）=x^2 x＜=0 1+2x x＞0 求积分∫0--＞2[f（x-1）]dx

Answer 1

f（x）=x^2 x＜=0

1+2x x＞0

∫0--＞2[f（x-1）]dx

令x-1=t

x=0,t=-1;

x=2,t=1

原式=∫(-1,1)f(t)dt

=∫(-1,0)t²dt+∫(0,1)(1+2t)dt

=t³/3|(-1,0)+(t+t²)|(0,1)

=0-(-1/3)+1+1-0

=1/3+2

=7/3