Question

求积分的题

Answer 1

f(x)
= 1/(1+e^x) : x<0
=1/(1+x) ; x≥0
let
u=x-1
du=dx
x=0, u=-1
x=2, u=1
∫(0->2) f(x-1) dx
=∫(-1->1) f(u) du
=∫(-1->0) du/(1+e^u) +∫(0->1) du/(1+u)
=ln2 +∫(-1->0) du/(1+e^u)
=ln2 -ln2 +ln(e+1)
=ln(e+1)
let
y=e^u
dy= e^u du
u=-1, y=1/e
u=0, y=1
∫(-1->0) du/(1+e^u)
=∫(1/e->1) dy/[y(1+y) ]
=[ ln|y/(y+1)|] |(1/e->1)
=ln(1/2) - ln(1/(e+1))
=-ln2 +ln(e+1)