Question

帮忙求一第三题，要详解

Answer 1

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