Question

计算积分,∫f(x-1)dx{∫上面为2,下面为0},其中f(x)=1/(1+x){x>=0},f

(x)=1/(1+e^x){x

Answer 1

令x-1=t
dx=dt
∫2 0 f(x-1)dx
=∫1 -1 f(t)dt
=∫0 -1 f(t)dt + ∫1 0 f(t)dt
=∫0 -1 dt/(1+e^t) + ∫1 0 dt/(1+t)
=[t-ln(1+e^t)]0 -1 + [ln(1+x)]1 0
=1+ln(1+1/e)+ln2
=1+ln(2+2/e)