(0,1)∫e^xdx=lim (n→∞)(1,n)∑e^(i/n)/n
注意e^(i/n)是个等比数列,公比为e^(1/n)
(1,n)∑e^(i/n)/n=1/n*[e^(n/n)-1]/[e^(1/n)-1]
=1/n*(e-1)/[e^(1/n-1)]
换元t=1/n
原式=lim (t→0) t(e-1)/(e^t-1)
洛必达法则
=lim (e-1)/e^t=e-1