Question

0到e 的定积分e ^√x

Answer 1

令√x=t,x=t^2,dx=2tdt.故S(0,e)e^√xdx=S(0,√e)e^t *2tdt=2S(0,√e)td(e^t) =2[te^t(0,√e)-S(0,√e)e^tdt ]=2[(t-1)e^t](0,√e)=2[(√春察e-1)e^(√e)]+2e^(√e)=2√e*e^(√则困e) =2e^[(√孙森念e)+1/2]