Question

高等数学。。在线等。。很急

Answer 1

令X=e^t，则t=lnx∈[0,1] 因为x∈[1，e]

∫t*e^td(e^t)=∫t*e^(2t)dt 再用分部积分法
=1/2∫td(e^2t)
=1/2[te^2t](从0到1）-1/2∫e^2tdt
=1/4e^2+1/4

注：上述积分符号的积分上下限为[0,1]，由于打不来，只能这样了，纯手打！

Answer 2

1/2e*2-e+1

Answer 3

分部积分呗

Answer 4

∫<1,e> xlnxdx = (1/2)∫<1, e>lnxdx^2
= (1/2)[x^2lnx]<1,e> - (1/2)∫<1,e>xdx
= (1/2)e^2 - (1/4)[x^2]<1,e>
= (1/2)e^2 - (1/4)(e^2-1) =(1/4)(e^2+1)