请教高数高手: ∫(π⼀2,0)xsinxdx,怎么解?

2024-11-07 21:02:04
有4个网友回答
网友(1):

∫(π/2,0)xsinxdx
=∫(π/2,0) x d (-cosx)

= -xcosx(代入积分限后,为0) + ∫(π/2,0) cosx dx

=1

网友(2):

∫[0→π/2] xsinx dx
= ∫[π/2→0] x dcosx
= xcosx |[π/2→0] + ∫[0→π/2] cosx dx
= [sinx] |[0→π/2]
= 1

网友(3):

网友(4):

楼上做的很好啊