8.计算j01 e
dx .
解:先用换元法:令
= t ,则 x = t2 ,dx = 2tdt ,且
当x = 0 时, t = 0 ;当 x = 1 时, t = 1 .
换元后分部积分, j01 e
dx = 2j01 tet dt = 2j01 td(et )= 2([tet ]null - j01 et dt)
= 2(e - [et ] )null = 2[e - (e - 1)]= 2
2.求定积分j14
dx
解:令
= t ,则 x = t2 ,dx = 2tdt .x = 1 不 t = 1, x = 4 不 t = 2
原式=2j12 dt -( 2)j12
dt =2 - 2[ln(1 + t)]null =2 + 2 ln