Question

大一高数定积分？

求第5题详解！

Answer 1

n∫(0->1) xf'(2x) dx =∫(0->2) xf'(x) dx
let
x=2u
dx = 2du
x=0, u=0
x=2, u=1
∫(0->2) xf'(x) dx
=∫(0->1) 2u .f'(2u) (2du)
=4∫(0->1) u .f'(2u) du
=4∫(0->1) x .f'(2x) dx
=>
n=4