Question

一道不定积分题 求解求详细

求k 要有过程我不是很会做求指导谢谢

Answer 1

=∫ (0,2) ∫(2,4) k(6-x-y)dxdy
=∫(0,2)(-kx^2/2+(6k-y)x) [2.4] dy
=∫(0,2) (-16k/2+(6k-y)4+4k/2-(6k-y)2)dy
=∫(0,2) (-8k+24k-4y+2k-12k+2y)dy
=∫(0,2) (6k-2y)dy
=6ky-y^2 [0,2]
=12k-4-0-0
=12k-4

12k-4=1
k=1/3