【答案】-4

11.已知f(x)={■(2x+1"," x"∈[-" 2"," 2"]," @1+x^2 "," x"∈(" 2"," 4"]," )┤ ∫_k^3▒ f(x)dx=40/3恒成立,求k的值.

　　【解析】分2

　　当2

　　整理得k3+3k+4=0,即k3+k2-k2+3k+4=0,∴(k+1)(k2-k+4)=0,解得k=-1.又2≤k<3,∴k=-1(舍去).

　　当-2≤k≤2时,∫_k^3▒f "(" x")" dx=∫_k^2▒ (2x+1)dx+∫_2^3▒ (1+x2)dx

　　=(x2+x)|■( ^2@ _k )┤+(x+x^3/3)|■( ^3@ _2 )┤

　　=(4+2)-(k2+k)+(3+9)-(2+8/3)

　　=40/3-(k2+k)=40/3,

　　∴k2+k=0,即k=0或k=-1.

　　综上所述,k=0或k=-1.