∫(2,0)(ax+b) dx=(0.5ax^2+bx)|(2,0)=2a+2b=1
即:a+b=0.5 (1)
又:∫(2,1)(ax+b)dx=(0.5ax^2+bx)|(2,1)=2a+2b-0.5a-b=1.5a-b=0.25
即:1.5a-b=0.25 (2)
解出:a=0.3 b=0.2
f(x)=0.3x+0.2 [0,2] 其它为0.
∫(2,0)(ax+b) dx=(0.5ax^2+bx)|(2,0)=2a+2b=1
即:a+b=0.5 (1)
又:∫(2,1)(ax+b)dx=(0.5ax^2+bx)|(2,1)=2a+2b-0.5a-b=1.5a-b=0.25
即:1.5a-b=0.25 (2)
解出:a=0.3 b=0.2
f(x)=0.3x+0.2 [0,2] 其它为0.