Question

设f(x)=ax+b-lnx,在【1,3】上f(x)＞=0,求常数a,b使∫(1,3)f(x)dx最小

Answer 1

简单分析一下，详情如图所示

Answer 2

f(x)=ax+b-lnx,依题意f(1)=a+b>=0,f(3)=3a+b-ln3>=0,g(a,b)=∫f(x)dx=[(1/2)ax^+bx-xlnx+x]|=4a+2b-3ln3+3,当a+b=0,3a+b=ln3,即a=(1/2)ln3,b=(-1/2)ln3时g(a,b)取最小值3-2ln3.