Question

高数,定积分求函数f(x)=∫(2t-1)/t²-t+1 dt 在[0,2]上的最大值和最小值.

Answer 1

∵f(x)=∫(2t-1)/(t²-t+1)dt
=∫d(t²-t+1)/(t²-t+1)
=[ln(t²-t+1)]│
=ln(x²-x+1)-ln1
∴f(x)=ln(x²-x+1)
∵f‘(x)=(2x-1)/(x²-x+1)
令f‘(x)=0,则x=1/2
∴函数f(x)在区间[0,2]上,只能在x=0,x=1/2,x=2三点上取得最大值和最小值
∵f(0)=ln(0²-0+1)=0
f(1/2)=ln((1/2)²-(1/2)+1)=ln(3/4)
f(2)=ln(2²-2+1)=ln3
又ln(3/4)