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)