∫f(x)= a？ lnx-x？ +ax, where x > 0

∴f'(x)=(a？ /x)-2x+a=-(x-a)(2x+a)/x

∫a > 0

∴ The monotonic increasing interval of f(x) is (0, a), and the monotonic decreasing interval of f (x) is (a, +∞).

②

Judging from the meaning of the question:

f( 1)=a- 1≥e- 1

That is, a≥e

It can be seen from ① that f(x) increases monotonically in [1, e].

Make e- 1 ≤ f (x) ≤ e? This applies to x∈[ 1, e]

As long as:

f( 1)=a- 1≥e- 1

f(e)=a？ -e？ +ae≤e？

Solution: a = e