HMMT 二月 2007 · CALC 赛 · 第 3 题

HMMT February 2007 — CALC Round — Problem 3

[ 4 ] Let a be a positive real number. Find the value of a such that the definite integral 2 ∫ a d x √ x + x a achieves its smallest possible value. x 2

解析

[ 4 ] Let a be a positive real number. Find the value of a such that the definite integral 2 ∫ a d x √ x + x a achieves its smallest possible value. √ Answer: 3 − 2 2 . Let F ( a ) denote the given definite integral. Then 2 ∫ a d d x 1 1 ′ F ( a ) = √ = 2 a · √ − √ . 2 2 d a x + x a + a a + a a √ √ √ √ ′ 2 Setting F ( a ) = 0, we find that 2 a + 2 a = a + 1 or ( a + 1) = 2. We find a = ± 2 − 1, and √ √ √ 2 because a > 0, a = ( 2 − 1) = 3 − 2 2. x 2