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HMMT 二月 2008 · CALC 赛 · 第 8 题

HMMT February 2008 — CALC Round — Problem 8

专题
Discrete Math / 离散数学
难度
L3
来源
HMMT

题目详情

  1. [ 7 ] Let T = dx . Evaluate e . 3 x 2 x x e + e − e + 1 0 1 ( ) 1 1 − 1+ 1 2 n 2 ( ) n 2 n
解析
  1. [ 7 ] Let T = dx . Evaluate e . 3 x 2 x x e + e − e + 1 0 11 x Answer: Divide the top and bottom by e to obtain that 4 ∫ ln 2 2 x x − x 2 e + e − e T = dx 2 x x − x e + e − 1 + e 0 2 x x − x 2 x x − x Notice that 2 e + e − e is the derivative of e + e − 1 + e , and so ( ) ( ) [ ] ln 2 1 11 2 x x − x T = ln | e + e − 1 + e | = ln 4 + 2 − 1 + − ln 2 = ln 0 2 4 T 11 Therefore, e = . 4 1 ( ) 1 1 − 1+ 1 2 n 2 ( ) n 2 n