HMMT 二月 2008 · CALC 赛 · 第 8 题
HMMT February 2008 — CALC Round — Problem 8
题目详情
英文原题
- [ 7 ] Let T = dx . Evaluate e .
3 x 2 x xe + e − e + 1
( )10
1 1 − 1+
1 2 n 2
( )
2 nn
解析
英文解析
- [ 7 ] Let T = dx . Evaluate e .
3 x 2 x xe + e − e + 1
11 x 0
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 − xe + e − 1 + e
2 x x − x 2 x x − x 0
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 = .
( )14
1 1 − 1+ 1 2 n 2
( )
2 nn