HMMT 二月 2008 · CALC 赛 · 第 3 题
HMMT February 2008 — CALC Round — Problem 3
题目详情
英文原题
- [ 4 ] Find all y > 1 satisfying x ln x dx = .
(ln(2 − x ))214
解析
英文解析
- [ 4 ] Find all y > 1 satisfying x ln x dx = .
√14
Answer: e Applying integration by parts with u = ln x and v = x , we get 21
∫ ∫2
y ∣ yy
1 1 1 1 1
∣
2 2 2
x ln x dx = x ln x − x dx = y ln y − y + .
∣
2 2 2 4 4
1 11
√
2 1 2 1
So y ln y = y . Since y > 1, we obtain ln y = , and thus y = e .
2 2
(ln(2 − x ))2