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