HMMT 二月 2007 · 冲刺赛 · 第 33 题
HMMT February 2007 — Guts Round — Problem 33
题目详情
- [ 18 ] Compute
∫
9 x + 42
dx.
5 2
x + 3 x + x
(No, your TI-89 doesn’t know how to do this one. Yes, the end is near.)1
10 HARVARD-MIT MATHEMATICS TOURNAMENT, 24 FEBRUARY 2007 — GUTS ROUNDth
英文原题
[ 18 ] Compute ∫ 2
1
9 x + 4
x 5 + 3 x 2 + x dx.
(No, your TI-89 doesn’t know how to do this one. Yes, the end is near.)
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10 th HARVARD-MIT MATHEMATICS TOURNAMENT, 24 FEBRUARY 2007 — GUTS ROUND
解析
英文解析
- [ 18 ] Compute
∫
9 x + 42
dx.
5 2
x + 3 x + x
(No, your TI-89 doesn’t know how to do this one. Yes, the end is near.)1
Answer: ln . We break the given integral into two pieces:80
∫ ∫ ∫23
2 2 2
4 4
9 x + 4 x + 3 x + 1 5 x + 6 x + 1
dx = 5 dx − dx.
5 2 5 2 5 2
x + 3 x + x x + 3 x + x x + 3 x + x
1 1 1
These two new integrals are easily computed; for, the first integrand reduces to 1 /x and the second is 11
′
of the form f ( x ) /f ( x ). We obtain
[ ]
5 2280
5 ln | x | − ln | x + 3 x + x | = ln 32 − ln 46 + ln 5 = ln
5 2 4231
Motivation. Writing f ( x ) = 9 x + 4 and g ( x ) = x + 3 x + x = x ( x + 3 x + 1), we wish to findthe antiderivative of f ( x ) /g ( x ). It makes sense to consider other rational functions with denominatorg ( x ) that have an exact antiderivative. Clearly, if the numerator were f ( x ) = x + 3 x + 1 or a 4
constant multiple, then we can integrate the function. Another trivial case is if the numerator were 1
′ 4
f ( x ) = g ( x ) = 5 x + 6 x + 1 or a constant multiple. Guessing that f ( x ) is a linear combination off ( x ) and f ( x ), we easily find that f ( x ) = 9 x + 4 = 5 f ( x ) − f ( x ).2
1 2 1 2
10 HARVARD-MIT MATHEMATICS TOURNAMENT, 24 FEBRUARY 2007 — GUTS ROUNDth