HMMT 二月 2002 · CALC 赛 · 第 6 题
HMMT February 2002 — CALC Round — Problem 6
题目详情
- Determine the positive value of a such that the parabola y = x + 1 bisects the area of 2 2 the rectangle with vertices (0 , 0) , ( a, 0) , (0 , a + 1), and ( a, a + 1).
解析
- Determine the positive value of a such that the parabola y = x + 1 bisects the area 2 2 of the rectangle with vertices (0 , 0) , ( a, 0) , (0 , a + 1), and ( a, a + 1). √ 3 Solution: 3 The area of the rectangle is a + a . The portion under the parabola has ∫ a 2 3 3 3 area x + 1 dx = a / 3 + a . Thus we wish to solve the equation a + a = 2( a / 3 + a ); 0 √ 2 dividing by a and rearranging gives a / 3 = 1, so a = 3. 1