HMMT 二月 2002 · CALC 赛 · 第 6 题
HMMT February 2002 — CALC Round — Problem 6
题目详情
英文原题
- 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
英文解析
- 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).
√
Solution: 3 The area of the rectangle is a + a . The portion under the parabola has 3
∫
2 3 3 3 aarea x + 1 dx = a / 3 + a . Thus we wish to solve the equation a + a = 2( a / 3 + a );
√0
dividing by a and rearranging gives a / 3 = 1, so a = 3.2 1