P 1920 55 假设将薄板沉到水中,深为h处,此时薄板的曲线方程为 h y 中=x·2y·dx=6 由题设知 dx=2×1920,即[x dx=640 5 (x-h)2+-(x-h) h 4.h=12 i.由图知抛物线方程为y=3 于是 dx 20-xx 6 1280 假设将薄板沉到水中,深为h处,此时薄板的曲线方程为 20+h-x 20+h-x dx=6 dx 由题设 20+h-x 5dx=2×1280,即 x(20+h-x)2dx=2560 +20 0+h)(20+h h -12+20+h=16.h=18 四.证明题 tf(o)dt 设f(x)为连续正值函数,证明当x≥0时函数d(x)= 单调增加 f(odt 证明.φ'(x) x/0M-M|(x(x=0/0 0 sA(dt f(odtdx x dp x y dx 5 2 6 2 3 = × × = 1920 0 20 5 2 5 6 5 6 2 20 5 0 2 3 = = × × = Ú dx x x p 假设将薄板沉到水中, 深为 h 处, 此时薄板的曲线方程为 5 3 x h y - = , dx x h dp x y dx x 5 2 6 - = × × = 由题设知 2 1920 5 6 20 = ¥ - Ú h+ h dx x h x , 即 640 5 20 = - Ú h+ h dx x h x ( ) 640 3 2 ( ) 5 2 5 1 20 2 3 2 5 ˙ = ˚ ˘ Í Î È - + - h+ h x h x h 4 3 = h , h = 12 ii. 由图知抛物线方程为 5 20 3 x y - = . 于是 dx x dp x y dx x 5 20 2 6 - = × × = 1280 0 20 (20 ) 3 2 5 120 0 20 (20 ) 5 2 5 6 20 5 6 2 3 2 20 5 0 = - = × × - - × - = Ú p x x dx x x 假设将薄板沉到水中, 深为 h 处, 此时薄板的曲线方程为 5 20 3 h x y + - = , dx h x dp x y dx x 5 20 2 6 + - = × × = 由题设知 2 1280 5 20 6 20 = ¥ + - Ú h+ h dx h x x , 即 (20 ) 2560 5 6 20 2 1 + - = Ú h+ h x h x dx h h h x 20 (20 ) 5 2 5 6 2 5 + × + - h h h h x 20 (20 )(20 ) 3 2 5 6 2 3 + - × + + - -12 + 20 + h = 16, h = 18 四. 证明题 1. 设 f(x)为连续正值函数, 证明当 x ³ 0 时函数 Ú Ú = x x f t dt tf t dt x 0 0 ( ) ( ) f ( ) 单调增加. 证明. 0 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ' ( ) 2 0 0 2 0 0 0 > ˙ ˚ ˘ Í Î È - = ˙ ˚ ˘ Í Î È ˙ ˚ ˘ Í Î È - = Ú Ú Ú Ú Ú x x x x x f t dt f x x t f t dt f t dt f x x f t dt tf t dt f x