设直线y=ax(a

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设直线y=ax(a
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设直线y=ax(a
设直线y=ax(a<1)与抛物线y=x2所围成的图形面积为s,它们与直线x=1围成的面积为T
(1)求函数u(a)=S+T的解析式 (2)若u(a)达到最小值,求a的值

设直线y=ax(a
三条线交点为(0,0),(a,a^2), (1,1),(1/a,1)
S = (0->a)积分(ax-x^2) = a^3 / 6
T = (a^2->1)积分(y/a - 根号(y))dy = 1/(2a)-a^3 / 6 - 2/3
u(a) = 1/2a - 2/3
a = 3/4 时,u(a)= 0