先求由y=x^2,y=0和x=1所围成的图形面积:
```````1(积分上限)
S1 =∫x^2 dx = 1/3
```````0(积分下限)
求由y=t,y=0,x=0,x=1所围成的长方形的面积:
S2 =t
求S1和S2图形相交的图形的面积:
````````t^(1/2)(积分上限)
S3 = ∫x^2 dx +(1-t^2)*t=t-(2/3)*t^(3/2)
````````0(积分下限)
则所求的面积为:
S = S1+S2-2*S3=1/3+(4/3)*t^(3/2)-t
求S的导数:S'=2*t^(1/2)-1
令S'=0 =》 唯一的驻点:t=1/4
当t0
即S的图象是先递减后递增
所以在t=1/4处取得最小值.
S=1/4
选B