y'=3x^2,
y'|(x=1)=3,
在(1,1)处切线方程:(y-1)/(x-1)=3,
即y=3x-2.
切线和X轴交点为(2/3,0)
S=∫[0,1]x^3dx-∫[2/3,1](3x-2)dx
=x^4/4[0,1]-(3x^2/2-2x] [2/3,1]
=1/4-(3/2-2-2/3+4/3)
=1/4-1/6
=1/12.
y'=3x^2,
y'|(x=1)=3,
在(1,1)处切线方程:(y-1)/(x-1)=3,
即y=3x-2.
切线和X轴交点为(2/3,0)
S=∫[0,1]x^3dx-∫[2/3,1](3x-2)dx
=x^4/4[0,1]-(3x^2/2-2x] [2/3,1]
=1/4-(3/2-2-2/3+4/3)
=1/4-1/6
=1/12.