二者的交点为原点和A(1,1)
不定积分为∫(√x - x)dx = (2/3)x^(3/2) - x²/2
所围成图形的面积为为0,1之间的定积分=[(2/3)1^(3/2) - 1²/2 ] - [(2/3)0^(3/2) - 0²/2]
= 2/3 - 1/2
= 1/6
计算由直线y＝x,曲线y2＝x所围成图形的面积
二者的交点为原点和A(1,1)
不定积分为∫(√x - x)dx = (2/3)x^(3/2) - x²/2
所围成图形的面积为为0,1之间的定积分=[(2/3)1^(3/2) - 1²/2 ] - [(2/3)0^(3/2) - 0²/2]
= 2/3 - 1/2
= 1/6