Question

由y=3-x^2,y=2x,所围成的面积.用定积分,

Answer 1

y=√x和y=x解得x=0,x=1 题目变成定积分∫[0,1](√x-x)dx=[2/3x^(3/2)-1/2x^2][0,1]=1/6 y=2x和y=3-x^2解得x=-3,x=1