∫(0~1)2|x^2-4|dx定积分

详细过程谢谢
2024-12-23 06:27:04
推荐回答(3个)
回答1:

0-1之间x^2-4<0,所以
∫(0~1)2|x^2-4|dx = ∫(0~1)2(4-x^2)dx = 8x-2x^3/3|(0,1) = 8-2/3 = 22/3

回答2:

积分上下限为[1,0],x^2-4<0,被积函数变成:-(X^2-4):
∫(0~1)2|x^2-4|dx
=- ∫(0~1) (x^2-4)dx
=- [x^3/3-4x]| (x:1,0) //: 上限代入1;下限代入0.
=-[1/3-4-0]
=4-1/3
=11/3

回答3:

∫(0~1)2|x^2-4|dx
=[(1/3)x^3-4x]|(0~1)
=-4+1/3-0
=-(11/3)