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更新1:
麻辣大大 我有個地方不理解 F(x)=∫(-1~0)x^2*dx+∫(0~x)2xdx =x^3/3(-1~0)+x^2(0~x) =(-1/3-0)+(x^2-0) 您的答案 =(0-(-1/3))+(x^2-0)我的理解 這樣算出來 答案是跟第3位大大一樣
微積分求解2
回答 (11)
2014-05-26 4:21 pm
✔ 最佳答案
If x<= 0, then
F(x)
= ∫(-1~x) y^2 dy
= [(1/3)*y^3] (-1~x)
= (1/3)(x^3 + 1)
If x > 0, then
F(x)
= ∫(-1~0) y^2 dy + ∫(0~x) 2y dy
= [(1/3)*y^3] (-1~0) + [y^2] (0~x)
= 1/3 + x^2
2014-06-21 10:33 pm
2014-06-20 9:46 pm
2014-06-11 12:51 am
2014-06-05 2:09 pm
2014-06-03 2:59 pm
2014-05-30 4:16 pm
2014-05-27 11:44 am
2014-05-26 3:04 pm
...............Sol
(1) x<－1
F(x)=0
(2) －1<=x<=0
F(x)= ∫(－1 to x)_y^2dy=y^3/3｜(－1 to x)
=(x^3/3)+1/3
(3) 0<x
F(x)= ∫(－1 to x)_f(y)dy
= ∫(－1 to 0)_f(y)dy+ ∫(0 to x)_f(y)dy
= ∫(－1 to 0)_y^2dy+ ∫(0 to x)_2ydy
=x^2－1/3
(1) x<－1
F(x)=0
(2) －1<=x<=0
F(x)= ∫(－1 to x)_y^2dy=y^3/3｜(－1 to x)
=(x^3/3)+1/3
(3) 0<x
F(x)= ∫(－1 to x)_f(y)dy
= ∫(－1 to 0)_f(y)dy+ ∫(0 to x)_f(y)dy
= ∫(－1 to 0)_y^2dy+ ∫(0 to x)_2ydy
=x^2－1/3
2014-05-26 2:27 pm
F(x)=∫(-1~0)x^2*dx+∫(0~x)2xdx=x^3/3(-1~0)+x^2(0~x)=(-1/3-0)+(x^2-0)=x^2-1/3
收錄日期: 2021-05-02 10:40:12
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20140526000010KK00810