or 0 + C?
or something else...?
駛唔駛 "+C"？...
個答案好似係 0、因為 ∫3 dx = 3*∫dx
但唔+C又好似唔make sense...
更新1:
駛唔駛+C？... 個答案好似係 0、因為 ∫3 dx = 3*∫dx 但唔+C又好似唔make sense...
Integration: ∫3 dx - 3*∫dx
2006-10-20 11:56 am
Is the answer 0?
or 0 + C?
or something else...?
駛唔駛 "+C"？...
個答案好似係 0、因為 ∫3 dx = 3*∫dx
但唔+C又好似唔make sense...
or 0 + C?
or something else...?
駛唔駛 "+C"？...
個答案好似係 0、因為 ∫3 dx = 3*∫dx
但唔+C又好似唔make sense...
回答 (4)
2006-10-20 12:06 pm
✔ 最佳答案
∫3 dx - 3*∫dx =∫3 dx - 3*∫1 dx
=3x+C1-3x-C2
=C1-C2
=C
應該加C
又
∫3 dx - 3*∫dx
=3*∫1 dx - 3*∫1 dx
=3(∫1 dx - ∫1 dx)
=3∫(1- 1) dx
=3C1
=C
2006-10-20 04:15:56 補充：
用文字說法就是因為不能肯定前面的常數等於後面的常數﹐相減後不能確定其值是0﹐所以要加C
2006-10-20 7:21 pm
∫3 dx - 3*∫dx
=∫3 dx - 3*∫1 dx
=3x+C1-3(x-C2) (where C1, C2 are arbitary constant)
=C1-3*C2
=C (since it is only an arbitary constant, we could change the variable)
=∫3 dx - 3*∫1 dx
=3x+C1-3(x-C2) (where C1, C2 are arbitary constant)
=C1-3*C2
=C (since it is only an arbitary constant, we could change the variable)
參考: A maths textbook
2006-10-20 7:01 pm
∫3 dx - 3*∫dx
=∫3 dx - 3*∫1 dx
=3x+C1-3x-C2
=C1-C2
=C
應該加C
又
∫3 dx - 3*∫dx
=3*∫1 dx - 3*∫1 dx
=3(∫1 dx - ∫1 dx)
=3∫(1- 1) dx
=3C1
=C
=∫3 dx - 3*∫1 dx
=3x+C1-3x-C2
=C1-C2
=C
應該加C
又
∫3 dx - 3*∫dx
=3*∫1 dx - 3*∫1 dx
=3(∫1 dx - ∫1 dx)
=3∫(1- 1) dx
=3C1
=C
參考: me
2006-10-20 12:09 pm
係要plus C架
因為如果work backwards
for eg C = 2,
d/dx 2 都係得返個零
所以唔可以排除非零既answer :D
因為如果work backwards
for eg C = 2,
d/dx 2 都係得返個零
所以唔可以排除非零既answer :D
收錄日期: 2021-04-23 12:34:43
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