Hi,
Can somebody answer and explain the following?
1) ∫(x^3/2 + 1/x - 12)dx
2) ∫(4 + 4Ø + Ø^2)dØ
Each 'x' is the symbol 'x', not a times sign.
Thanks.
Mathematics - Integration?
2016-06-22 9:56 am
回答 (2)
2016-06-22 1:28 pm
✔ 最佳答案
1) ∫(x^3/2 + 1/x - 12)dx = ∫x^(3/2) dx + 1∫ dx/x -12 ∫ dx
= x^(3/2 +1) /(3/2 +1) + ln |x| - 12 x
= (2/5) x^(5/2)+ ln |x| - 12 x + C
2) ∫(4 + 4Ø + Ø^2)dØ
∫ 4 dØ + 4∫ Ø dØ + ∫Ø^2 dØ
= 4 Ø + (4)(1/2) Ø^2 + (1/3) Ø^3
= 4Ø + 2Ø^2 + (1/3)Ø^3 + C
2016-06-22 10:03 am
1)
∫[x^(3/2) + (1/x) - 12] dx
= ∫x^(3/2) dx + ∫(1/x) dx - ∫12 dx
= (2/5) x^(5/2) + ln|x| - 12 x + C
====
2)
∫(4 + 4Ø + Ø^2)dØ
= ∫4 dØ + ∫4Ø dØ + ∫Ø^2 dØ
= 4 Ø + 4*(1/2)*Ø^2 + (1/3) Ø^3 + C
= 4 Ø + 2 Ø^2 + (1/3) Ø^3 + C
∫[x^(3/2) + (1/x) - 12] dx
= ∫x^(3/2) dx + ∫(1/x) dx - ∫12 dx
= (2/5) x^(5/2) + ln|x| - 12 x + C
====
2)
∫(4 + 4Ø + Ø^2)dØ
= ∫4 dØ + ∫4Ø dØ + ∫Ø^2 dØ
= 4 Ø + 4*(1/2)*Ø^2 + (1/3) Ø^3 + C
= 4 Ø + 2 Ø^2 + (1/3) Ø^3 + C
收錄日期: 2021-04-18 15:11:50
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