integration question?
回答 (2)
2008-01-16 11:30 pm
✔ 最佳答案
π/2∫3sin2x dx
0
π/2
=3∫sin2x dx
0
π/2
=[-3/2(cos2x)+C]
0
=--3/2 cos(2 x (π/2))-(-3/2)cos 0
= -3/2 cosπ-(-3/2)cos 0
= -3/2(-1) -(-3/2)(1)
= 3/2 + 3/2
= 3
2008-01-15 9:39 pm
∫3sin2x dx
=3∫sin2x dx
=(-3/2)(cos2x)+C
So the area under the curve 3sin2x between x=0 ,x=π/2
=(-3/2)(cosπ)-(-3/2)(cos0)
=(3/2)+(3/2)
=3
=3∫sin2x dx
=(-3/2)(cos2x)+C
So the area under the curve 3sin2x between x=0 ,x=π/2
=(-3/2)(cosπ)-(-3/2)(cos0)
=(3/2)+(3/2)
=3
收錄日期: 2021-04-25 16:58:04
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20080115000051KK01006