Complete the square on the quadratic (21-4x- x^2) and hence show that
integration of
limit 0-->-2
(2/ (21-4x- x^2)) dx = (2/5)tanh-1(2/5)
唔明點樣complete the square之後轉到去tanh到~
上面所有x係代數
想問一題數學題有關Complete the square同 tanh既~
2008-08-20 8:30 am
回答 (2)
2008-08-24 10:52 pm
I guess your problem is you do not know the definition of the inverse hyberbolic function tanh-1(x)
Definition is as follows:
tanh-1(x) = 0.5 * [ ln |1+x| - ln|1-x| ]
where |x| <1
Now you should have no problem to complete this question
My answer is as follows:
http://www.box.net/shared/a64ail07qs
However, I noticed that my solution is negative of your required solution, so I want to know whether integration limit should be (-2 to 0) instead of (0 to -2)
Definition is as follows:
tanh-1(x) = 0.5 * [ ln |1+x| - ln|1-x| ]
where |x| <1
Now you should have no problem to complete this question
My answer is as follows:
http://www.box.net/shared/a64ail07qs
However, I noticed that my solution is negative of your required solution, so I want to know whether integration limit should be (-2 to 0) instead of (0 to -2)
2008-08-20 9:21 am
-(x^2)-4x+21=-(x+2)^2+25
SO,
S2/(-(x+2)^2+25))dx=2S(1/(-(x+2)^2+5^2)dx
Let 5sinY=x+2
5cosYdY=dx
SO,
2S(1/(-(x+2)^2+5^2)dx=2S(5cosY/25(cosY)^2)dY
=(2/5)SsecYdY
我唔係好知TAN H個H係邊黎- -'
REMARK:S=INTEGRATION個符號,即係果個蛇仔-3-'
2008-08-20 01:28:43 補充：
=(2/5)SsecYdY=(2/5)[ln|secY+tanY|](0to-2)
再REPLACE返個Y做X就應該得@.@
SO,
S2/(-(x+2)^2+25))dx=2S(1/(-(x+2)^2+5^2)dx
Let 5sinY=x+2
5cosYdY=dx
SO,
2S(1/(-(x+2)^2+5^2)dx=2S(5cosY/25(cosY)^2)dY
=(2/5)SsecYdY
我唔係好知TAN H個H係邊黎- -'
REMARK:S=INTEGRATION個符號,即係果個蛇仔-3-'
2008-08-20 01:28:43 補充：
=(2/5)SsecYdY=(2/5)[ln|secY+tanY|](0to-2)
再REPLACE返個Y做X就應該得@.@
收錄日期: 2021-05-03 11:38:54
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