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Integration question
How do I do part (b). I've been stuck at it for hours, even after looking at all the different types of integration formulas...
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1st June 2012, 06:47 PM

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Re: Integration question
This.
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1st June 2012, 10:58 PM

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Re: Integration question
I doubt you are correct...
You cannot divide by 18x1
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1st June 2012, 11:09 PM

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Re: Integration question
Originally Posted by .Memo I doubt you are correct...
You cannot divide by 18x1 I don't understand what you mean by that..
y ÷ 18a^1
= y ÷ (18/a)
= y x (a/18)
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2nd June 2012, 12:06 AM

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Re: Integration question
If I'm not wrong, I think,
∫((16x²9)½/x
Let u=16x²9, du/dx=32x, xdx=1/32du
∫((16x²9)½/x dx = 1/32∫1/u du
1/32 ln u + C
1/32 ln (16x²9) + C
Hope that helps 
 
2nd June 2012, 12:23 AM

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Re: Integration question
(a) ∫1/e^x+2 dx
Let u=e^x+2, du=e^x dx
∫1/e^x+2 dx = ∫du/u
=ln u² + C
=ln (e^x+2)² + C

 
2nd June 2012, 12:35 AM

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Re: Integration question
wow after reading the answers i suppose i have to declare i totally dun understand anymore.all pass back to teacher.
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2nd June 2012, 01:05 AM

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Re: Integration question
Originally Posted by Shohie If I'm not wrong, I think,
∫((16x²9)½/x
Let u=16x²9, du/dx=32x, xdx=1/32du ∫((16x²9)½/x dx = 1/32∫1/u du
1/32 ln u + C
1/32 ln (16x²9) + C
Originally Posted by Shohie (a) ∫1/e^x+2 dx
Let u=e^x+2, du=e^x dx ∫1/e^x+2 dx = ∫du/u
=ln u² + C
=ln (e^x+2)² + C I have no idea how you got this.
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2nd June 2012, 01:19 AM

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Re: Integration question
Originally Posted by .Memo I have no idea how you got this. Precisely, Shohie, I think you may have substituted incorrectly.
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2nd June 2012, 11:32 AM

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Re: Integration question
Originally Posted by .Memo I have no idea how you got this. this is how we answer integration in business maths..I can't rmb how you guys in secondary answer it..maybe you help answering if you know the correct answer?

 
2nd June 2012, 11:33 AM

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Re: Integration question
Originally Posted by Betakuwe Precisely, Shohie, I think you may have substituted incorrectly. Then why don't you give the correct answer If I answer wrongly.. 
 
2nd June 2012, 05:59 PM

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Re: Integration question
Tough nut to crack, I don't know if I done it correctly.
Tried using integration by parts but became too complex...
Used trigo substitution in the end.
Pardon me if any stupid mistakes, its been 2.5 years since I last touched maths. Thanks to NS
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Last edited by aforl; 2nd June 2012 at 06:01 PM.

 
2nd June 2012, 06:15 PM

#13 (permalink)
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Re: Integration question
Originally Posted by Shohie this is how we answer integration in business maths..I can't rmb how you guys in secondary answer it..maybe you help answering if you know the correct answer?
∫((16x²9)½/x
Let u=16x²9, du/dx=32x, xdx=1/32du
∫((16x²9)½/x dx = 1/32∫1/u du
1/32 ln u + C
1/32 ln (16x²9) + C What happened to the denominator x and the sqrt?
Anyway this is wrong, differentiating 1/32 ln (16x²9) definitely does not give (16x²9)^½/x.
This is no secondary school maths though, I believe TS is in JC.
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Last edited by aforl; 2nd June 2012 at 06:17 PM.

 
2nd June 2012, 08:25 PM

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Re: Integration question
Here is my solution... Not so sure if i am correct > 
 
2nd June 2012, 09:00 PM

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Re: Integration question
Originally Posted by megaman123 Here is my solution... Not so sure if i am correct > ermmmm... how do you get the formula?
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2nd June 2012, 10:17 PM

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Re: Integration question
Hahas, my method kinda complicated, but i no want use substitution, cos i find it a little cheap (but it is a very good and useful technique!!!).

 
2nd June 2012, 11:45 PM

#17 (permalink)
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Re: Integration question
Wow, I like your solution more than mine, megaman.
But wonder if which is correct? Looks extremely different
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Last edited by aforl; 2nd June 2012 at 11:45 PM.

 
3rd June 2012, 12:12 AM

#18 (permalink)
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Re: Integration question
I tried to check the answer using a GC...
so i try putting lower limit 2, upper limit 3. *type directly from the question*
answer i got is ... 3.807601993
take note we cannot use 0.75 < x < 0.75 otherwise we will get an error. *can't square root negative no*
assuming megaman123's answer is correct.
just substituting 3 in his answer, we get an error already... o.o
sec1 (12) = error
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3rd June 2012, 12:22 AM

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Re: Integration question
Oops, i made a bad bad mistake, my answer is not correct. But differentiating all our solutions don't give us back to (16x^29)^0.5/x
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3rd June 2012, 12:26 AM

#20 (permalink)
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Re: Integration question
Megaman solution:
Sec Inv (4x/3),
Domain of x is −1 ≤ 4x/3 ≤ 1
So, 3/4 ≤ x ≤3/4.
However the square root means anything between x cannot be 0.75 < x < 0.75..
Means the only 2 possible values is 0.75 and 0.75? Lol?
Someone check mine too?? (I dont have a GC with me)
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Last edited by aforl; 3rd June 2012 at 12:32 AM.

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