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Old 27th May 2012, 08:17 PM   #1 (permalink)
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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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Old 1st June 2012, 06:47 PM   #2 (permalink)
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Re: Integration question

This.

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Old 1st June 2012, 10:58 PM   #3 (permalink)
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Re: Integration question

I doubt you are correct...

You cannot divide by 18x-1

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Old 1st June 2012, 11:09 PM   #4 (permalink)
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Re: Integration question

Originally Posted by .Memo View Post
I doubt you are correct...

You cannot divide by 18x-1
I don't understand what you mean by that..

y 18a^-1
= y (18/a)
= y x (a/18)

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Old 2nd June 2012, 12:06 AM   #5 (permalink)
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Re: Integration question

If I'm not wrong, I think,
∫((16x-9)/x
Let u=16x-9, du/dx=32x, xdx=1/32du
∫((16x-9)/x dx = 1/32∫1/u du
1/32 ln |u| + C
1/32 ln (16x-9) + C

Hope that helps

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Old 2nd June 2012, 12:23 AM   #6 (permalink)
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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 u + C
=ln (e^x+2) + C

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Old 2nd June 2012, 12:35 AM   #7 (permalink)
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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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Old 2nd June 2012, 01:05 AM   #8 (permalink)
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Re: Integration question

Originally Posted by Shohie View Post
If I'm not wrong, I think,
∫((16x-9)/x
Let u=16x-9, du/dx=32x, xdx=1/32du
∫((16x-9)/x dx = 1/32∫1/u du
1/32 ln |u| + C
1/32 ln (16x-9) + C
Originally Posted by Shohie View Post
(a) ∫1/e^x+2 dx
Let u=e^x+2, du=e^x dx
∫1/e^x+2 dx = ∫du/u
=ln u + C
=ln (e^x+2) + C
I have no idea how you got this.

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Old 2nd June 2012, 01:19 AM   #9 (permalink)
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Re: Integration question

Originally Posted by .Memo View Post
I have no idea how you got this.
Precisely, Shohie, I think you may have substituted incorrectly.

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Old 2nd June 2012, 11:32 AM   #10 (permalink)
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Re: Integration question

Originally Posted by .Memo View Post
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?

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Old 2nd June 2012, 11:33 AM   #11 (permalink)
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Re: Integration question

Originally Posted by Betakuwe View Post
Precisely, Shohie, I think you may have substituted incorrectly.
Then why don't you give the correct answer If I answer wrongly..

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Old 2nd June 2012, 05:59 PM   #12 (permalink)
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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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Old 2nd June 2012, 06:15 PM   #13 (permalink)
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Re: Integration question

Originally Posted by Shohie View Post
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?

∫((16x-9)/x
Let u=16x-9, du/dx=32x, xdx=1/32du
∫((16x-9)/x dx = 1/32∫1/u du
1/32 ln |u| + C
1/32 ln (16x-9) + C
What happened to the denominator x and the sqrt?

Anyway this is wrong, differentiating 1/32 ln (16x-9) definitely does not give (16x-9)^/x.

This is no secondary school maths though, I believe TS is in JC.

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Old 2nd June 2012, 08:25 PM   #14 (permalink)
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Re: Integration question

Here is my solution... Not so sure if i am correct >

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Old 2nd June 2012, 09:00 PM   #15 (permalink)
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Re: Integration question

Originally Posted by megaman123 View Post
Here is my solution... Not so sure if i am correct >
ermmmm... how do you get the formula?

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Old 2nd June 2012, 10:17 PM   #16 (permalink)
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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!!!).

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Old 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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Old 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
sec-1 (12) = error

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Old 3rd June 2012, 12:22 AM   #19 (permalink)
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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^2-9)^0.5/x

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Old 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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