C4 Integration Help - 2 Questions Watch

TheKevinFang
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https://onedrive.live.com/redir?resi...hint=folder%2c

Hi, I'm stuck on Question 11 and Question 4 attached above...

I would like tips on how to complete/do question 11, but for question 4, if you could give a solution to it, as I get negative ln values at the end, that would be appreciated.

Thanks.
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B_9710
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For Q11  \displaystyle y^2 = \frac{x^2}{x^3+4}

You seem to have claimed that  (a+b)^{1/2} = a^{1/2} + b^{1/2} which is not true.
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B_9710
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Have you remembered that  \displaystyle \int \frac{f'(x)}{f(x)} dx = \ln |f(x)| + c \neq \ln \left (f(x) \right ) + c .
You need the absolute value inside the log function.
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stochasym
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May I ask sth about Q4? In order to calculate the area of the shaded region, did you use something like this  A=\int_{3/2}^{5/2}\left | \frac{1}{(X-1)(X-3)}+\frac{4}{3} \right |dx?
thanks.
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ItsMJ
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(Original post by depymak)
May I ask sth about Q4? In order to calculate the area of the shaded region, did you use something like this  A=\int_{3/2}^{5/2}\left | \frac{1}{(X-1)(X-3)}+\frac{4}{3} \right |dx?
thanks.
Well I don't know of any way to solve that integrand in that specific form. What I did was convert that rational function into a partial fraction, which allows us to integrate it ( integral of 1/(x-1) is just ln(x-1). Then I used the y=f(x), (f(x)= -4/3) to get the intersection points which are also the limits of our integral to find the area. Then just integrate away. Although I'm not sure if my answer is correct as OP given any answers. Hope this helped.
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TheKevinFang
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(Original post by B_9710)
Have you remembered that  \displaystyle \int \frac{f'(x)}{f(x)} dx = \ln |f(x)| + c \neq \ln \left (f(x) \right ) + c .
You need the absolute value inside the log function.
Ahh okay, so when I get negative values of ln when substituting in x values, I would just use the absolute values?

Thanks.
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Zacken
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(Original post by TheKevinFang)
Ahh okay, so when I get negative values of ln when substituting in x values, I would just use the absolute values?

Thanks.
Yes.
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Ayman!
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(Original post by TheKevinFang)
Ahh okay, so when I get negative values of ln when substituting in x values, I would just use the absolute values?

Thanks.
Nope, not the absolute values of x, you take the absolute value of the argument of the logarithm. You substitute the values for x in as you would normally, but if the argument of your logarithm is less than 0, then you take the absolute value since the argument of a logarithm can't be less than 0.
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TheKevinFang
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(Original post by aymanzayedmannan)
Nope, not the absolute values of x, you take the absolute value of the argument of the logarithm. You substitute the values for x in as you would normally, but if the argument of your logarithm is less than 0, then you take the absolute value since the argument of a logarithm can't be less than 0.
Ahh thanks, that's what I meant but I'm pretty bad at typing maths ideas in my head
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TheKevinFang
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(Original post by Zacken)
Yes.
Hi, I've completed the question by getting area = 0.235 (3s.f.)

One question though,

I get area = 4/3 - (-ln(3) + ln (1))

Would that just be 4/3 - (ln (3/1))?

If I have -ln(3) would I just use the absolute value also? (as opposed to something like ln(-3) )

Thanks in advance.
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B_9710
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(Original post by TheKevinFang)
Hi, I've completed the question by getting area = 0.235 (3s.f.)

One question though,

I get area = 4/3 - (-ln(3) + ln (1))

Would that just be 4/3 - (ln (3/1))?

If I have -ln(3) would I just use the absolute value also? (as opposed to something like ln(-3) )

Thanks in advance.
I'm not sure if fully understand what we mean by the absolute value inside the log (if you do understand and I have misread what you posted then I apologise).
If you have

 \displaystyle I= \int_{-2}^{1} \frac{1}{x} dx then you get

 \displaystyle I = \left  [\ln |x| \right ]^{1}_{-2} = \ln (1) - \ln |-2| = 0- \ln(2) .
So the area would be  \ln 2 .
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AmazingAmirah786
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(Original post by TheKevinFang)
https://onedrive.live.com/redir?resi...hint=folder%2c

Hi, I'm stuck on Question 11 and Question 4 attached above...

I would like tips on how to complete/do question 11, but for question 4, if you could give a solution to it, as I get negative ln values at the end, that would be appreciated.

Thanks.
Out of interest, which book are those questions from ?
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Zacken
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(Original post by TheKevinFang)
Hi, I've completed the question by getting area = 0.235 (3s.f.)

One question though,

I get area = 4/3 - (-ln(3) + ln (1))

Would that just be 4/3 - (ln (3/1))?

If I have -ln(3) would I just use the absolute value also? (as opposed to something like ln(-3) )

Thanks in advance.
Nopes, you leave -ln(3) as it is, you would change ln(-3) into ln(3).
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stochasym
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Yes,thank you
(Original post by ItsMJ)
Well I don't know of any way to solve that integrand in that specific form. What I did was convert that rational function into a partial fraction, which allows us to integrate it ( integral of 1/(x-1) is just ln(x-1). Then I used the y=f(x), (f(x)= -4/3) to get the intersection points which are also the limits of our integral to find the area. Then just integrate away. Although I'm not sure if my answer is correct as OP given any answers. Hope this helped.
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