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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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.

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

Thanks.

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#4

May I ask sth about Q4? In order to calculate the area of the shaded region, did you use something like this ?

thanks.

thanks.

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#5

(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 ?

thanks.

May I ask sth about Q4? In order to calculate the area of the shaded region, did you use something like this ?

thanks.

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(Original post by **B_9710**)

Have you remembered that .

You need the absolute value inside the log function.

Have you remembered that .

You need the absolute value inside the log function.

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#7

(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.

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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#8

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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(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.

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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(Original post by **Zacken**)

Yes.

Yes.

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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#11

(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.

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.

If you have

then you get

.

So the area would be .

Last edited by B_9710; 2 years ago

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#12

(Original post by **TheKevinFang**)

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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.

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

Thanks.

1

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reply

Report

#13

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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#14

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.

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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