Integration of logs (1 Viewer)

[Avenger6]

I am struggling with these two questions:

I am unsure of how to integrate e^2y in question 11 and am stuck on the second half of question 12, as in I can show that they equal (x-6)/(x-1) but am struggling when it comes to integrating.

Thanks in advance for any help.

BTW sorry about the gap/indentation in question 11.

 

[lyounamu]

I am struggling with these two questions:

I am unsure of how to integrate e^2y in question 11 and am stuck on the second half of question 12, as in I can show that they equal (x-6)/(x-1) but am struggling when it comes to integrating.

Thanks in advance for any help.

BTW sorry about the gap/indentation in question 11.

11. y = ln x
Then x = e^y
And x^2 = e^2y

V = pi ∫_1^3▒〖e^2y dy〗
= pi [e^2y/2]3 1 (this means 3 at the top and 1 at the bottom, sorry!)
= pi (e^6/2 - e^2/2) units ^3

12. You can easily finish the first part of question. So I will do second part.


∫▒〖(x-6)/(x-1) dx= ∫▒〖1-5/(x-1)〗〗 dx= x-5 ln⁡〖(x-1)+c〗

Therefore, the answer is x - 5 ln(x-1) +c

I am sorry if my answer is not clear. I will post a new one to clear all things up if you wish.

 

[Avenger6]

Thanks for the response...question 11 is alot more clear now, however the answer in the book is (pi/2).e^2(e^4-1). I just need a bit of clarification on q12...it seems you've added the 1 to ∫▒〖(x-6)/(x-1) dx should it then be ∫▒ (x-5)/(x-1) dx or is there something that i'm missing when it comes to integrating that to get x-5 ln⁡(x-1)+c???

Thanks again.

 

[lyounamu]

Thanks for the response...question 11 is alot more clear now. I just need a bit of clarification on q12...it seems you've added the 1 to ∫▒〖(x-6)/(x-1) dx should it then be ∫▒ (x-5)/(x-1) dx or is there something that i'm missing when it comes to integrating that to get x-5 ln⁡(x-1)+c???

Thanks again.

I didn't.

If you look at the first question, (x-6)/(x-1) = (x-1)/(x-1) - 5/(x-1)

So, you just substitute that instead of (x-6)/(x-1). You can find the integral of 1 (which is x) and you can also find the integral of 5/(x-1) which is (5 ln(x-1))

 

[foram]

Thanks for the response...question 11 is alot more clear now, however the answer in the book is (pi/2).e^2(e^4-1). I just need a bit of clarification on q12...it seems you've added the 1 to ∫▒〖(x-6)/(x-1) dx should it then be ∫▒ (x-5)/(x-1) dx or is there something that i'm missing when it comes to integrating that to get x-5 ln⁡(x-1)+c???

Thanks again.

because (x-6)/(x-1) = 1 - 5/(x-1)
int. (x-6)/(x-1) = int. 1 - 5/(x-1) = int.1 - int.(5/x-1)
= x - 5ln(x-1) + c

 

[Avenger6]

Ah ok, that makes more sense, i wasn't breaking it up into two parts. Have you guys got any idea on how the book came up with (pi/2).e^2(e^4-1) as the answer for question 11???

 

[lyounamu]

Ah ok, that makes more sense, i wasn't breaking it up into two parts. Have you guys got any idea on how the book came up with (pi/2).e^2(e^4-1) as the answer for question 11???

They took out half out and basically tidied up.

 

[Avenger6]

but shouldn't it be (pi/2).e^2(e^3-1)???

 

[lyounamu]

but shouldn't it be (pi/2).e^2(e^3-1)???

No, it should be 4 because e^4 . e^2 = e^6 not e^8

 

[Avenger6]

ahk, my mistake lol...