Hey there! Sign in to join this conversationNew here? Join for free
x Turn on thread page Beta
    • Thread Starter
    Offline

    2
    ReputationRep:
    How would I integrate 4 / x(4lnx + 3)^2, with the limits of e and 1. I know that dy/dx of lnx= 1/x but don't know how to integrate it?

    Also, when using modulus such as l 4x -3 l < l 2x + 1 l. I can't seem to get the right answer. Is the method squaring both sides when there are inequalities of modulus, right?
    Offline

    0
    ReputationRep:
    Do you mean differentiation? Integration isn't in C3. What question/paper is this?
    For the 2nd question, I wouldn't square it cos I dunno if you can but also it loses some of the answers if gets any of them.
    l4x-3l = 4x-3 AND -4x+3
    this is the same for the 2nd side: l2x+1l = 2x+1 AND -2x-1
    therefore the are 4 combinations of the overall equation... long, i know
    hope that was right/helpful
    GOOD LUCK TOMORROW!
    Offline

    12
    ReputationRep:
    Could you link towards the question itself? You might have overlooked a small comment or whatever. I personally always prefer to get the question at hand, since I've seen people misread stuff too many times to puzzle my head over something that might have been a lot simpler.
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by phen)
    Could you link towards the question itself? You might have overlooked a small comment or whatever. I personally always prefer to get the question at hand, since I've seen people misread stuff too many times to puzzle my head over something that might have been a lot simpler.
    http://www.ocr.org.uk/download/pp_07..._jun_l_gce.pdf

    Questions 2 and 8iii. For the first part, I think I know understand that you make the different possible combinations, but when does squaring both sides work and when does it not?
    Offline

    11
    ReputationRep:
    (Original post by definite_maybe)
    How would I integrate 4 / x(4lnx + 3)^2, with the limits of e and 1. I know that dy/dx of lnx= 1/x but don't know how to integrate it?
    If the question's \displaystyle \frac{4}{x(4\mathrm{ln}(x) + 3)^2}, try a substitution of \displaystyle u=(4\mathrm{ln}(x) + 3).
    • PS Helper
    Offline

    14
    PS Helper
    Have a look at your answer to part 8(i) for a massive hint.

    (Original post by james.h)
    If that's the question, try a substitution of \displaystyle u=(4\mathrm{ln}(x) + 3).
    Integration by substitution doesn't come up in the OCR syllabus until C4; however there's enough information needed to do the integration in the first part of the question.
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by nuodai)
    Have a look at your answer to part 8(i) for a massive hint.


    Integration by substitution doesn't come up in the OCR syllabus until C4; however there's enough information needed to do the integration in the first part of the question.
    With the 4lnx-3/4lnx+3, to get the value integrated do you divide by 6 or multiply by 6? Both times I get the wrong answer?

    Also for the modulus, do you only use squared when it is =, and not for inequalities?
    Offline

    0
    ReputationRep:
    Yo,

    im new to this and im not sure how to type it all out...but here goes.

    V = pi ∫(e to 1) y^2 dx

    = pi ∫(e to 1) 4/x(4lnx+3)^2 dx

    = pi/6 ∫ (e to 1) 24/x(4lnx+3)^2 dx

    = pi/6 [4lnx-3/4lnx+3] (e to 1)

    = pi/6 ((1/7 - (-1))

    = pi/6 x (1/7 + 1)

    = pi/6 x 8/7 (Cancel This Down)

    = 4pi/21
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by Bhav_Singh)
    Yo,

    im new to this and im not sure how to type it all out...but here goes.

    V = pi ∫(e to 1) y^2 dx

    = pi ∫(e to 1) 4/x(4lnx+3)^2 dx

    = pi/6 ∫ (e to 1) 24/x(4lnx+3)^2 dx

    = pi/6 [4lnx-3/4lnx+3] (e to 1)

    = pi/6 ((1/7 - (-1))

    = pi/6 x (1/7 + 1)

    = pi/6 x 8/7 (Cancel This Down)

    = 4pi/21
    Thank you. Any help with the modulus?
    Offline

    0
    ReputationRep:
    If you draw them both out, it helps.

    But ill just show you how to get the answer but then just try drawing them out and getting the answer.

    4x-3 = 2x+1

    2x = 4

    x = 2


    2x+1 = 3-4x

    6x = 2

    x = 1/3


    Therefore, 1/3 < x < 2
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by Bhav_Singh)
    If you draw them both out, it helps.

    But ill just show you how to get the answer but then just try drawing them out and getting the answer.

    4x-3 = 2x+1

    2x = 4

    x = 2


    2x+1 = 3-4x

    6x = 2

    x = 1/3


    Therefore, 1/3 < x < 2
    I udnerstand how to get them by graph method, but what I don't understand is in which situations you can use squaring to solve
    Offline

    0
    ReputationRep:
    Lol, i dont know. I dont do OCR anyway, I do MEI - so its a bit different.

    gl with your C3 exam tomorrow. X
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: January 19, 2010
Poll
Are you going to a festival?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.