Turn on thread page Beta
    • Thread Starter
    Offline

    10
    ReputationRep:
    Integrate x/x-2 in the range from 3 to 4.
    So far I've only been taught how to integrate via substation so I thought u = x-2 and tried doing the usual but it doesn't seem to work out. Is there an alternative method?
    • Study Helper
    Offline

    15
    Study Helper
    (Original post by MrToodles4)
    Integrate x/x-2 in the range from 3 to 4.
    So far I've only been taught how to integrate via substation so I thought u = x-2 and tried doing the usual but it doesn't seem to work out. Is there an alternative method?
    The substitution should work - post your working, and someone can check it out.

    Alternatively, you can just perform the polynomial/synthetic division, if you've covered it (don't know what module that is), and then integrate.
    Offline

    6
    ReputationRep:
    (Original post by MrToodles4)
    Integrate x/x-2 in the range from 3 to 4.
    So far I've only been taught how to integrate via substation so I thought u = x-2 and tried doing the usual but it doesn't seem to work out. Is there an alternative method?
    You could also write the fraction x/(x-2) as 1 + 2/(x-2). If you subtract 2 and then add 2 to the numerator it might be easier to see why.
    • Thread Starter
    Offline

    10
    ReputationRep:
    (Original post by akaash13)
    You could also write the fraction x/(x-2) as 1 + 2/(x-2). If you subtract 2 and then add 2 to the numerator it might be easier to see why.
    How did you get this?
    Offline

    21
    ReputationRep:
    If you post your working out or what you have been doing, we can see where the confusion is and can then help you.
    Offline

    20
    ReputationRep:
    (Original post by MrToodles4)
    How did you get this?
    x/{x - 2} is the same as {x - 2 + 2}/{x - 2}
    Offline

    6
    ReputationRep:
    (Original post by MrToodles4)
    How did you get this?


    Left out the dx because there was not enough space. :/
    The u-substitution should also work. Just remember to change the limits and you should get (u+2)/u = 1 + 2/u.
    • Thread Starter
    Offline

    10
    ReputationRep:
    (Original post by ghostwalker)
    The substitution should work - post your working, and someone can check it out.

    Alternatively, you can just perform the polynomial/synthetic division, if you've covered it (don't know what module that is), and then integrate.
    So I wrote out the equation as x(x-2)^-1. And let u=x-2. Therefore du/dx = 1. dx/du = 1.

    so x*u^-1. to get x in terms of u i did u+2=x
    So (u+2)*u^-1. I multiplied out the bracket to get 1+2u^-1.
    If this is integrated this is the same as u+2lnu. And then I put in the new ranges from intergral u=1 and u= 2. (originally x=3 and x=4) so 2+2ln2 - 1+2ln1 however this does not give me the correct answer of 1+2ln2. Does the 1 not need to be integrated into u? is that the mistake Im making here?
    • Community Assistant
    • Study Helper
    Offline

    20
    ReputationRep:
    Community Assistant
    Study Helper
    (Original post by MrToodles4)
    So I wrote out the equation as x(x-2)^-1. And let u=x-2. Therefore du/dx = 1. dx/du = 1.

    so x*u^-1. to get x in terms of u i did u+2=x
    So (u+2)*u^-1. I multiplied out the bracket to get 1+2u^-1.
    If this is integrated this is the same as u+2lnu. And then I put in the new ranges from intergral u=1 and u= 2. (originally x=3 and x=4) so 2+2ln2 - 1+2ln1 however this does not give me the correct answer of 1+2ln2. Does the 1 not need to be integrated into u? is that the mistake Im making here?
    2+2ln2 - 1+2ln1

    1+2ln2

    These are the same thing since ln(1) = 0.

    For future questions please post all your working at the start so that we know where you're stuck.
    Offline

    11
    ReputationRep:
    Note: (a+b) - (c+d) = a + b - c - d, (note the bold emphasis).
    Offline

    21
    ReputationRep:
    (Original post by MrToodles4)
    So I wrote out the equation as x(x-2)^-1. And let u=x-2. Therefore du/dx = 1. dx/du = 1.

    so x*u^-1. to get x in terms of u i did u+2=x
    So (u+2)*u^-1. I multiplied out the bracket to get 1+2u^-1.
    If this is integrated this is the same as u+2lnu. And then I put in the new ranges from intergral u=1 and u= 2. (originally x=3 and x=4) so 2+2ln2 - 1+2ln1 however this does not give me the correct answer of 1+2ln2. Does the 1 not need to be integrated into u? is that the mistake Im making here?
    let u = x-2 , x = u +2
    du/dx = 1
    dx=du
    then
    (U+2)/U Du
    = U/2 + 2/U
    =2u^-1 + 2lnU
    solve from there
    • Thread Starter
    Offline

    10
    ReputationRep:
    (Original post by Notnek)
    2+2ln2 - 1+2ln1

    1+2ln2

    These are the same thing since ln(1) = 0.

    For future questions please post all your working at the start so that we know where you're stuck.
    Oh, I see. Thank you very much . There is one other question which was similar:
    • Thread Starter
    Offline

    10
    ReputationRep:
    (Original post by simon0)
    Note: (a+b) - (c+d) = a + b - c - d, (note the bold emphasis).
    What bold emphasis?
    Offline

    11
    ReputationRep:
    (Original post by MrToodles4)
    What bold emphasis?
    Sorry, it is the last minus sign.

    What you have done looks fine except for the final part but it is fine in this case since ln(1)=0.
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: November 2, 2017

University open days

  • Sheffield Hallam University
    City Campus Postgraduate
    Wed, 17 Oct '18
  • Northumbria University
    All faculties Undergraduate
    Wed, 17 Oct '18
  • Staffordshire University
    Nursing and Midwifery Undergraduate
    Wed, 17 Oct '18
Poll
Who is most responsible for your success at university
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

Equations

Best calculators for A level Maths

Tips on which model to get

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.