Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    2
    ReputationRep:
    given that:

    "Integration sign with upper limit 3, lower limit 1" (x^2 -2x + K) dx = 8 2/3

    find the value of K.



    *******sorry for poor formatting, I have no idea how to make it look snazzy
    • Very Important Poster
    Online

    21
    ReputationRep:
    Very Important Poster
    (Original post by theguywhosaidhi)
    given that:

    "Integration sign with upper limit 3, lower limit 1" (x^2 -2x + K) dx = 8 2/3

    find the value of K.



    *******sorry for poor formatting, I have no idea how to make it look snazzy
    What have you tried so far?
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by SeanFM)
    What have you tried so far?
    integrating then plugging in the value of x for 3 and 1, and then subtracting the 2 equations but then the k values cancel since they're subtracted from one another. I'm not sure where to start.
    • Very Important Poster
    Online

    21
    ReputationRep:
    Very Important Poster
    (Original post by theguywhosaidhi)
    integrating then plugging in the value of x for 3 and 1, and then subtracting the 2 equations but then the k values cancel since they're subtracted from one another. I'm not sure where to start.
    I suspect that your integral of k isn't correct - what have you got in your working in the bits where you are just about to evaluate the integral? i.e [.... + .... + ...] between 3 and 1.
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by SeanFM)
    I suspect that your integral of k isn't correct - what have you got in your working in the bits where you are just about to evaluate the integral? i.e [.... + .... + ...] between 3 and 1.
    (x^3/3 - x^2 + kx) with upper 3 and lower 1

    then: (9 - 9+ kx) - (1/3 - 1 + kx) = 8 2/3

    = kx + 2/3 - kx = 8 2/3
    • Very Important Poster
    Online

    21
    ReputationRep:
    Very Important Poster
    (Original post by theguywhosaidhi)
    (x^3/3 - x^2 + kx) with upper 3 and lower 1

    then: (9 - 9+ kx) - (1/3 - 1 + kx) = 8 2/3

    = kx + 2/3 - kx = 8 2/3
    Ah - I see, your integral of k is correct - the problem lies in the evaluation.

    Remember, just like you've done for the x^3/3 and the x^2, you can put in x = 3 and x =1 next to the k.
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by SeanFM)
    I suspect that your integral of k isn't correct - what have you got in your working in the bits where you are just about to evaluate the integral? i.e [.... + .... + ...] between 3 and 1.
    I SEE WHERE I WENT WRONG DAMN IT! thanks anyways i didn't sub in x value in for the kx

    thank you for the help!
    • Very Important Poster
    Online

    21
    ReputationRep:
    Very Important Poster
    (Original post by theguywhosaidhi)
    I SEE WHERE I WENT WRONG DAMN IT! thanks anyways i didn't sub in x value in for the kx

    thank you for the help!
    Well done it is very good (genuinely) that you spotted it yourself.
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by SeanFM)
    Well done it is very good (genuinely) that you spotted it yourself.
    too nice/polite man aw i know you probably have stuff to do. but.... when a binomial is to the power of n, how would you expand it? I don't how you can use the 'nCr' button to get the factorial, the only info I'm given is that n is greater or equal to 2 (and an integer)
    Offline

    22
    ReputationRep:
    (Original post by theguywhosaidhi)
    too nice/polite man aw i know you probably have stuff to do. but.... when a binomial is to the power of n, how would you expand it? I don't how you can use the 'nCr' button to get the factorial, the only info I'm given is that n is greater or equal to 2 (and an integer)
    Look in your formula booklet.
    • Very Important Poster
    Online

    21
    ReputationRep:
    Very Important Poster
    (Original post by theguywhosaidhi)
    too nice/polite man aw i know you probably have stuff to do. but.... when a binomial is to the power of n, how would you expand it? I don't how you can use the 'nCr' button to get the factorial, the only info I'm given is that n is greater or equal to 2 (and an integer)
    No worries even if I cannot help, there are quite a few active posters in the Maths forum so someone can jump in

    nCr on your calculator gives you the value of \frac{n!}{(n-r)!(r!)}, so if it's not satisfactory to leave it as nCr (I don't know if it is or not) then that's another way of going about it. Eg 3C2 is \frac{3!}{(3-2)!(2!)} =\frac{3*2*1}{(1)(2*1)} = 3.
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by Zacken)
    Look in your formula booklet.
    so I'm going to have n's everywhere in my expansion, or am I missing something.

    1st expansion. (n over 1) then 1
    2nd: (n over 2) 1 + 1/4
    3rd: (n over 3) 1 + 1/4^2
    Offline

    22
    ReputationRep:
    (Original post by theguywhosaidhi)
    so I'm going to have n's everywhere in my expansion, or am I missing something.

    1st expansion. (n over 1) then 1
    2nd: (n over 2) 1 + 1/4^2 ---> since n is greater equal to 2? idek
    3rd: (n over 3) 1 + 1/4^3
    You haven't told us the original expression you want to expand, but it looks wrong.

    You will have n's everywhere, yes.
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by Zacken)
    You haven't told us the original expression you want to expand, but it looks wrong.

    You will have n's everywhere, yes.
    damn, I'm retarded, original eq: (1 + 1/4x)^n

    i changed my answer a little in the edit
    Offline

    22
    ReputationRep:
    (Original post by theguywhosaidhi)
    damn, I'm retarded, original eq: (1 + 1/4x)^n

    i changed my answer a little in the edit
    Okay, so the formula booklet says

    (1+x)^n = 1 + nx + n(n-1)/2 x^2 + n(n-1)(n-2)/3! x^3 + ...

    So in your case, just replace all the x's with x/4.
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by Zacken)
    Okay, so the formula booklet says

    (1+x)^n = 1 + nx + n(n-1)/2 x^2 + n(n-1)(n-2)/3! x^3 + ...

    So in your case, just replace all the x's with x/4.
    thank you!
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    Would you like to hibernate through the winter months?
    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
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • 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

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