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    (Original post by lam12)
    We've just finished the trapezium rule in integration today, not much left to go but we haven't started D1. However we can decide to do M1, what would people advise?


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    I did D1 last year as an AS Further Maths module, and I'm doing M1 this year as an A2 Maths module.
    I really enjoyed doing D1 and got 100 UMS in the exam. It involves a lot of algorithms so if you're good at following methods and like working out values with diagrams rather than lots of formulae then you will probably enjoy it.
    I found M1 quite difficult at first because it was unlike normal maths, but I am finding it easier now I have got the hang of using the relevant formulae and drawing the diagrams correctly.
    I would highly recommend examsolutions.com for useful videos that explain the difficult topics very clearly.
    I would advise that you look through the syllabus for M1 and D1 if you haven't yet made your decision and see which best suits your strengths and interests.
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    I need help with this question, I dont get why when integrating the term [ (3+2x)^7] / 7

    it is ( 3 + 2x)^8 / ( 7 x 16)

    Where is the 16 from? Shouldn't it be / (7 x 8 )


    http://gyazo.com/42b1c7f62c1e5d6b747c20c7917d4353



    Also I am stuck on question 3e on the c4 heinmann book. This is my working up to now but it's like the book has gone from DV/dx to v = sec^2x where I have gone to tan^2x. Picture of working attached below
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    Name:  ImageUploadedByStudent Room1424659830.222636.jpg
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Size:  181.1 KB


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    EDIT: Ive figured out what I did wrong here, thanks for the help
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    (Original post by Hudl)
    I need help with this question, I dont get why when integrating the term [ (3+2x)^7] / 7

    it is ( 3 + 2x)^8 / ( 7 x 16)

    Where is the 16 from? Shouldn't it be / (7 x 8 )


    http://gyazo.com/42b1c7f62c1e5d6b747c20c7917d4353



    Also I am stuck on question 3e on the c4 heinmann book. This is my working up to now but it's like the book has gone from DV/dx to v = sec^2x where I have gone to tan^2x. Picture of working attached below
    To integrate, add one to power, then divide by new power and divide by derivative of thing that was to power (derivative was 2)
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    (Original post by Random1357)
    To integrate, add one to power, then divide by new power and divide by derivative of thing that was to power (derivative was 2)
    Not for the 2sec^2xtanx integral it is a site integral and will be done by trying y = tan^2x therefore dy/dx is 2sec^2xtanx and this 2sec^2xtan x integrates to tan^2x


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    (Original post by Hudl)
    Not for the 2sec^2xtanx integral it is a site integral and will be done by trying y = tan^2x therefore dy/dx is 2sec^2xtanx and this 2sec^2xtan x integrates to tan^2x


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    Or more formally by substitution: I didn't bother looking at the attached hence just explained the former.
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    (Original post by Random1357)
    Or more formally by substitution: I didn't bother looking at the attached hence just explained the former.
    It's a by parts question


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    (Original post by Hudl)
    It's a by parts question


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    No: that's a clumsy method if it even works: inspection for a general Q and substitution for being formal (e.g. Proof)


    Oh sorry, just actually read it

    Yeah use both
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    You have 2 problems: 1) your working is very hard to follow: I used to be bad, trust me, learning to be neat is essential.

    2) the integral of tan^2(x) is integral of (sec^2-1) is tan(x)-x+C
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    guys if you had this bracket : -2( 2x - 1 )^-1

    Can I change it to (-1 + 2x) then expand etc or not?



    (binomial question)
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    (Original post by Mary562)
    guys if you had this bracket : -2( 2x - 1 )^-1

    Can I change it to (-1 + 2x) then expand etc or not?



    (binomial question)
    If you mean can I write as -2[(-1) + (2x)]^(-1) then that is a true statement but useless as can't be expanded!

    We could do: 2[(1) + (-2x)]^(-1), valid for IxI<1/2 by pulling out a factor of (-1)^(-1)=-1
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    (Original post by Random1357)
    If you mean can I write as -2[(-1) + (2x)]^(-1) then that is a true statement but useless as can't be expanded!

    We could do: 2[(1) + (-2x)]^(-1), valid for IxI<1/2 by pulling out a factor of (-1)^(-1)=-1
    Yep that is what I'm saying - thanks the bottom one is correct

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    Can you integrate Ln^2 x using a chain rule method or site integral method

    chain rule method

    (Ln^3 x)/3 * x


    Site integral method

    Try y = Ln^3 x
    dy/dx = 3Ln^2 x * 1/x

    Therefore legit y and thus answer of integral is x/3Ln^3 x


    Ive seen the by parts answer but why doesnt the chain rule / site integral method work?
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    (Original post by Hudl)
    Can you integrate Ln^2 x using a chain rule method or site integral method

    chain rule method

    (Ln^3 x)/3 * x


    Site integral method

    Try y = Ln^3 x
    dy/dx = 3Ln^2 x * 1/x

    Therefore legit y and thus answer of integral is x/3Ln^3 x


    Ive seen the by parts answer but why doesnt the chain rule / site integral method work?
    If the bit you gain when you differentiate isn't a constant, you can't "undo" it when you want to integrate.
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    (Original post by tiny hobbit)
    If the bit you gain when you differentiate isn't a constant, you can't "undo" it when you want to integrate.
    Yeah thanks a lot, whats wrong with the chain rule method...why doesnt that work because you're allowed to differentiate the term in the brackets and even if it gives you a non linear term like x^2 you have to do 1/x^2 as you're integrating.
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    (Original post by Hudl)
    Yeah thanks a lot, whats wrong with the chain rule method...why doesnt that work because you're allowed to differentiate the term in the brackets and even if it gives you a non linear term like x^2 you have to do 1/x^2 as you're integrating.
    If you differentiate your answer, using the product or quotient rule, you won't get the thing you are trying to integrate.
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    Hey im just starting doing maths paper, can someone give me a estimate on how many papers i should do to fully master the syllabus, and how many times i should re do the papers. Thank you guys
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    (Original post by john122334455)
    Hey im just starting doing maths paper, can someone give me a estimate on how many papers i should do to fully master the syllabus, and how many times i should re do the papers. Thank you guys
    I'm also doing A2 at the moment but am using same process as last year which worked out well (got full 300 UMS). 1 past paper every other day starting in February right up until the exams, do them daily as of May 1st - this includes Solomon Papers and Edexcel ones, swapping between the two in blocks of 2 weeks. Seems a bit excessive when you consider this will total to about 90 past papers but I self teach so have daily free periods in which to do them. Personally, I would never re-do past papers - I don't see the point, surely you'll subconciously remember answer/method? For me at least there is more than enough papers available to not re-do any but what works for you, works for you!!
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    (Original post by Gilo98)
    I'm also doing A2 at the moment but am using same process as last year which worked out well (got full 300 UMS). 1 past paper every other day starting in February right up until the exams, do them daily as of May 1st - this includes Solomon Papers and Edexcel ones, swapping between the two in blocks of 2 weeks. Seems a bit excessive when you consider this will total to about 90 past papers but I self teach so have daily free periods in which to do them. Personally, I would never re-do past papers - I don't see the point, surely you'll subconciously remember answer/method? For me at least there is more than enough papers available to not re-do any but what works for you, works for you!!
    did u do any other subjects? lool
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    (Original post by PinkElephant16)
    did u do any other subjects? lool
    No, I was in Yr 10 last year so had no GCSEs - doing the same this year though
 
 
 
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