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    (Original post by Imaad)
    there is an official mark shecem for the mock , ill scan it in later today and post it, prob in the evenin say 11ish
    That'd be brilliant.
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    :cool:

    nice1
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    I have done a page of the 4 extra integrals with proofs by differentiation.

    You may like to take a look:

    http://153rd.com/sam/as/C4/integrals2

    An exam question could be to prove one of them.
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    here we go i have hosted them on my space

    http://imaad.pwp.blueyonder.co.uk/mockanswer1.jpg

    http://imaad.pwp.blueyonder.co.uk/mockanswer2.jpg

    hope you can all read it properly , i think you can anyway
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    (Original post by samdavyson)
    I have done a page of the 4 extra integrals with proofs by differentiation.

    You may like to take a look:

    http://153rd.com/sam/as/C4/integrals2

    An exam question could be to prove one of them.

    for proving cot x, couldnt you integrate cosx/sinx , using the identity that integral of f '(x)/f(x) = lnx ?
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    (Original post by Hash)
    for proving cot x, couldnt you integrate cosx/sinx , using the identity that integral of f '(x)/f(x) = lnx ?
    Yes.
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    So edexcel made a mock paper and a specimen paper?
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    yes
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    i cant do rate of change differentiation questions please help!
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    (Original post by Econ4m1t)
    i cant do rate of change differentiation questions please help!
    Usually they involve establishing an equation, differentiating some stuff, and using a bit of chain rule.

    If it helps, this used to be on P3, so have a look through some past P3 papers and find such questions to work though. http://math.mdsalih.com/Data/index.p...el+Mathematics
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    Can someone explain question 3 on paper A1 please?

    I get upto ∫cosθ/(cos²θ )^3/2 dθ and then i get stuck.. is what i've done wrong?

    Thanks!
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    I'm having trouble with part C of this question. Keep in mind that the integral of 1/(2x+1)(x-2)dx is lnk[(x-2)/(2x+1)]^1/5
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    wow
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    (Original post by ntrik)
    I'm having trouble with part C of this question. Keep in mind that the integral of 1/(2x+1)(x-2)dx is lnk[(x-2)/(2x+1)]^1/5
    You have to do:
    ∫1/10y dy = ∫ 1/(2x+1)(x-2) dx

    you'll get a constant obviously so you need to plug in the given x and y values to find its value

    so 1/10ln(10y) = lnk[(x-2)/(2x+1)]^1/5 assuming your previous answer is correct

    1/10ln(10) = lnk[(3-2)/(2(3)+1)]^1/5

    lnk = ln[10^(7/2)]

    this looks so wrong :rolleyes:
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    (Original post by violet)
    Can someone explain question 3 on paper A1 please?

    I get upto ∫cosθ/(cos²θ )^3/2 dθ and then i get stuck.. is what i've done wrong?

    Thanks!
    Remember: am/n = nth root of a to the power of m.

    So:

    ∫cosθ/(cos²θ )^3/2 dθ = ∫cosθ/(cos³θ) dθ = ∫1/(cos²θ) dθ

    => ∫sec²θ dθ => Standard Integral. ( => tanθ + C)
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    (Original post by ntrik)
    I'm having trouble with part C of this question. Keep in mind that the integral of 1/(2x+1)(x-2)dx is lnk[(x-2)/(2x+1)]^1/5
    As featherflare says you need to separate the variables as in any differential equation.

    Here rearrange to:

    1/10y (dy/dx) = 1/(2x+1)(x-2)

    And integrate both sides with respect to x:


    ∫1/10y (dy/dx) dx = ∫ 1/(2x+1)(x-2) dx

    => ∫1/10y dy = ∫ 1/(2x+1)(x-2) dx

    Then use the integral from earlier in the question for the RHS and the LHS comes to 1/10 ln(y) (or 1/10 ln(10y) ).

    The different results depend on how you integrate it. The constant takes care of the difference.
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    (Original post by Featherflare)
    You have to do:
    ∫1/10y dy = ∫ 1/(2x+1)(x-2) dx

    you'll get a constant obviously so you need to plug in the given x and y values to find its value
    Edit - Argh crap, I think I made a silly mistake...

    Okay, here is what I have

    ∫1/10y dy = ∫ 1/(2x+1)(x-2) dx
    (1/10)lny = lnk[(x-2)/(2x+1)]^1/5
    y^1/10 = k[(x-2)/(2x+1)]^1/5

    x = 3 and y = 1

    1 = k(1/7)^1/5
    k = 1.476 (3 dp)

    I'm pretty sure that wrong because I'm not sure how I can get to y = 49[(x-2)/(2x+1)]² which is the answer.
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    (Original post by Featherflare)
    so 1/10ln(10y) = lnk[(x-2)/(2x+1)]^1/5 assuming your previous answer is correct
    But is it?

    4 (a). 1 = A(x - 2) + B(2x + 1)

    Put x = 2: 1 = 5B, B = 1/5
    Put x = -1/2: 1 = -5/2A, A = -2/5

    (b) ∫[(-2/5)/(2x + 1)] + [(1/5)/(x - 2)]dx

    => -2/10ln(2x + 1) + 1/5ln(x - 2)

    => -1/5ln(2x + 1) + 1/5ln(x - 2)

    => ln(2x + 1)-1/5 + ln(x - 2)1/5

    => ln[(x - 2)1/5/(2x + 1)1/5]

    =>ln[(x - 2)/(2x + 1)1/5]

    Yes
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    plz if a Q. asks for volume of a curve rotated 360 or 180 , what do we use
    pie or 2pie for anyone plz answer me any1 now
    and thanks in advance
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    Well, i've just finished partial fractions, so I can help people with those, lol.
    Although it's a very short/easy chapter so noone will need my help.
 
 
 
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