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    (Original post by imusti96)
    Same i got 5/3
    1/4 and 5/3 I got

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    (Original post by elliemoriarty)
    I got 5/3 for q
    dont say that!
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    can someone post how to do the proof on the second part of the reduction formula question???????
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    (Original post by piddym7)
    can someone post how to do the proof on the second part of the reduction formula question???????
    The -sin^n-1 x cos x under limit [pi/2,0]will become 0 eventually regardless of n since cos(pi/2)=0=sin(0), so Ln will left with (n-1)/n Ln-2 eventually, and then easy stuff

    Not a hard question but strange, it will definitely hit some less able students comparing to 2009 one, (see my UMS comparison above)
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    Does anybody know how to do the revolution area part? I don't know think it is like the normal ones...Do you just differentiate the whole thing with the square root?
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    8/105 for reduction part c???
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    (Original post by trooken)
    1/4 and 5/3 I got

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    how?
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    (Original post by blankk)
    Does anybody know how to do the revolution area part? I don't know think it is like the normal ones...Do you just differentiate the whole thing with the square root?
    it's C4, not FP3 (i hope this message will not hit you so much....)
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    (Original post by Ilovemaths96)
    8/105 for reduction part c???
    Yes!
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    (Original post by Chazatthekeys)
    Yes!
    Pls help me with a MS if you can remember some answers
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    (Original post by piddym7)
    can someone post how to do the proof on the second part of the reduction formula question???????
    Let  J_n = \int^{\pi/2}_{0} \sin^{n} x \ dx = -\frac{1}{n}\sin^{n-1} x \cos x |^{\frac{\pi}{2}}_{0} + \frac{n-1}{n}J_{n-2}

    Therefore

     J_n = \frac{n-1}{n}J_{n-2} . Also  J_1 = \int^{\frac{\pi}{2}}_{0} \sin x \dx = 1

    So  J_n = \frac{(n-1)(n-3)(n-5)...4.2}{n(n-2)(n-4)...5.3}J_1 \implies J_n = \frac{(n-1)(n-3)(n-5)...4.2}{n(n-2)(n-4)...5.3} .



    Think that was it. My memory after an exam isn't very good.
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    (Original post by Ilovemaths96)
    how?
    Don't fully remember the question, but it was the integral of y², multiplied by π
    You had to use one of the integrals from the formula sheet that don't involve hyperbolic or trigonometric functions, just partial fractions, although you didn't have to do the partial fractions yourself

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    (Original post by cogito.)
    it's C4, not FP3 (i hope this message will not hit you so much....)
    But you need to do the square root of 1+(dy/dx)^2 thing don't you...
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    anyone feel sure about their answer of the triangle area in hyperbola?
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    (Original post by Chazatthekeys)
    I got OQR=1

    Had like Q(e^t, e^t) and R(e^-t, -e^-t) and find OQ and OR by Pythagoras and then area = 1 therefore independent of variable 't'
    Thank GOD!!!
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    (Original post by blankk)
    But you need to do the square root of 1+(dy/dx)^2 thing don't you...
    It asked volume of solid not the surface area, sorry
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    Expected boundaries anyone?
    I found the paper was fairly hard, made a few stupid mistakes but hope it's enough for an A.
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    (Original post by cogito.)
    Pls help me with a MS if you can remember some answers
    I remember the answers, but not so much the questions! Lol
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    (Original post by cogito.)
    It asked volume of solid not the surface area, sorry
    If I put 2pi instead of pi for the volume of revolution formula, do you think that's only one mark lost? (The integration and everything else was correct).
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    (Original post by cogito.)
    It asked volume of solid not the surface area, sorry
    Omg should've read it more carefully...thank you
 
 
 
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