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    (Original post by TH3-FL45H)
    If it asks me to differentiate something like 1/((x+2)^2)

    Can I use substitution (u=x+2) or would that not get any marks because it's not in the mark scheme

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    I'd like to know this too actually. I'm sure it's okay if you just sub back in your substitution afterwards but I'd still like to confirm this.
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    Really confused. trying to do this integral: 1/(3x+3). Pretty sure the answer is (1/3)*ln(3x+3) but surely if you take out 1/3 you get 1/(x+1) and then as the top is the differetial of the bottom it would be (1/3)*ln(x+1). Any help appreciated. Good luck everyone!
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    (Original post by Nurishment)
    ~

    Can someone explain b) to me. I don't quite fully understand why you take out a factor of 1000 . Why wouldn't you rearrange 7100 to = 8-9x and solve for x and sub that back in?
    BTW theres an error in the MS for that Question. Its supposed to be 10³√7.1 not 10³√71. Try typing on the calc.
    Now to explain i "think" x should always be less than 1. Someone correct me.

    Dont know why but if subbing 7100 gives you the wrong answer compared to when you type ³√7100 on calculator try turing it to surds.
    ³√7100 = ³√7.1 × ³√1000 = 10³√7.1
    Equate 10³√(9-8x) = 10³√7.1
    9-8x = 7.1
    x = 0.1
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    (Original post by jamesoneil98)
    Really confused. trying to do this integral: 1/(3x+3). Pretty sure the answer is (1/3)*ln(3x+3) but surely if you take out 1/3 you get 1/(x+1) and then as the top is the differetial of the bottom it would be (1/3)*ln(x+1). Any help appreciated. Good luck everyone!
    ln(3x + 3) = ln3 + ln(x + 1)
    All you're doing with the first case is adding a specific constant bit
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    Who reckons this paper will be at the same difficulty or harder as the c3 paper?
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    \displaystyle\int^\frac{\pi}{4}_  0 \pi x^2(1+sin2x)\ dx

    For integration by parts to work in this question would you need to expand out the brackets first then solve them as two separate integrals? Like this:

    \displaystyle\int^\frac{\pi}{4}_  0 \pi x^2\ dx + \displaystyle\int^\frac{\pi}{4}_  0 \pi x^2(sin2x)\ dx

    I tried to to this question last night without expanding first and got most of the answer correct but the first term was missing, guessing this is why...
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    (Original post by Bealzibub)
    for what shapes do we need t know how to work out volume, area etc.??

    volume of polyhedron

    and circumference of mars

    standard procedure
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    (Original post by sj97)
    http://www.examsolutions.net/a-level...8&solution=8.2 I don't really understand what he's doing here, could you please explain?
    This is what i did....Name:  Untitled.png
Views: 96
Size:  9.5 KB
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    What are the "R" papers by the way? I always seem to find them harder, how come?
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    (Original post by grn)
    \displaystyle\int^\frac{\pi}{4}_  0 \pi x^2(1+sin2x)\ dx

    For integration by parts to work in this question would you need to expand out the brackets first then solve them as two separate integrals? Like this:

    \displaystyle\int^\frac{\pi}{4}_  0 \pi x^2\ dx + \displaystyle\int^\frac{\pi}{4}_  0 \pi x^2(sin2x)\ dx

    I tried to to this question last night without expanding first and got most of the answer correct but the first term was missing, guessing this is why...
    You would need to integrate by parts twice for that wouldn't you?
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    (Original post by 13 1 20 8 42)
    ln(3x + 3) = ln3 + ln(x + 1)
    All you're doing with the first case is adding a specific constant bit
    but what have i done wrong? one of the 2 ways is wrong as they dont equal each other.
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    (Original post by jamesoneil98)
    but what have i done wrong? one of the 2 ways is wrong as they dont equal each other.
    It's not wrong, it's just a specific case. Normally you would put a constant of integration, right? If you take out the constant in the 3x + 3 version, you can then amalgamate that with some arbitrary constant to get another arbitrary constant; i.e. if you write the integral as

    (1/3)ln(3x + 3) + c this is 1/3(ln(3) + ln(x + 1)) + c = (1/3)ln3 + (1/3)ln(x + 1) + c

    Now this can be written as (1/3)ln(x + 1) + k because the c and the (1/3)ln3 are constants

    If you put the answer (1/3)ln(x + 1) this is more correct to be honest (although having a plus c is always the most correct)
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    When a gradient is parallel to the x axis and you have eg dy/dx = (3x+ y)/(2y+1) is the 3x + y = 0 or then 2y+1? thanks
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    (Original post by jamesoneil98)
    Really confused. trying to do this integral: 1/(3x+3). Pretty sure the answer is (1/3)*ln(3x+3) but surely if you take out 1/3 you get 1/(x+1) and then as the top is the differetial of the bottom it would be (1/3)*ln(x+1). Any help appreciated. Good luck everyone!
    Both are equivalent because the 1/3ln(3x+3) is 1/3(ln3+ln(x+1))
    When you put your limits in you have ln3 and -ln3 so the answer is the same for both
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    (Original post by WillGood)
    BTW theres an error in the MS for that Question. Its supposed to be 10³√7.1 not 10³√71. Try typing on the calc.
    Now to explain i "think" x should always be less than 1. Someone correct me.

    Dont know why but if subbing 7100 gives you the wrong answer compared to when you type ³√7100 on calculator try turing it to surds.
    ³√7100 = ³√7.1 × ³√1000 = 10³√7.1
    Equate 10³√(9-8x) = 10³√7.1
    9-8x = 7.1
    x = 0.1
    I'm beginning to grasp why this works now. How Edexcel thinks about putting questions like these in, hurt me.
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    (Original post by WillGood)
    BTW theres an error in the MS for that Question. Its supposed to be 10³√7.1 not 10³√71. Try typing on the calc.
    Now to explain i "think" x should always be less than 1. Someone correct me.

    Dont know why but if subbing 7100 gives you the wrong answer compared to when you type ³√7100 on calculator try turing it to surds.
    ³√7100 = ³√7.1 × ³√1000 = 10³√7.1
    Equate 10³√(9-8x) = 10³√7.1
    9-8x = 7.1
    x = 0.1
    But if you solve for x you get 0.2375?
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    (Original post by Anon-)
    What are the "R" papers by the way? I always seem to find them harder, how come?
    they are built and made for the asians
    Spoiler:
    Show
    forreal though
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    (Original post by bg9876)
    When a gradient is parallel to the x axis and you have eg dy/dx = (3x+ y)/(2y+1) is the 3x + y = 0 or then 2y+1? thanks
    If the gradient is equal to 0, the top line (numerator) is equated to 0. If a line is parallel to the y-axis then the gradient is undefined, and you only get an undefined value when 0 is on the bottom of a fraction, therefore you equate 2y+1= 0 (denominator)
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    (Original post by sj97)
    But if you solve for x you get 0.2375?
    probably meant 8 - 9x
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    (Original post by 13 1 20 8 42)
    It's not wrong, it's just a specific case. Normally you would put a constant of integration, right? If you take out the constant in the 3x + 3 version, you can then amalgamate that with some arbitrary constant to get another arbitrary constant; i.e. if you write the integral as

    (1/3)ln(3x + 3) + c this is 1/3(ln(3) + ln(x + 1)) + c = (1/3)ln3 + (1/3)ln(x + 1) + c

    Now this can be written as (1/3)ln(x + 1) + k because the c and the (1/3)ln3 are constants

    If you put the answer (1/3)ln(x + 1) this is more correct to be honest (although having a plus c is always the most correct)
    got it thanks!
 
 
 
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