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    I was doing a past paper, AQA Core 4 (June 2014) and on question 2)b) I got the right answer, but my constant of integration was a lowercase c, rather than the capital C in the markscheme.

    Would I lose marks for this?

    Why is it sometimes a lowercase c and sometimes a capital C or does it not matter?

    Thanks
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    Does not matter of course... Although to be very explicit you may want to specify that C/ c / A / B... is a constant.
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    (Original post by MathsALevel123)
    I was doing a past paper, AQA Core 4 (June 2014) and on question 2)b) I got the right answer, but my constant of integration was a lowercase c, rather than the capital C in the markscheme.

    Would I lose marks for this?

    Why is it sometimes a lowercase c and sometimes a capital C or does it not matter?

    Thanks
    Any kind of 'c' is fine if you're sticking to the convention which I'd recommend.

    But you could really use anything e.g.

    \displaystyle \int 2x \ dx = x^2 + elephant

    as long as you specify that elephant is a constant.
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    (Original post by hassassin04)
    Does not matter of course... Although to be very explicit you may want to specify that C/ c / A / B... is a constant.
    Thanks for that. I have another question as well:

    On question 8)b, I had to integrate -4/(1+x)^2

    I put -4x/1+x but the answer is 4/1+x

    Why is this?

    Thanks
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    (Original post by notnek)
    Any kind of 'c' is fine if you're sticking to the convention which I'd recommend.

    But you could really use anything e.g.

    \int 2x \ dx = x^2 + elephant

    as long as you specify that elephant is a constant.
    Nice
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    (Original post by MathsALevel123)
    Thanks for that. I have another question as well:

    On question 8)b, I had to integrate -4/(1+x)^2

    I put -4x/1+x but the answer is 4/1+x

    Why is this?

    Thanks
    -4/(1+x)^2 = -4(1+x)^-2
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    (Original post by MathsALevel123)
    Thanks for that. I have another question as well:

    On question 8)b, I had to integrate -4/(1+x)^2

    I put -4x/1+x but the answer is 4/1+x

    Why is this?

    Thanks
    \displaystyle \int -4(1+x)^{-2} \, \mathrm{d}x = -4\int (1+x)^{-2} \, \matrm{d}x = -4\left(\frac{(1+x)^{-2 + 1}}{-2 + 1}\right)
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    (Original post by hassassin04)
    -4/(1+x)^2 = -4(1+x)^-2
    That's really helpful, thanks. I was stuck on that for ages...
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    (Original post by Zacken)
    \displaystyle \int -4(1+x)^{-2} \, \mathrm{d}x = -4\int (1+x)^{-2} \, \matrm{d}x = -4\left(\frac{(1+x)^{-2 + 1}}{-2 + 1}\right)
    But if it was say 4 x (1+2x) ^ -2, it would be different because of the 2 and reversing the chain rule???
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    (Original post by MathsALevel123)
    But if it was say 4 x (1+2x) ^ -2, it would be different because of the 2 and reversing the chain rule???
    Yes, you would need to divide by 2.
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    (Original post by Zacken)
    Yes, you would need to divide by 2.
    Thanks. If it's ok, could you check my post called "not sure about these answers" please?

    Thanks
 
 
 
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