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Integration constant

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 :smile:
Does not matter of course... Although to be very explicit you may want to specify that C/ c / A / B... is a constant.
Reply 2
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 :smile:

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.

2x dx=x2+elephant\displaystyle \int 2x \ dx = x^2 + elephant

as long as you specify that elephant is a constant.
(edited 8 years ago)
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 :smile:
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.

2x dx=x2+elephant\int 2x \ dx = x^2 + elephant

as long as you specify that elephant is a constant.


Nice :smile:
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 :smile:


-4/(1+x)^2 = -4(1+x)^-2
Reply 6
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 :smile:


Unparseable latex formula:

\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)

Original post by hassassin04
-4/(1+x)^2 = -4(1+x)^-2


That's really helpful, thanks. I was stuck on that for ages...
Original post by Zacken
Unparseable latex formula:

\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???
Reply 9
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.
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 :smile:

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