# stuck on ocr c3 specimen paperWatch

#1
stuck on ocr c3 specimen paper

i've done it all, just question 4 part ii (ii)
you find the integral of pi times y^2 which is 1/(4x+1)

i don't know if i'm just having a really stupid block, but in the answers when they integrated that they said it is 0.25ln(4x+1), why is it 0.25?!?? surely it'd just be ln(4x+1) ?

the question and mark scheme is here:
http://www.ocr.org.uk/Data/publicati..._Maths_SAM.pdf
0
9 years ago
#2
there is a 4x.

you've given no justification why 'surely' it shouldn't have the 0.25.
0
9 years ago
#3
lils270391, use the chain rule.

y=ln(4x+1) where
4x+1=u ==> du/dx = 4
y=lnu ==> dy/du = 1/u
dy/dx=dy/du * du/dx
dy/dx = 4 * 1/u = 4/u = 4/(4x+1).

So we got 4/(4x+1)..and we want 1/(4x+1). We are therefore a factor of four out! To rectify this, we simply multiply ln(4x+1) by 0.25 or one quarter to remove the *4 term.
0
#4
thank you for clearing that up, i'm glad i noticed that question because i always thought that if an integral was 1 divided by an expression it was just ln(expression)
0
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