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AQA MFP3 - 16th June 2011

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Reply 60
Original post by vladtheimpaler
I thought it went really well! A lot easier than FP2. The last one you did the area bound by the sin curve - the area bound by the tan curve, and i got a=1/2 and b=-3/4
Got y = (Ax^4)/4 + Ax + x^2 + B for the differential equation
-2 for the limit
(2e^3)/9 for another limit


Exactly the same as you! :smile:
Original post by JJBSussat
That was solid - what did everyone get for the last differential one? The one where you had to substitute...

I took the u over to the one side and the (x^3 + 1...) over to the other.

So then I had: lnu =3x^2(ln(x^3+1) )/3x^2 + c
u = A(x^3 + 1)
y = Int( Ax^3 + A1 -2x)
y = (A/4)X^4 +Ax -x^2 + B

I think I had it completely wrong... did anybody else do that?

I think I got (A/4)X^4 + Ax + x^2 + B. I'm fairly sure x^2 was a positive sign because u=(dy/dx)-2x
Reply 62
Original post by vladtheimpaler
I thought it went really well! A lot easier than FP2. The last one you did the area bound by the sin curve - the area bound by the tan curve, and i got a=1/2 and b=-3/4
Got y = (Ax^4)/4 + Ax + x^2 + B for the differential equation
-2 for the limit
(2e^3)/9 for another limit


I got both of those as well. What took me a while was finding

ddx(2cos2x+sin2x)\frac{d}{dx}\left(\frac{2}{\cos^2 x+\sin 2x}\right)

I was integrating one of the trig. functions instead of differentiating so I was struggling on that for a while, but I got that near the end.
Reply 63
Original post by Devp93
Yeh me too :biggrin: was very happy when my calculator told me it was the same as what I worked out :smile:. Almost forgot how to integrate tan^2 x tho :s-smilie: good thing I had 45 mins for the last question :smile:.


I was trying to find the integral of tan2x\tan^2 x in the formula book for a while and then I saw sec2x\sec^2 x in there and it dawned on me :biggrin:
Original post by gillyhl
I got both of those as well. What took me a while was finding

ddx(2cos2x+sin2x)\frac{d}{dx}\left(\frac{2}{\cos^2 x+\sin 2x}\right)

I was integrating one of the trig. functions instead of differentiating so I was struggling on that for a while, but I got that near the end.

I used the quotient rule on that so I got -2(2cos2x-2sinxcosx)/((cosx)^2+sin2x))^2
Reply 65
I also used that, but I was, basically, doing it wrong xD
Reply 66
Original post by Unkempt_One
I think I got (A/4)X^4 + Ax + x^2 + B. I'm fairly sure x^2 was a positive sign because u=(dy/dx)-2x


YAAAY - Actually, I think I got positive too - I just cudnt remember:smile: How was that 8 marks though?
Original post by JJBSussat
YAAAY - Actually, I think I got positive too - I just cudnt remember:smile: How was that 8 marks though?

Not sure. I did a C4-style differential equation solving stage, which involved an integration by substitution, and then a C2 integration. It felt like 8 marks to me though, because it was quite long.
Reply 68
Original post by Unkempt_One
Not sure. I did a C4-style differential equation solving stage, which involved an integration by substitution, and then a C2 integration. It felt like 8 marks to me though, because it was quite long.


But we got the same answer?
Reply 69
I didn't use substitution because it the integrand was in the form f(x)f(x)\frac{f'(x)}{f(x)}
Original post by gillyhl
I got both of those as well. What took me a while was finding

ddx(2cos2x+sin2x)\frac{d}{dx}\left(\frac{2}{\cos^2 x+\sin 2x}\right)

I was integrating one of the trig. functions instead of differentiating so I was struggling on that for a while, but I got that near the end.


Which question is that for? Defo don't recgonise that D:
Original post by gillyhl
I didn't use substitution because it the integrand was in the form f(x)f(x)\frac{f'(x)}{f(x)}

Yeah, but remembering standard integrals . . . pfft
Reply 72
Loved this exam!! Strangely alot easier than Core 4!
Original post by JJBSussat
But we got the same answer?

Apparently. I can't quite follow what you did.
I thought the exam was ok, there were a couple of hard bits but not as many as in fp2!
What were you supposed to do for Q4 where it had the integral of sinxsin2x?
Original post by Igiveup12345
I thought the exam was ok, there were a couple of hard bits but not as many as in fp2!
What were you supposed to do for Q4 where it had the integral of sinxsin2x?

It works out as the integral of 2((sinx)^2)cosx, as sin2x=2sinxcosx So you use the substitution u=sinx and go from there.
Reply 76
Original post by Igiveup12345
I thought the exam was ok, there were a couple of hard bits but not as many as in fp2!
What were you supposed to do for Q4 where it had the integral of sinxsin2x?


2sin^2 xcosx and then substitution :smile:
Reply 77
Original post by SmileyGurl13
Which question is that for? Defo don't recgonise that D:


It was for the limits question where you had to differentiate twice.
Thank you :smile: I think I did something along those lines, or at least I got halfway there so I might get some marks! :s-smilie:
Reply 79
Haha ok - well I'll try to explain better:

(x^3 + 1)du/dx = (3x^2)u

(1/u)(du/dx) = (3x^2)/(x^3 + 1) by dividing u down and dividing by (x^3 + 1).

lnu = ((3x^2) ln (x^3 + 1)) / 3x^2 + c Integrating. Divide the log by the differential of (x^3 +1) which is (3x^2) which cancels with the first bit.

ln u = ln (x^3 + 1) + c
u = e^ (ln (x^3 + 1) + c)
u = (x^3 + 1)e^c e^c = A
u = A(x^3 + 1)
u = dy/dx - 2x from beginning of q
dy/dx = u + 2x
dy/dx = A(x^3 + 1) + 2x integrate
y = (A/4)x^4 + Ax + x^2 + b

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