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# Edexcel C4 Revision Thread (28/06 14:00) watch

there is an official mark shecem for the mock , ill scan it in later today and post it, prob in the evenin say 11ish
That'd be brilliant.

2. nice1
3. I have done a page of the 4 extra integrals with proofs by differentiation.

You may like to take a look:

http://153rd.com/sam/as/C4/integrals2

An exam question could be to prove one of them.
4. here we go i have hosted them on my space

hope you can all read it properly , i think you can anyway
5. (Original post by samdavyson)
I have done a page of the 4 extra integrals with proofs by differentiation.

You may like to take a look:

http://153rd.com/sam/as/C4/integrals2

An exam question could be to prove one of them.

for proving cot x, couldnt you integrate cosx/sinx , using the identity that integral of f '(x)/f(x) = lnx ?
6. (Original post by Hash)
for proving cot x, couldnt you integrate cosx/sinx , using the identity that integral of f '(x)/f(x) = lnx ?
Yes.
7. So edexcel made a mock paper and a specimen paper?
8. yes
10. (Original post by Econ4m1t)
Usually they involve establishing an equation, differentiating some stuff, and using a bit of chain rule.

If it helps, this used to be on P3, so have a look through some past P3 papers and find such questions to work though. http://math.mdsalih.com/Data/index.p...el+Mathematics
11. Can someone explain question 3 on paper A1 please?

I get upto ∫cosθ/(cos²θ )^3/2 dθ and then i get stuck.. is what i've done wrong?

Thanks!
12. I'm having trouble with part C of this question. Keep in mind that the integral of 1/(2x+1)(x-2)dx is lnk[(x-2)/(2x+1)]^1/5
Attached Images

13. wow
14. (Original post by ntrik)
I'm having trouble with part C of this question. Keep in mind that the integral of 1/(2x+1)(x-2)dx is lnk[(x-2)/(2x+1)]^1/5
You have to do:
∫1/10y dy = ∫ 1/(2x+1)(x-2) dx

you'll get a constant obviously so you need to plug in the given x and y values to find its value

1/10ln(10) = lnk[(3-2)/(2(3)+1)]^1/5

lnk = ln[10^(7/2)]

this looks so wrong
15. (Original post by violet)
Can someone explain question 3 on paper A1 please?

I get upto ∫cosθ/(cos²θ )^3/2 dθ and then i get stuck.. is what i've done wrong?

Thanks!
Remember: am/n = nth root of a to the power of m.

So:

∫cosθ/(cos²θ )^3/2 dθ = ∫cosθ/(cos³θ) dθ = ∫1/(cos²θ) dθ

=> ∫sec²θ dθ => Standard Integral. ( => tanθ + C)
16. (Original post by ntrik)
I'm having trouble with part C of this question. Keep in mind that the integral of 1/(2x+1)(x-2)dx is lnk[(x-2)/(2x+1)]^1/5
As featherflare says you need to separate the variables as in any differential equation.

Here rearrange to:

1/10y (dy/dx) = 1/(2x+1)(x-2)

And integrate both sides with respect to x:

∫1/10y (dy/dx) dx = ∫ 1/(2x+1)(x-2) dx

=> ∫1/10y dy = ∫ 1/(2x+1)(x-2) dx

Then use the integral from earlier in the question for the RHS and the LHS comes to 1/10 ln(y) (or 1/10 ln(10y) ).

The different results depend on how you integrate it. The constant takes care of the difference.
17. (Original post by Featherflare)
You have to do:
∫1/10y dy = ∫ 1/(2x+1)(x-2) dx

you'll get a constant obviously so you need to plug in the given x and y values to find its value
Edit - Argh crap, I think I made a silly mistake...

Okay, here is what I have

∫1/10y dy = ∫ 1/(2x+1)(x-2) dx
(1/10)lny = lnk[(x-2)/(2x+1)]^1/5
y^1/10 = k[(x-2)/(2x+1)]^1/5

x = 3 and y = 1

1 = k(1/7)^1/5
k = 1.476 (3 dp)

I'm pretty sure that wrong because I'm not sure how I can get to y = 49[(x-2)/(2x+1)]² which is the answer.
18. (Original post by Featherflare)
But is it?

4 (a). 1 = A(x - 2) + B(2x + 1)

Put x = 2: 1 = 5B, B = 1/5
Put x = -1/2: 1 = -5/2A, A = -2/5

(b) ∫[(-2/5)/(2x + 1)] + [(1/5)/(x - 2)]dx

=> -2/10ln(2x + 1) + 1/5ln(x - 2)

=> -1/5ln(2x + 1) + 1/5ln(x - 2)

=> ln(2x + 1)-1/5 + ln(x - 2)1/5

=> ln[(x - 2)1/5/(2x + 1)1/5]

=>ln[(x - 2)/(2x + 1)1/5]

Yes
19. plz if a Q. asks for volume of a curve rotated 360 or 180 , what do we use
pie or 2pie for anyone plz answer me any1 now
20. Well, i've just finished partial fractions, so I can help people with those, lol.
Although it's a very short/easy chapter so noone will need my help.

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