AQA A2 Mathematics MPC4 Core 4 - 9th June 2015 [Discussion &amp; unofficial markscheme]

Original post by tiny hobbit

Oh I didn't realise, I guess that figures.

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Reply 43

707456

Original post by CD223

Usually if you have 1 over a function you can make the whole thing to the power of minus one. I.E: 1/(x^2-3x-8) would become (x^2-3x-8)^-1 then you can use integration by substitution.

If you have x over a function then check to see if you can notice if the function on top is the derivative (or a multiple of the derivative) of the bottom. If so, it is ln of whatever is on the bottom. Or a multiple of ln of whatever was on the bottom.

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How wud u use subsitution for that(what wud u take as u)?
Wud u need to factorise it first or something?

Reply 44

707456

Original post by CD223

Oh I didn't realise, I guess that figures.

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Oh okay i understand.now!!! Thank you :biggrin:

for.ur help

Reply 45

How would you integrate sinxe^cosx? :|

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Reply 46

Original post by 707456

Oh okay i understand.now!!! Thank you :biggrin:

for.ur help

That's okay! Any more questions, don't hesitate to ask! :smile:

Reply 47

Original post by saad97

How would you integrate sinxe^cosx? :|

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As far as I know, with e you multiply the coefficient by the reciprocal of the derivative of the original power.

EG: Integrating 2e^4x with respect to x
= (1/4)2e^4x = (1/2)e^4x + C

In your case it follows that the integral of (sinx)e^(cosx) would be:

(sinx)/(-sinx) [e^cosx] + C

Which cancels to (-1)[e^cosx] + C

http://m.wolframalpha.com/input/?i=integrate+sinx*e%5Ecosx&x=0&y=0

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(edited 9 years ago)

Reply 48

tiny hobbit

Original post by CD223

This only works because what you've got in front of the power of e is the derivative of the power (give or take a constant factor). If you had, say, cos x in front of e^(cos x), you'd be stuffed.

Reply 49

Original post by tiny hobbit

This only works because what you've got in front of the power of e is the derivative of the power (give or take a constant factor). If you had, say, cos x in front of e^(cos x), you'd be stuffed.

Sorry, can I ask why that wouldn't just be (cosx/-sinx)e^cosx?

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Reply 50

tiny hobbit

Original post by CD223

Sorry, can I ask why that wouldn't just be (cosx/-sinx)e^cosx?

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Try differentiating that, using product and quotient. Does it get back to where you started?

You can only do this sort of "compensating" with a constant.

Reply 51

Original post by tiny hobbit

Try differentiating that, using product and quotient. Does it get back to where you started?

You can only do this sort of "compensating" with a constant.

Oh right! So many flaws in my maths ability lol. An A level seems to have done nothing 🙈😂

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Reply 52

Original post by tiny hobbit

Try differentiating that, using product and quotient. Does it get back to where you started?

You can only do this sort of "compensating" with a constant.

Thank you though, you've been really helpful. Forgive me if not, but are you taking the exam this year? (I kinda assume not given how good you are at maths!)

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Reply 53

tiny hobbit

Original post by CD223

Thank you though, you've been really helpful. Forgive me if not, but are you taking the exam this year? (I kinda assume not given how good you are at maths!)

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No, it was a few decades back.

Reply 54

Original post by tiny hobbit

No, it was a few decades back.

Ah right - didn't mean to patronise you! Well thank you so much for you help, it is greatly appreciated.

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Reply 55

Hi guys, just a quick modelling with differential equations.

Mould is spreading through a piece of cheese of mass 1kg in such a way that at a time t days, x kg of the cheese has become mouldy where: -

dx/dt = 2x(1-x)

Given that initially 0.01 kg of the cheese is mouldy.

a) Show that x = [e^(2t)] / [99+e^(2t)]

I have integrated and ended up with:
ln(x/1-x) = 2t + C.
Subbing in x when t=0 gave me that C=1/99.
How would I rearrange to find x and get it in the form the question wants?
I raised both sides by e and got
x/1-x = e^2t x e^c.

Thank you. :smile:

Reply 56

Original post by saad97

How did you integrate it?

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Reply 57

Original post by CD223

How did you integrate it?

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ImageUploadedByStudent Room1426436639.274845.jpg

Reply 58

Original post by saad97

Oh right! That's as far as I got with it aha! Let me have a second look at it. Thanks for posting your written work - that makes it a whole lot clearer! :smile:

Reply 59