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# C4 - Integrating e^x Tweet

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1. C4 - Integrating e^x
Mark Scheme

2011 Jan Paper

Can any maths G tell me why for Q6 c) Integrating (e^2x − 2)^2 directly, instead of expanding it is wrong? I added one to the power and divided by 3 and then divided by 2 as e^2x integrates to e^2x/2
2. Re: C4 - Integrating e^x
That isn't a rule, it's true for things like (x-a)^n dx as d/dx [ 1/(n+1) * (x-a)^(n+1)] = (x-a)^n. (n =/= -1)

But

d/dx 1/6 * (e^(2x) - 2)^3 = e^(2x) * (e^(2x) - 2)^2
3. Re: C4 - Integrating e^x
(Original post by Mr Einstein)
Mark Scheme

2011 Jan Paper

Can any maths G tell me why for Q6 c) Integrating (e^2x − 2)^2 directly, instead of expanding it is wrong? I added one to the power and divided by 3 and then divided by 2 as e^2x integrates to e^2x/2
What you have done does not work. Why do you think it should?
4. Re: C4 - Integrating e^x
(Original post by Allofthem)
That isn't a rule, it's true for things like (x-a)^n dx as d/dx [ 1/(n+1) * (x-a)^(n+1)] = (x-a)^n. (n =/= -1)

But

d/dx 1/6 * (e^(2x) - 2)^3 = e^(2x) * (e^(2x) - 2)^2
Ty for a constructive answer, though I'm not quite sure what you're saying 1/6 * (e^(2x) - 2)^3 doesn't equal e^(2x) * (e^(2x) - 2)^2.

(Original post by james22)
What you have done does not work. Why do you think it should?
I know..... that's why I posted.

I know how to get the right answer I was just asking for an explanation of why it is wrong, so why do you think it shouldn't?
5. Re: C4 - Integrating e^x
(Original post by Mr Einstein)
I know..... that's why I posted.

I know how to get the right answer I was just asking for an explanation of why it is wrong, so why do you think it shouldn't?
The reason it doesn't work is it gives the wrong answers. There is no more to it, you just cannot do that.

I've just realized that you might have been taught integration badly, it is common for people to think this. You can do this when you are integrating f(ax+b) but not in any other situation. There is no reverse chain rule (which is what it looks like you are doing). It only really works with f(ax+b) because if you substitute u=ax+b you get the integral of 1/af(u)du.