AQA C3 What type of integration is this question asking for?

Hi, would anybody be able to tell me the integration topic or method to solve this question described by the mark scheme, as I didn't have a clue how to answer the question and need more practise.
The question is 8a from the AQA C3 January 2013 paper here:

And the mark scheme being

I usually just read the mark scheme and work it out but its using k for some reason as well as the exponential 1 just disappearing, if anyone can tell me the integration type this answer requires it will be greatly appreciated, thank you

Reply 1

Kevin De Bruyne

21

Original post by Sayless

'Recognition'. You know that the differential of e^(kx) = ke^(kx) so you 'work backwards'.

Also, how can you write x^(1-a) as x^(...) * x^(...)? Then for e^(1-a) it will be the same logic.

Original post by Sayless

x

Reply 2

Zacken

22

Original post by Sayless

x

As per SeanFM - you should recognise that

\displaystyle \int f'(x) e^{f(x)} \, \mathrm{d}x = e^{f(x)} + c

Reply 3

Sayless

OP

12

Original post by SeanFM

'Recognition'. You know that the differential of e^(kx) = ke^(kx) so you 'work backwards'.

Also, how can you write x^(1-a) as x^(...) * x^(...)? Then for e^(1-a) it will be the same logic.

yeah so i simplify it to:
integral of e x e^(-2x)
now what do i do I don't know how to approach this types of questions even after I simplify it to that,
I know how to do e^2x or 2e^4x differentiation and integration but I don't know when there are 2e's timesed together

Reply 4

Zacken

22

Original post by Sayless

yeah so i simplify it to:
integral of e x e^(-2x)
now what do i do I don't know how to approach this types of questions even after I simplify it to that,
I know how to do e^2x or 2e^4x differentiation and integration but I don't know when there are 2e's timesed together

e

is just a number so you need only integrate

e \int_0^{\ln 2} e^{-2x} \, \mathrm{d}x

.

Standard Patterns

Original post by Sayless

yeah so i simplify it to:
integral of e x e^(-2x)
now what do i do I don't know how to approach this types of questions even after I simplify it to that,
I know how to do e^2x or 2e^4x differentiation and integration but I don't know when there are 2e's timesed together

e is a constant and it does not depend on x.

Reply 7

Sayless

OP

12

Original post by Zacken

e

is just a number so you need only integrate

e \int_0^{\ln 2} e^{-2x} \, \mathrm{d}x

.

ah right, I thought that if I took e out of the integrand it would mean I would have to take e out of e^(-2x)
didn't think it worked that way but now it makes sense if its timsing, when I did it I did
integral of e/e^(2x)
which i couldnt make sense of

Reply 8

Zacken

22

Original post by Sayless

ah right, I thought that if I took e out of the integrand it would mean I would have to take e out of e^(-2x)
didn't think it worked that way but now it makes sense if its timsing, when I did it I did
integral of e/e^(2x)
which i couldnt make sense of

Have you managed to solve the problem now?