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# Does anyone understand this question? watch

1. https://gyazo.com/701ffd648e171c0bb050226b67458d5a

Doesn't make any sense to me.
2. (Original post by Louisb19)
https://gyazo.com/701ffd648e171c0bb050226b67458d5a

Doesn't make any sense to me.
Are you told what f(x) or f(t) are?
3. (Original post by multiratiunculae)
Are you told what f(x) or f(t) are?
Nope.
4. (Original post by Louisb19)
Nope.
Hint:

Integrate the identity between 1 and -1 and observe that f(-x) is just a reflection of f(x) in the y-axis.
5. (Original post by Louisb19)
Nope.
Is this an a level question or university?
6. (Original post by multiratiunculae)
Is this an a level question or university?
C1 and C2 question
7. (Original post by Louisb19)
C1 and C2 question
I'd appreciate a solution when you get there!
8. (Original post by Indeterminate)
Hint:

Integrate the identity between 1 and -1 and observe that f(-x) is just a reflection of f(x) in the y-axis.
I tried integrating both sides but I was left with something extremely messy. and also you get like integral(1, - 1) of 3x^2 * integral(1, -1) of f(t)dtdx which has thrown me off.
9. (Original post by multiratiunculae)
I'd appreciate a solution when you get there!
https://gyazo.com/c6d92504a60b795768d619ea4e2587e8

Doesn't make any sense to me.
10. (Original post by Louisb19)
I tried integrating both sides but I was left with something extremely messy. and also you get like integral(1, - 1) of 3x^2 * integral(1, -1) of f(t)dtdx which has thrown me off.
How can you integrate a function which you haven't seen?
11. (Original post by Louisb19)
I tried integrating both sides but I was left with something extremely messy. and also you get like integral(1, - 1) of 3x^2 * integral(1, -1) of f(t)dtdx which has thrown me off.
You need to realise that the definite integral inside the bracket on the RHS is JUST a constant, so all you have to do is integrate the term on the outside

After that, it's simple. Just note the relationship between f(x) and f(-x) and collect multiples of

and you're done
12. (Original post by multiratiunculae)
How can you integrate a function which you haven't seen?
Yeah I honestly have no remote idea what is going on in that solution let alone the question itself. Where does the 12 even come from in the first place. I understand the integral(1, -1) of f(x) = integral(1, -1) of f(-x) however beyond that I'm very confused. Where does du come from?
13. (Original post by Indeterminate)
You need to realise that the definite integral inside the bracket on the RHS is JUST a constant, so all you have to do is integrate the term on the outside

After that, it's simple. Just note the relationship between f(x) and f(-x) and collect multiples of

and you're done
https://scontent-lhr3-1.xx.fbcdn.net...f5&oe=562C0931

I got this far however I'm confused again.
14. (Original post by Louisb19)
https://scontent-lhr3-1.xx.fbcdn.net...f5&oe=562C0931

I got this far however I'm confused again.
You shouldn't have called it c because it's the same as all the other multiples of the integral.

All you have to do is treat it as a constant when doing the integration and then rearrange to find its value
15. (Original post by Indeterminate)
You shouldn't have called it c because it's the same as all the other multiples of the integral.

All you have to do is treat it as a constant when doing the integration and then rearrange to find its value
But t could be any number so I don't see how integral(1, -1) of the integral(1 , -1) f(t) dt dx = integral(1, -1) f(x)dx
16. If the answer is k=6 then I know the missing step.
17. (Original post by dpm)
If the answer is k=6 then I know the missing step.
18. (Original post by Louisb19)
But t could be any number so I don't see how integral(1, -1) of the integral(1 , -1) f(t) dt dx = integral(1, -1) f(x)dx
The person who wrote the question was just trying to be awkward. The fact is that f is the same function.

For example, it doesn't matter if i write

or .

They're the same. You should re-label it before starting the integration.

As I said, you have to treat that bracket as a constant and just worry about integrating the the term

(Original post by Louisb19)
Nope, it's not 6
19. (i)Firstly, I did exactly the same as Loiusb19 and used his work to confirm.
(ii) in the given function, sub in x=1
(iii) in the given function sub in x=-1
(iv) solve simultaneously to conclude that f(1)=f(-1)
(v) re-arrange the original function to make the integral the subject, using x=1 and step (iv) and letting f(1)=f(-1)=k
(vi) this should lead to (6-k)/1
(vii) set this equal to 12-2k (as seen in Louisb19's work, albeit with different variables)
(viii) .... this gives k=6
(ix) I now see that this does not lead to the integral being what I said !!!!!
(x) I will post it anyway, simply because I think that steps (ii) - (iv) are relevant!!
20. (Original post by Indeterminate)
The person who wrote the question was just trying to be awkward. The fact is that f is the same function.

For example, it doesn't matter if i write

or .

They're the same. You should re-label it before starting the integration.

As I said, you have to treat that bracket as a constant and just worry about integrating the the term

Nope, it's not 6

Yeah you are right, I really wasn't thinking there. For some reason my brain didn't like it however it makes complete sense now! I wish I could rep you again however I can't :/

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