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    https://gyazo.com/701ffd648e171c0bb050226b67458d5a

    Doesn't make any sense to me.
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    (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?
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    (Original post by multiratiunculae)
    Are you told what f(x) or f(t) are?
    Nope.
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    (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.
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    (Original post by Louisb19)
    Nope.
    Is this an a level question or university?
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    (Original post by multiratiunculae)
    Is this an a level question or university?
    C1 and C2 question
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    (Original post by Louisb19)
    C1 and C2 question
    I'd appreciate a solution when you get there!
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    (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.
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    (Original post by multiratiunculae)
    I'd appreciate a solution when you get there!
    https://gyazo.com/c6d92504a60b795768d619ea4e2587e8

    Doesn't make any sense to me.
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    (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?
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    (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

    \displaystyle \int_{-1}^{1} f(x) \ dx

    and you're done
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    (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?
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    (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

    \displaystyle \int_{-1}^{1} f(x) \ dx

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

    I got this far however I'm confused again.
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    (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
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    (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
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    If the answer is k=6 then I know the missing step.
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    (Original post by dpm)
    If the answer is k=6 then I know the missing step.
    The answer is 4, please tell anyway!
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    (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

    f(t)=t^2 or f(x) = x^2 .

    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 3x^2 term

    (Original post by Louisb19)
    Yeah it is! Please tell
    Nope, it's not 6
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    (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!!
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    (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

    f(t)=t^2 or f(x) = x^2 .

    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 3x^2 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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