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.

Reply 8

Louisb19

Original post by multiratiunculae

I'd appreciate a solution when you get there!

https://gyazo.com/c6d92504a60b795768d619ea4e2587e8

Doesn't make any sense to me.

Reply 9

multiratiunculae

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?

Reply 10

Indeterminate

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 :smile:

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 :smile:

Reply 11

Louisb19

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?

Reply 12

Louisb19

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 :smile:

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 :smile:

https://scontent-lhr3-1.xx.fbcdn.net/hphotos-xft1/v/t34.0-12/12170456_437849129744407_124324348_n.jpg?oh=c41d27e5c3774cb14c86e4cafe73f6f5&oe=562C0931

I got this far however I'm confused again.

Reply 13

Indeterminate

Original post by Louisb19

https://scontent-lhr3-1.xx.fbcdn.net/hphotos-xft1/v/t34.0-12/12170456_437849129744407_124324348_n.jpg?oh=c41d27e5c3774cb14c86e4cafe73f6f5&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 :smile:

Reply 14

Louisb19

Original post by Indeterminate

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

Reply 15

dpm

If the answer is k=6 then I know the missing step.

Reply 16

Louisb19

Original post by dpm

If the answer is k=6 then I know the missing step.

The answer is 4, please tell anyway!

(edited 8 years ago)

Reply 17

Indeterminate

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

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 :smile:

Nope, it's not 6 :smile:

Reply 18

dpm

(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!!

Reply 19

Louisb19

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

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 :smile:

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 :/