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C2 integration

Ugfurdytntitled.png

Hey, I was wondering if someone could help me with this question. The answer to part (i) is k=4

In the second part i got 4/2root2 but the mark scheme says that it is wrong. In the mark scheme they change infinity to 4, but why?
What is the answer on the MS?
Original post by Appazap


Hey, I was wondering if someone could help me with this question. The answer to part (i) is k=4

In the second part i got 4/2root2 but the mark scheme says that it is wrong. In the mark scheme they change infinity to 4, but why?


If you mean you got 422\dfrac{4}{2\sqrt{2}} then this is fine because it is equal to 2\sqrt{2} which is the answer.

As for why they change to 4, I don't know. Post the mark scheme.
Reply 3
Original post by TheTechnocrat
What is the answer on the MS?


just root 2
I think it is most likely a misprinted
Original post by Appazap
just root 2


You just need to rationalise the denominator.
Reply 6
Original post by RDKGames
If you mean you got 422\dfrac{4}{2\sqrt{2}} then this is fine because it is equal to 2\sqrt{2} which is the answer.

As for why they change to 4, I don't know. Post the mark scheme.


oh i feel like an idiot!! have they they just rationalised the denominator then?


also mark scheme Ugfurdytntitled.png
Original post by Appazap
oh i feel like an idiot!! have they they just rationalised the denominator then?


also mark scheme Ugfurdytntitled.png



Yes.
Original post by Appazap
oh i feel like an idiot!! have they they just rationalised the denominator then?


also mark scheme


Yes they rationalised the denominator.

What they do in MS is they turn the integral into a proper one with a limit of kk (which has a different value than in the prev part), then they evaluate the integral in terms of k, then they see what the limit of it is when kk \rightarrow \infty.
Reply 9
Original post by TheTechnocrat
Yes.


Original post by RDKGames
Yes they rationalised the denominator.

What they do in MS is they turn the integral into a proper one with a limit of kk (which has a different value than in the prev part), then they evaluate the integral in terms of k, then they see what the limit of it is when kk \rightarrow \infty.


Thank you all!!:smile:

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