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# Integration CORE 2 maths question - help me!

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1. given that:

"Integration sign with upper limit 3, lower limit 1" (x^2 -2x + K) dx = 8 2/3

find the value of K.

*******sorry for poor formatting, I have no idea how to make it look snazzy
2. (Original post by theguywhosaidhi)
given that:

"Integration sign with upper limit 3, lower limit 1" (x^2 -2x + K) dx = 8 2/3

find the value of K.

*******sorry for poor formatting, I have no idea how to make it look snazzy
What have you tried so far?
3. (Original post by SeanFM)
What have you tried so far?
integrating then plugging in the value of x for 3 and 1, and then subtracting the 2 equations but then the k values cancel since they're subtracted from one another. I'm not sure where to start.
4. (Original post by theguywhosaidhi)
integrating then plugging in the value of x for 3 and 1, and then subtracting the 2 equations but then the k values cancel since they're subtracted from one another. I'm not sure where to start.
I suspect that your integral of k isn't correct - what have you got in your working in the bits where you are just about to evaluate the integral? i.e [.... + .... + ...] between 3 and 1.
5. (Original post by SeanFM)
I suspect that your integral of k isn't correct - what have you got in your working in the bits where you are just about to evaluate the integral? i.e [.... + .... + ...] between 3 and 1.
(x^3/3 - x^2 + kx) with upper 3 and lower 1

then: (9 - 9+ kx) - (1/3 - 1 + kx) = 8 2/3

= kx + 2/3 - kx = 8 2/3
6. (Original post by theguywhosaidhi)
(x^3/3 - x^2 + kx) with upper 3 and lower 1

then: (9 - 9+ kx) - (1/3 - 1 + kx) = 8 2/3

= kx + 2/3 - kx = 8 2/3
Ah - I see, your integral of k is correct - the problem lies in the evaluation.

Remember, just like you've done for the x^3/3 and the x^2, you can put in x = 3 and x =1 next to the k.
7. (Original post by SeanFM)
I suspect that your integral of k isn't correct - what have you got in your working in the bits where you are just about to evaluate the integral? i.e [.... + .... + ...] between 3 and 1.
I SEE WHERE I WENT WRONG DAMN IT! thanks anyways i didn't sub in x value in for the kx

thank you for the help!
8. (Original post by theguywhosaidhi)
I SEE WHERE I WENT WRONG DAMN IT! thanks anyways i didn't sub in x value in for the kx

thank you for the help!
Well done it is very good (genuinely) that you spotted it yourself.
9. (Original post by SeanFM)
Well done it is very good (genuinely) that you spotted it yourself.
too nice/polite man aw i know you probably have stuff to do. but.... when a binomial is to the power of n, how would you expand it? I don't how you can use the 'nCr' button to get the factorial, the only info I'm given is that n is greater or equal to 2 (and an integer)
10. (Original post by theguywhosaidhi)
too nice/polite man aw i know you probably have stuff to do. but.... when a binomial is to the power of n, how would you expand it? I don't how you can use the 'nCr' button to get the factorial, the only info I'm given is that n is greater or equal to 2 (and an integer)
11. (Original post by theguywhosaidhi)
too nice/polite man aw i know you probably have stuff to do. but.... when a binomial is to the power of n, how would you expand it? I don't how you can use the 'nCr' button to get the factorial, the only info I'm given is that n is greater or equal to 2 (and an integer)
No worries even if I cannot help, there are quite a few active posters in the Maths forum so someone can jump in

nCr on your calculator gives you the value of , so if it's not satisfactory to leave it as nCr (I don't know if it is or not) then that's another way of going about it. Eg 3C2 is = = 3.
12. (Original post by Zacken)
so I'm going to have n's everywhere in my expansion, or am I missing something.

1st expansion. (n over 1) then 1
2nd: (n over 2) 1 + 1/4
3rd: (n over 3) 1 + 1/4^2
13. (Original post by theguywhosaidhi)
so I'm going to have n's everywhere in my expansion, or am I missing something.

1st expansion. (n over 1) then 1
2nd: (n over 2) 1 + 1/4^2 ---> since n is greater equal to 2? idek
3rd: (n over 3) 1 + 1/4^3
You haven't told us the original expression you want to expand, but it looks wrong.

You will have n's everywhere, yes.
14. (Original post by Zacken)
You haven't told us the original expression you want to expand, but it looks wrong.

You will have n's everywhere, yes.
damn, I'm retarded, original eq: (1 + 1/4x)^n

i changed my answer a little in the edit
15. (Original post by theguywhosaidhi)
damn, I'm retarded, original eq: (1 + 1/4x)^n

i changed my answer a little in the edit
Okay, so the formula booklet says

(1+x)^n = 1 + nx + n(n-1)/2 x^2 + n(n-1)(n-2)/3! x^3 + ...

So in your case, just replace all the x's with x/4.
16. (Original post by Zacken)
Okay, so the formula booklet says

(1+x)^n = 1 + nx + n(n-1)/2 x^2 + n(n-1)(n-2)/3! x^3 + ...

So in your case, just replace all the x's with x/4.
thank you!

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