Integration CORE 2 maths question - help me! Watch

theguywhosaidhi
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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
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Kevin De Bruyne
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(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?
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theguywhosaidhi
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(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.
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Kevin De Bruyne
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(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.
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theguywhosaidhi
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(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
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Kevin De Bruyne
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(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.
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theguywhosaidhi
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(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!
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Kevin De Bruyne
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(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.
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theguywhosaidhi
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(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)
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Zacken
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(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)
Look in your formula booklet.
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Kevin De Bruyne
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(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 \frac{n!}{(n-r)!(r!)}, 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 \frac{3!}{(3-2)!(2!)} =\frac{3*2*1}{(1)(2*1)} = 3.
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theguywhosaidhi
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(Original post by Zacken)
Look in your formula booklet.
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
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Zacken
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(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.
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theguywhosaidhi
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(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
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Zacken
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(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.
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theguywhosaidhi
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(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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