# Integration CORE 2 maths question - help me! Watch

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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

"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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#2

(Original post by

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

**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

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(Original post by

What have you tried so far?

**SeanFM**)What have you tried so far?

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#4

(Original post by

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.

**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.

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(Original post by

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.

**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.

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

= kx + 2/3 - kx = 8 2/3

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#6

(Original post by

(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

**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

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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**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.

thank you for the help!

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#8

(Original post by

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!

**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!

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(Original post by

Well done it is very good (genuinely) that you spotted it yourself.

**SeanFM**)Well done it is very good (genuinely) that you spotted it yourself.

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#10

(Original post by

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)

**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)

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#11

**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)

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.

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(Original post by

Look in your formula booklet.

**Zacken**)Look in your formula booklet.

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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#13

(Original post by

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

**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 will have n's everywhere, yes.

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(Original post by

You haven't told us the original expression you want to expand, but it looks wrong.

You will have n's everywhere, yes.

**Zacken**)You haven't told us the original expression you want to expand, but it looks wrong.

You will have n's everywhere, yes.

i changed my answer a little in the edit

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#15

(Original post by

damn, I'm retarded, original eq: (1 + 1/4x)^n

i changed my answer a little in the edit

**theguywhosaidhi**)damn, I'm retarded, original eq: (1 + 1/4x)^n

i changed my answer a little in the edit

(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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(Original post by

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.

**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.

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