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.
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.
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 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
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)
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)
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 (n−r)!(r!)n!, 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 (3−2)!(2!)3! =(1)(2∗1)3∗2∗1 = 3.