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January 25th, 2009, 07:01 AM   #1
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Riemann Sum help please

I'm stuck on this Riemann sum and I have no idea what I'm doing wrong:
y= 3x^2 + 2 on [1,2]

Ive gotten Delta X = 1/n
xi = 1 + 1/n

the equation i used was - 3[(1+(1/n)i)^2 + 2] [1/n]
i simplified that down into this by distributing- [3 + 3/n^2 i^2 +2] [1/n]
that turns into - (3/n) + (3/n^3)i^2 + (2/n)
that then becomes (3/n) and (2/n) can be taken out of the equation to become 5... and I replace i^2 with (n(n+1)(2n+1))/6 and i multiply it with 3/n^3
I take the limit of that as it approaches infinity and i get 1.
3+2+1 is 6, but the when I take the anti derivative of this it is 9?!?!?!?
I need to know where I went wrong!!! and would someone write this out for me step by step from the beginning if you have the time?
Thanks!
pranavpuck is offline  
 
January 25th, 2009, 12:54 PM   #2
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Re: Riemann Sum help please

(1+i/n)^2 = 1 + 2i/n + (i/n)^2

In you calculation you left out the middle term.
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January 31st, 2009, 02:47 PM   #3
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Re: Riemann Sum help please

Definite Integral is defined with the infinite Riemann Sum:


Your problem is this:

f(x) = 3x^2 + 2; a=1; b=2





That there is your problem...

Simplify...










Now you need to know how to interpret summations... rewrite them in a simplified algebraic form...







Evaluate the limit...




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