easy limit question.

**Evan.Kimia** · March 28th 2010, 11:45 AM

Problem:

Correct me if i am wrong, but i think i know that the delta x would be 2/n according to (b-a)/n which would also give 2i/n as our right interval. I would then factor out the 2/n and 2i/n and get 4/n^2. So i would end solving the limit of n -> infinity of
$\frac{4}{n^{2}}\left( \frac{n\left( n+1 \right)}{2} \right)^{2}$

trouble is im not sure how to evaluate this limit because the whole thing wrapped around the exponent is throwing me off.

**Plato** · March 28th 2010, 12:13 PM

$\sum {\left( {\frac{{2i}} {n}} \right)^3 \left( {\frac{2} {n}} \right) = \frac{{16}} {{n^4 }}\sum {i^3 } }$
$\frac{{16}} {{n^4 }}\left( {\frac{{n(n + 1)}} {2}} \right)^2 = \frac{{16}} {{n^4 }}\left( {\frac{{n^4 + 2n^3 + n^2 }} {4}} \right) \to ? $

**Evan.Kimia** · March 28th 2010, 12:40 PM

Ahhh. I see. The limit would be 4. Cant believe i missed that. Anyways, i see I made a mistake. I do have one question for you. when you take the $\left( \frac{2i}{n} \right)^{3}$ as a constant multiplied by the sum of f(x) why would the value of i be cubed if that too was a constant? im assuming its not but im not quite sure. thank you though.

**Plato** · March 28th 2010, 12:50 PM

$\left( {\frac{{2i}} {n}} \right)^3 = \left( {\frac{2} {n}} \right)^3 i^3$

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