# real analysis – existence of an integral by selecting a dissection

So I answer a question about the real analysis. I answered the question, but I'm not sure I did it right. Could someone look at my work and let me know if it's going in the right direction?

The question is: by choosing a suitable preparation from $$(1, 3)$$ , show that $$int ^ 3_1 2x ^ 2 -1 dx$$ exists and finds its value.

So, to prove it exists, I think I have to prove the lower sum $$s_d$$ and the upper sum $$s ^ u$$ are equal?

I decided for my suitable preparation $$(1,1+ frac {2i} {n}, …, 1+ frac {2i-1} {n}, 3)$$

I have the feeling that I chose the preparation incorrectly.

I work from there $$m_i =$$ inf $$f ( frac {1 + 2 (i-1)} {n}, frac {1 + 2i} {n})$$ = $$f ( frac {2i-1} {n})$$

and $$M_i =$$ sup $$f ( frac {1 + 2 (i-1)} {n}, frac {1 + 2i} {n})$$ = $$f ( frac {1 + 2i} {n})$$

I would use that to train $$s_d$$ and $$s ^ u$$ but I get very confusing answers, I was then given the tip $$sum ^ n_ {i = 1} i ^ 2 = frac {n ^ 3} {3} + frac {n ^ 2} {2} + frac {n} {6}$$ So I know that I have to get a similar equation.

Any help would be appreciated.

Posted on Categories Articles