Using the midpoint rule to show that the definite integral of f(x)=x^2 on [0,1] is 1/3

Question

I have a question that asks the following

Let $f(x)=x^2$ on [0,1]. If we use the midpoint rule $M_n$ with $n$ subintervals of width $\Delta x= \frac{1}{n}$ so that $$M_n= lim_{n\to∞}\sum_{i=1}^n \Delta xf(\frac{(i-1) \Delta x + i \Delta x } {2} )= \sum_{i=1}^n \frac{1}{n} \frac {(\frac {i}{n} - \frac{1}{n} + \frac {i}{n})^2} {4} $$ show that $\lim_{n\to∞} M_n= \frac{1}{3}$.

I have attached a picture of my work so far and am unable to further simplify to get the desired answer of $\frac{1}{3}$. Any help appreciated!

Answer 1

Note $$\sum_{i=1}^n(2i-1)^2=4\sum_{i=1}^n i^2 -4\sum_{i=1}^n i + \sum_{i=1}^n 1=\frac{2}{3}n(n+1)(2n+1)-2n(n+1)+n$$ $$=\frac{4}{3}n^3-\frac{1}{3}n$$ Now dividing by $4n^3$ gives $$\frac{1}{3}-\frac{1}{12n^2}.$$ Taking the limit as $n\to \infty$ gives $\frac{1}{3}$.