I'm asked to find the Riemann integral that the following limit equals:
$$ \lim_{n\rightarrow\infty}\sum_{i=1}^n\frac{1}{2n}\frac{5}{7-(\frac{i-1}{\sqrt{n(i-1)}})^4} $$
as $$ \lim_{n\rightarrow\infty}\sum_{i=1}^nf(x_i^*)\Delta{x}= \int_b^af(x)dx $$
I know that my $\Delta{x}$ = $\frac{1}{2n}$
$$\lim_{n\rightarrow\infty}\sum_{i=1}^n\frac{1}{2n}\frac{5}{7-(\frac{i-1}{\sqrt{n(i-1)}})^4} = \lim_{n\rightarrow\infty}\sum_{i=1}^n\Delta{x}\frac{5}{7-\frac{i^4+1}{{n^2(i-1)^2}}}$$ $$ = \lim_{n\rightarrow\infty}\sum_{i=1}^n\Delta{x}\frac{5}{7-\frac{i^4+1}{{n^2(i^2-2i+1)}}}$$ $$ = \lim_{n\rightarrow\infty}\sum_{i=1}^n\Delta{x}\frac{5}{7-\frac{i}{{n^2(-2)}}}$$ $$ = \lim_{n\rightarrow\infty}\sum_{i=1}^n\Delta{x}\frac{5}{7-(\frac{-i}{{n}}\frac{1}{2n})}$$ $$ = \lim_{n\rightarrow\infty}\sum_{i=1}^n\Delta{x}\frac{5}{7-(\frac{-i}{{n}}\Delta{x})}$$
I'm stuck at this point... Does it look correct so far? If so, how should I proceed in finding my $f(x)$ and my upper and lower limits of integration?
Hints: Something happened in your calculation, I'm not sure what. I would note
$$\frac{i-1}{\sqrt {n(i-1)}} = \sqrt {\frac{i-1}{n}}.$$
So when you raise that to the $4$th power you get
$$\left (\frac{i-1}{n}\right)^2$$
You might also consider $\Delta x = \frac{1}{n}.$