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?

  • $\begingroup$ You didn't write a sum. Also, we don't convert a Riemann sum into an integral. We take the limit of Riemann sums to get an integral. $\endgroup$ – zhw. May 27 '18 at 17:43
  • $\begingroup$ I just left out the limit part for simplicity sake of writing, as it is not needed when converting the sum to a definite integral (ie. finding a, b and f(x)) - at least to my understanding of how to solve these questions. $\endgroup$ – miweo May 27 '18 at 17:49

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 $\Delta x = \frac{1}{n}.$

  • $\begingroup$ why did you decide $\Delta{x} = \frac{1}{n}$? $\endgroup$ – miweo May 27 '18 at 18:58
  • $\begingroup$ Try it with $1/n$ rather than $1/2n$ and see what happens. $\endgroup$ – zhw. May 28 '18 at 1:05

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