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For this function, I know that the a value is 1 and that the $\Delta x$ is $\frac{1}{n}$. I'm not sure how to go from there in order to express this as a definite integral. Can anyone please help me out?

$$\lim_{n\to \infty}\frac{1}{n}\sum_{i=1}^n\frac{3}{1+(\frac{i}{n})^2}$$

winonator
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  • Welcome to MSE! This was a very fun question to work with $\ddot\smile$ You may already know this, but if you find an answer helpful or a question interesting, click the little up arrow to up-vote it. This will give the author a nice reward of points. If you feel an answer totally resolved a question of yours, give it the $\color{green}{\checkmark}$ to thank the author with a slightly larger sum of points. Anyways, thanks for joining, and I look forward to your future contributions! – gen-ℤ ready to perish Jan 22 '18 at 02:22
  • @dg123, see https://math.stackexchange.com/questions/469885/the-limit-of-a-sum-sum-k-1n-fracnn2k2 – lab bhattacharjee Jan 22 '18 at 02:29

2 Answers2

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One way to express the Riemann sum as a definite integral is

$$\int_a^bf(x)\,dx = \lim_{n\to\infty}\sum_{i=1}^{n} f(a+i\,\Delta x)\,\Delta x$$ where $$\Delta x =\frac{b-a}n$$

Taking $$J=\lim_{n\to\infty}\frac1n\sum_{i=1}^n\frac3{1+\left(\frac in\right)^2}$$ and making the substitution $g(u)=1/(1+u^2)$ with some rearrangement gives

$$J=3\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^ng\!\left(\frac in\right)$$

Take $\Delta x=1/n$, and at this point you can probably see that $a=0$ and $b=1$:

$$\begin{align} J&=3\lim_{n\to\infty}\frac{1-0}{n}\sum_{i=1}^n g\!\left( 0+i\frac{1-0}{n}\right) \\ &=3\lim_{n\to\infty}\Delta x\sum_{i=1}^n g(a+i\,\Delta x) \end{align}$$

and thusly you get $$J=3\int_0^1 \frac{dx}{1+x^2}$$

This graph clearly suggests that this is in fact a correct reformulation of $J$.

gen-ℤ ready to perish
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  • Is there a way to do this without substitution? I haven't learned that yet. – winonator Jan 22 '18 at 16:49
  • @dg123 The substitution isn’t important, and you don’t have to make it; I’m merely trying to make the pattern more visible. Also, this substitution in particular isn’t too convoluted, so I think it should be fine if it’s for an assignment. Are you still confused? – gen-ℤ ready to perish Jan 22 '18 at 16:51
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    How do we know that a is 0 and b is 1? – winonator Jan 22 '18 at 16:52
  • @dg123 It’s a little bit of trial and error and recognizing patterns. You said yourself that $\Delta x=1/n$, so when I plugged that in, it immediately became apprentices that $a=0$ and that $b-a=1$. Part of recognizing patterns is to identify key components and substitute them with symbols from the formula. Also note that the purpose of this site isn’t to give you an answer exactly as you or your instructor might want but rather to illuminate the technique so that you learn. You may have to do a bit of rearranging and working on your own. – gen-ℤ ready to perish Jan 22 '18 at 17:01
  • What if I take the summand in the third equation and rearrange it to $\frac{1}{9+(\frac{3i}{n})^2}.\frac{3}{n}.9$? You could say b-a=3, and the result is an integral with different limits multiplied by nine. You could express this as an integral in infinite ways. Why am I wrong? – harry Apr 06 '21 at 11:07
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You want to see the $n^{th}$ term in that sequence that you are taking the limit from as the $n^{th}$ Riemann sum. The $\frac{1}{n}$ tells you that you divide your interval in $n$ equal parts. Then, by definition of Riemann sums, each term in the sum must be the value of your function either at the beginning or at the end of each part of the interval in which you divided it. The thing that varies term by term is what is inside the square, so this already suggests that this is the $x$. And now it reminds to compute the limts of integration. In the last term of the sum, you get $\frac{n}{n}=1$, so take this as the upper limit (hence we have a right Riemann sum). This makes the value in the first term, $\frac{1}{n}$, the value of $x$ at the right end of our first segment of interval (of length $\frac{1}{n}$). This shows that the left integration limit must be $0$. So we have:

$$\int_{0}^{1}\frac{3}{1+x^{2}}= \lim_{n\to \infty} \frac{1}{n}\sum_{i=1}^{n}\frac{3}{1+(\frac{i}{n})^{2}}$$

Pedro
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  • How did you get that? – winonator Jan 22 '18 at 01:43
  • I can try to explain, let me edit the answer – Pedro Jan 22 '18 at 01:45
  • How do we know that a must be zero? isn't a equal to 1 cuz if of the $ 1 + (\frac{i}{n})^2$ corresponding to $x_i = a + i\Delta(x)$? – winonator Jan 22 '18 at 16:56
  • We know its zero because we know its an upper sum (we evaluate on the right endpoint of the piece of interval) and the first term in the sum is 1/n. Since the length of the piece of interval is 1/n, the starting point was 1/n-1/n=0 – Pedro Jan 22 '18 at 17:06