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Calculate the integral by Riemann $\displaystyle \int _{0}^{1}\sqrt{x} \, dx$

$$\displaystyle a_{k} =0+k\cdot \frac{1}{n} =\frac{k}{n}$$

$$\displaystyle \Delta x=\frac{1-0}{n} =\frac{1}{n}$$

\begin{align*} \lim_{n\rightarrow \infty }\sum_{k=1}^{n}\left(\sqrt{\frac{k}{n}}\right) \cdotp \frac{1}{n} & = \lim_{n\rightarrow \infty }\sum_{k=1}^{n}\left(\frac{\sqrt{k}}{\sqrt{n}}\right) \cdotp \frac{1}{n} \\[1ex] & = \lim_{n\rightarrow \infty }\sum_{k=1}^{n}\left(\frac{1}{n\sqrt{n}} \cdotp \sqrt{k}\right) \\[1ex] & = \lim_{n\rightarrow \infty }\frac{1}{n \sqrt{n}} \cdotp \sum_{k=1}^{n}\left(\sqrt{k}\right) \end{align*}

I am stuck on the sum of $\sqrt k$ maybe have another way to solve this question?

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  • $\begingroup$ Can you use, that root is integrable? Then you'll be able to choose specific points for partition. $\endgroup$
    – zkutch
    May 23, 2022 at 17:02
  • $\begingroup$ Does this answer your question? For Riemann sums involving square roots, why do we let $c_{i} = \frac{i^{2}}{ n^{2}}$? $\endgroup$
    – user170231
    May 23, 2022 at 17:02
  • $\begingroup$ @user170231 well thanks but I tried right now and I didn't succeed I write $\dfrac{k^2}{n^2}$ and calculate the sum and the limit and I got an answer of 0.5 with is wrong $\endgroup$
    – Epsilon1.2
    May 23, 2022 at 17:13
  • $\begingroup$ You can use the method here to do all general power integrals (continuity needed for irrational powers) $\endgroup$
    – FShrike
    May 23, 2022 at 17:27

6 Answers 6

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As suggested in the linked question and other answers posted here, you need not use equally-spaced subintervals in the partition of $[0,1]$. Instead, consider the sequence of intervals

$$\left\{\left[\left(\frac{i-1}n\right)^2, \left(\frac in\right)^2\right] \,\bigg| \, 1 \le i \le n \right\}_{n\in\Bbb N}$$

each with length $\frac{i^2}{n^2} - \frac{(i-1)^2}{n^2} = \frac{2i - 1}{n^2}$.

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Then the (right-endpoint) Riemann sum is

$$\sum_{i=1}^n \sqrt{\left(\frac in\right)^2} \frac{2i-1}{n^2} = \sum_{i=1}^n \frac{2i^2-i}{n^3} = \frac{(n+1)(4n-1)}{6n^2}$$

As $n\to\infty$, the sum converges to the definite integral, and we have

$$\int_0^1 \sqrt x \, dx = \lim_{n\to\infty} \frac{(n+1)(4n-1)}{6n^2} = \frac23$$

which agrees with the known antiderivative result,

$$\int_0^1 x^{\frac12} \, dx = \frac23 x^{\frac32}\bigg|_0^1 = \frac23$$

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In the spirit of Reimann sums, you do not need to make the sequence $a_k$ arithmetic, you can use any sequence that covers the interval.

In this case, since you need to take the square root, it will make things simpler if you make $a_k$ a quadratic sequence...

$$a_k = 0 + \frac {k^2 }{n^2}$$

Then $\sqrt{a_k}$ simplifies to $\frac kn$ , but your $\Delta x$ will now depend of $k$... $$ \Delta x_k = a_{k+1}-a_k = \frac {2k+1}{n^2}$$

So $$\int_0^1 \sqrt x dx = \lim_{n \to \infty} \sum_{k=1}^n a_k \Delta x_k $$

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

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

Now you can use the formulas...

$$ \sum_{k=1}^n k^2 = \frac n6 (2n+1)(n+1) \text{ and } \sum_{k=1}^n k=\frac n2(n+1) $$

to obtain the required result

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  • $\begingroup$ Note that it is not quite enough that the partition cover the interval. You also need the length of the largest subinterval to converge to $0$ as $n$ grows to $\infty$. That does happen here with the largest subinterval being the last one, which is $(2n-1)/n^2$ wide. $\endgroup$
    – 2'5 9'2
    May 23, 2022 at 17:59
  • $\begingroup$ That was what I was trying to evoke using the term "cover" - but I accept that my use of the term was not correct. Perhaps an alternate formulation of what constitutes a valid Reimann partition could be that every open cover of the sequence is also an open cover of the interval. $\endgroup$
    – WW1
    May 23, 2022 at 18:22
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I thought it might be instructive to present a way forward that uses creative telescoping and simple estimates. To that end we proceed.


Note that we can write

$$\begin{align} n^{3/2}-1&=\sum_{k=1}^n \left((k+1)^{3/2}-k^{3/2}\right)\\\\ &=\frac32 \sum_{k=1}^n \left(\sqrt{k} +O(k^{-1/2})\right)\tag1 \end{align}$$

Rearranging $(1)$ reveals

$$\frac1{n^{3/2}}\sum_{k=1}^n \sqrt k=\frac23 \left(1-n^{-3/2}\right)+\frac1{n^{3/2}}\underbrace{\sum_{k=1}^n O(k^{-1/2})}_{\le O(n)}$$

Letting $n\to \infty$ yields the coveted result

$$\lim_{n\to\infty}\frac1{n^{3/2}}\sum_{k=1}^n \sqrt k=\frac23$$

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Here is a direct way to look at this if you would like to stick with a Riemann sum that uses equally spaced subintervals. The strategy is, in small doses, to replace the irrational $\sqrt{k}$ values with nearby integers.

$$\begin{align} \lim_{n\to\infty}\frac{1}{n\sqrt{n}}\sum_{k=1}^n\sqrt{k} &=\lim_{n\to\infty}\frac{1}{n^2\sqrt{n^2}}\sum_{k=1}^{n^2}\sqrt{k} \end{align}$$

Already this might give you pause. But for one thing, as $n\to\infty$, so does $n^2$. The sum on the right is like the sum on the left but between one value of $n$ and the next, you add

$$\frac{1}{n^2\sqrt{n^2}}\left(\sqrt{(n-1)^2+1}+\sqrt{(n-1)^2+2}+\cdots+\sqrt{(n-1)^2+(2n-1)}\right)$$

That is, you add $2n-1$ terms that are each less than or equal to $n$, and the result is divided by $n^3$. So this tail that you add with each step, all by itself, converges to $0$. So the two sums have the same limit. Starting again:

$$\begin{align} \lim_{n\to\infty}\frac{1}{n\sqrt{n}}\sum_{k=1}^n\sqrt{k} &=\lim_{n\to\infty}\frac{1}{n^2\sqrt{n^2}}\sum_{k=1}^{n^2}\sqrt{k}\\ &=\lim_{n\to\infty}\frac{1}{n^3}\left(\sqrt{1}+\sqrt{2}+\cdots+\sqrt{n^2}\right)\\ &\leq\lim_{n\to\infty}\frac{1}{n^3}\left(\sqrt{1}+\sqrt{4}+\sqrt{4}+\sqrt{4}+\sqrt{9}+\sqrt{9}+\cdots+\sqrt{n^2}\right)\\ \end{align}$$ Here we have one $\sqrt{1}$s, then three $\sqrt{4}$s, then five $\sqrt{9}$s, and so on.

$$\begin{align} &=\lim_{n\to\infty}\frac{1}{n^3}\left(1+3(2)+5(3)+\cdots+(2n-1)n\right)\\ &=\lim_{n\to\infty}\frac{1}{n^3}\sum_{k=1}^n(2k-1)k\\ &=\lim_{n\to\infty}\frac{1}{n^3}\left(2\frac{n(n+1)(2n+1)}{6}-\frac{n(n+1)}2\right)\\ &=\lim_{n\to\infty}\left(\frac13(1+1/n)(2+1/n)-\frac{(1+1/n)}{2n}\right)\\ &=\frac23 \end{align}$$

So we have an upper bound and the limit you are after is at most $\frac23$. (Of course we know independently that the limit actually is $\frac23$.)

Do you see how to modify this to get a lower bound? Hint: $1,1,1,2,2,2,2,2,3,3,3,3,3,3,3,\ldots$.

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Maybe this will help you (i hope this is true).
First let's make a change of variable: $t=\sqrt{x}$ so we get $\int_{0}^{1}\sqrt{x}dx=\int_{0}^{1}2t^2dt$

Now lets compute: $\int_{0}^{1}2t^2dt$ using Riemman integrable.
Thus by doing exactly as you did we get: $lim_{n \to \infty } \frac{2}{n^3} \sum_{k=1}^{n}k^2$
And we know that the sum of squares of n natural numbers can be calculated using the formula $[n(n+1)(2n+1)] / 6$. So: $lim_{n \to \infty } \frac{2}{n^3} [n(n+1)(2n+1)] / 6 = 2/3$
This is correct because: $\int_{0}^{1}\sqrt{x}dx = 2/3$

So we did exactly what it was asked:

Calculate the integral by Riemann $\int_{0}^{1}\sqrt{x}dx$

But by calculating integral by Riemman on a simplest function.

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  • $\begingroup$ This is not a clear answer $\endgroup$
    – FShrike
    May 23, 2022 at 17:27
  • $\begingroup$ @FShrike why? how can i improve it? $\endgroup$
    – X0-user-0X
    May 23, 2022 at 17:31
  • $\begingroup$ They want square roots $\endgroup$
    – FShrike
    May 23, 2022 at 17:34
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    $\begingroup$ @FShrike It is not say in the question that we have to calculate the Riemman sum directly on the root function given. In the question it is only asked to compute the given integrable using a Riemman integrable, and this is exactly what we did. If not the op must be more precise. $\endgroup$
    – X0-user-0X
    May 23, 2022 at 17:38
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If the answer upper isn't satisfying here an oher.

Choose a partition as follow: $x_0=0 , x_1 = 1/n^2 ,x_2=(2^2)/n^2 ,...,x_n=n^2/n^2=1$ so $x_k-x_{k-1}=\Delta x_k=\frac{2k-1}{n^2}$.
$\sum_{k=1}^{n}\frac{2k-1}{n^2}\frac{k}{n}=\sum_{k=1}^{n}\frac{2k^2-k}{n^3}$
And we know a formula for each of those sums.

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