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At the opening of a chapter on Riemann Sums and definite integrals, my book gives the following example problem:

Consider the region bounded by the graph of $$f(x)=\sqrt{x}$$ and the x-axis for $0≤x≤1$. Evaluate the limit: $$\lim\limits_{n \to ∞} \sum_{i=1}^n f(c_i)\Delta x_i$$ Where $c_i$ is the right endpoint of the partition given by $c_i = i^2/n^2$ and $\Delta x_i$ is the width of the $i$th interval.

Solution

The width of the $i$th subinterval is $$\Delta x_i ={i^2 \over n^2} - {(i - 1)^2 \over n^2}$$

Two questions concerning this:

a) Why is $c_i$ defined a $i^2/n^2$ ?

b) Why is $\Delta x_i$ defined as ${i\,^2 \over n^2} - {(i - 1)^2 \over n^2}$ ?

(For the purposes the question I have not included the rest of the question, nor have I included the associated graph).

2 Answers 2

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It appears to be that the x-Interval (0,1) has been divided into sub intervals $(c_{i-1},c_i)$ for $i=1,\ldots,n$.

a) The $c_i$ are defined in such a way as to make $f(c_i)$ easy to evaluate (to $\frac{i}{n}$), probably in order to make the evaluation of the limit $$\lim\limits_{n \to ∞} \sum_{i=1}^n f(c_i)\Delta x_i$$ not so technically demanding.

b) $\Delta x_i$ as defined is the length of the above mentioned interval $(c_{i-1},c_i)$. That makes sense as Rieman Sums of a function are sums of products of interval length ($\Delta x_i$) and a function value from that interval ($f(c_i)$).

Answering questions from OP:

Please use the below answer from Paramanand Singh if you need to understand what the Rieman Sums are. I assumed above that you basically know them, but are unsure how they apply to this problem.

The values $\frac{i^2}{n^2}$ are just a convenient way to partition the interval [0,1]. They come from "having solved lots of similar problems and thus having a repertoire of tools for them" on the part of the textbook author. You are, I think, not supposed yet to come up with them yourself, you are just supposed to follow along and solve the problem as stated. Later when learning about integrals you will see why this approach was chosen, what alternatives there were, etc.

The length given for $\Delta x_i$ is just the normal formula for the length of an interval: upper boundary - lower boundary. Since the upper boundary is $c_i=\frac{i^2}{n^2}$ and the lower boundary is $c_{i-1}=\frac{(i-1)^2}{n^2}$ you get the stated formula for $\Delta x_i$

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    I've spend the past hour searching but I'm still confused. Besides the convenience of the defining $c_i$ as $i^2/n^2$, where did the value come from. How was $\Delta x_i$ determined given that the length of the sub interval is variable? – 2017-02-01
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    I added my answers to to the above to my original answers – 2017-02-01
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It appears that you are not familiar with the definition of a Riemann sum. For clarity I repeat the definition here. Let $f$ be a bounded function on interval $[a, b] $. A set $P$ of the form $$P=\{x_{0},x_{1},x_{2},\ldots,x_{n}\}$$ is said to be a partition of interval $[a, b] $ if $$a=x_{0}Riemann sum for $f$ over partition $P$. The expression $(x_{i} - x_{i-1})$ is also denoted by $\Delta x_{i} $.

In the current question the interval under consideration is $[0,1]$. The points $t_{i} $ are chosen to match the right end points so that $t_{i} =x_{i} $. Now the question chooses $x_{i} =i^{2}/n^{2}$. This is OK because we require $x_{0}=a=0,x_{n}=b=1$ (both of these are true here) and also that $x_{i-1}

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    Ok, this is making better sense now. Just to clarify: **a)** $i^2/n^2$ works because when $i = 0$ $x_0 = {0^2 \over n^2} = 0$ and when $i = n$, $x_n = {n^2 \over n^2}=1$ which fits with the definition of a Riemann sum **b)** ∆$x$ is defined as ${i\,^2 \over n^2} - {(i - 1)^2 \over n^2}$ since ∆$x$ is $(x_i - x_{i-1})$ and $x_i = {i\,^2 \over n^2}$ and $x_{i-1} = {(i - 1)^2 \over n^2}$ **c)** Finally, it is perfectly acceptable (although less convenient) to define $c_i$ as ${(i - 1)^2 \over n^2}$ (left endpoint form) or trapezoidal form. Is there anything wrong with these statements? – 2017-02-01
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    @Theseus: whatever you mention is fully correct. The beauty of the Riemann integral is that when it exists, the choice of $c_{i}$ does not matter as long as $c_{i} \in [x_{i-1},x_{i}]$, so you can take $c_{i} =(i-1)^{2}/n^{2}$ and still get the same answer $2/3$. – 2017-02-01
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    I have a question. Why do we bother working with unequal intervals when having equal intervals makes the partial sum simpler to evaluate? Whether they are equal or unequal, they all tend to zero during the computation of the integral. – 2017-09-29
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    @R004: well the evaluation of the limit of sum may not be easy in all cases when we deal with intervals of equal length. In fact for the common function like $x^{n} $ it is much easier to keep the points of partition in geometric progression. – 2017-09-29
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    @R004: Also note that the Riemann sum involves two choices: partitions as well as tags. We can't fix both of them and one of these choices must remain arbitrary. The general definition keeps both of them arbitrary. – 2017-09-29