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\newtheorem{theorem}{Theorem}
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\newtheorem{conclusion}[theorem]{Conclusion}
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\lhead{\color{blue} \large Lecture Notes}
\chead{\color{black} \Large Riemann Sums}
\rhead{\small page   \ \thepage}
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\lfoot{\footnotesize \copyright  \; Hidegkuti,  2019}
\rfoot{\footnotesize Last revised: April 10, 2019}
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We have several reasons to want to know the area under the graph of a
function. \ One reason is because this is how we obtaim area formulas. \ For
example, the area of a circle is found by doubling the area under the graph
of the function $f\left( x\right) =\sqrt{r^{2}-x^{2}}$. \ \ Another example
is when we need to multiply the dependent and independent variables ($x$ and 
$y$) as the next example shows.

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\textbf{Example 1.} \ Suppose that an object is moving along a vertical
line. \ The graph shown depicts the velocity function $v\left( t\right) $ of
the object. \ We know about the location function, $s\left( t\right) $ that $%
s\left( t\right) =5$. \ What is the location of the object at $t=10$?\vspace{%
0.06in}

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\textbf{Solution:} \ This problem is quite easy, it is just strangely
phrased. \ The graph is showing a velocity function that is constant on
intervals. \ We also call a function like this a step-function.

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During\vspace{0.06in} the first two seconds, the object has a constant
velocity of $2\dfrac{\unit{m}}{\unit{s}}$: \ \ \ \ \ \ \ \ \ $\ v\left(
t\right) =2\dfrac{\unit{m}}{\unit{s}}$ on $\left[ 0\unit{s},2\unit{s}\right] 
$

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On this interval, the distance traveled is $s=vt=2\dfrac{\unit{m}}{\unit{s}}%
\cdot 2\unit{s}=4\unit{m}$, thus the object traveled $4$ meters upward
between $t=0\unit{s}$ and $t=2\unit{s}$. \ If $s\left( 0\right) =5$, then $%
s\left( 2\right) =5+4=9$. \ 

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Next, between $t=2\unit{s}$ and $t=5\unit{s}$, the object has a constant
velocity of $1\dfrac{\unit{m}}{\unit{s}}$: \ \ \ $\ v\left( t\right) =1%
\dfrac{\unit{m}}{\unit{s}}$ on $\left[ 2\unit{s},5\unit{s}\right] .$

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On this interval, the distance traveled is\vspace{0.06in} $s=vt=1\dfrac{%
\unit{m}}{\unit{s}}\cdot 3\unit{s}=3\unit{m}$, thus the object traveled $3$
meters upward between $t=2\unit{s}$ and $t=5\unit{s}$. \ If $s\left(
2\right) =9$, then $s\left( 5\right) =9+3=12$.

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During the next second, on $\left[ 5\unit{s},6\unit{s}\right] $, the object
is at rest. \ Thus $s\left( 6\right) =12+0=12$.

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Next, between $t=6\unit{s}$ and $t=8\unit{s}$, the object has a constant
velocity of $-2\dfrac{\unit{m}}{\unit{s}}$ . \ The negative velocity
indicates that the object is moving downward. \ \ \ \ $\ v\left( t\right) =-2%
\dfrac{\unit{m}}{\unit{s}}$ on $\left[ 6\unit{s},8\unit{s}\right] .$

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On this interval, the distance traveled is\vspace{0.06in} $s=vt=-2\dfrac{%
\unit{m}}{\unit{s}}\cdot 2\unit{s}=-4\unit{m}$, thus the object traveled $4$
meters downward on $\left[ 6\unit{s},8\unit{s}\right] $. \ If $s\left(
6\right) =12$, then $s\left( 8\right) =12-4=8$.

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Finally, between $t=8\unit{s}$ and $t=10\unit{s}$, object has a constant
velocity of $4\dfrac{\unit{m}}{\unit{s}}$: \ \ \ \ $\ v\left( t\right) =4%
\dfrac{\unit{m}}{\unit{s}}$ on $\left[ 8\unit{s},10\unit{s}\right] .$

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On this interval, the distance traveled is\vspace{0.06in} $s=vt=4\dfrac{%
\unit{m}}{\unit{s}}\cdot 2\unit{s}=8\unit{m}$, thus the object traveled $8$
meters upward on $\left[ 8\unit{s},10\unit{s}\right] $. \ If $s\left(
8\right) =8$, then $s\left( 10\right) =8+8=16$. \ Thus \fbox{$s\left(
10\right) =16$}.\vspace{0.06in}\vspace{0.06in}

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This example shows that the distance traveled has a geometric
interpretation, we are adding (signed) areas of rectangles. \ So we again
have a reason for wanting to find the area under the graph. \ 

In this sense, the area under the graph represents multiplication. \ Every
time we need to multiply the dependent and independent variables (time and
velocity in this case) and the dependent variable is not a constant, then
the product becomes the area under the graph.

It is difficult to imagine an object whose velocity changes in no time from $%
-2\dfrac{\unit{m}}{\unit{s}}$ to $4\dfrac{\unit{m}}{\unit{s}}$. \ Our
intuition suggests that the velocity graph of an object should be
continuous. \ However, the following construction will be a method of
finding the area under graphs of functions that are not necessarily
continuous; so this concept of area will be more general than the area known
in geometry.

\pagebreak

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Recall first the definition of a bounded function.\vspace{0.1in}

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\textbf{Definition}: \ $\ $A function is \textbf{bounded above} if there
exists a real number $B$ such that for all $x$ in domain, $f\left( x\right)
\leq B$. \ We say that $B$ is an upper bound for $f$. \ \vspace{0.1in}%
\newline
Similarly, a function is \textbf{bounded below} if there exists a real
number $S$ such that for all $x$ in domain, $f\left( x\right) \geq S$. \ We
say that $S$ is a lower bound for $f$.\vspace{0.1in}\newline
A function is \textbf{bounded} if it is bounded from above and from below.

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Suppose that $f$ is a function bounded on a closed interval $\left[ a,b%
\right] $. \ Boundedness is a nexessary but not sufficient condition on the
existence of area under the graph. \ In other words, we will not even try to
find area under the graph if $f$ is not bounded on $\left[ a,b\right] $.

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\textbf{Definition}: \ $\ $A \textbf{partition} of $\left[ a,b\right] $ is a
finite set of numbers within the interval, listed in an increasing order. \
We usually denote a partition of $\left[ a,b\right] $ by $\left\{
a=x_{0},x_{1},x_{2},...,x_{n}=b\right\} .$ \ Partitions divide a single
interval into several smaller intevals. \ In the case of $\left\{
a=x_{0},x_{1},x_{2},...,x_{n}=b\right\} $, the partition creates $n$
sub-intervals. \ We often denote the length of the $k$th sub-interval by $%
\Delta x_{k}$.

\vspace{0.1in} 
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Consider the five-unit long interval $\left[ 7,12\right] $. \ The partition $%
P=\left\{ 7,8,9,10,11,12\right\} $, defined by six numbers, partitions the
interval $\left[ 7,12\right] $ into five smaller intervals: $\left[ 7,8%
\right] $ , $\left[ 8,9\right] $ , $\left[ 9,10\right] $ ,$\left[ 10,11%
\right] ,$ and $\left[ 11,12\right] $ . \ Because we need six numers to
define the five intervals, it is smart to start counting the numbers in the
partition starting with zero. \ Assuming that each sub-interval has the same
length, $\Delta x_{k}=\dfrac{5}{5}=1$ for $k=1,2,3,4,5$. 
\begin{equation*}
\left\{ 7,8,9,10,11,12\right\} \text{ where }7=x_{0}\text{, }8=x_{1}\text{, }%
9=x_{2}\text{, }10=x_{3}\text{, }11=x_{4}\text{, and }12=x_{5}
\end{equation*}%
The same interval can be divided (or partitioned) into ten smaller
intervals. \ Assuming that each sub-interval has the same length, $\Delta
x_{k}=\dfrac{5}{10}=\dfrac{1}{2}$ for $k=1,2,3,...,10$.%
\begin{eqnarray*}
\left\{ x_{0},x_{1},x_{2},...,x_{9},x_{10}\right\} &=&\left\{
7,7.5,8,...,11.5,12\right\} \\
x_{k} &=&7+0.5k\text{ for }k=1,2..,10
\end{eqnarray*}
or into $100$ smaller intervals: $\Delta x_{k}=\dfrac{5}{100}=0.05$ for $%
k=1,2,3,...,100$.%
\begin{eqnarray*}
\left\{ x_{0},x_{1},x_{2},...,x_{99},x_{100}\right\} &=&\left\{
7,7.05,7.1,7.15...,11.5,12\right\} \\
x_{k} &=&7+0.05k\text{ where }k=0,1,2,...,100
\end{eqnarray*}

The partitions we have seen divided the interval into smaller intervals of
the same length. \ This is not necessary but often useful.\vspace{0.1in}

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\textbf{Definition}: \ $\ $A partition that defines sub-intervals of the
same length is called a \textbf{uniform partition}. \ In such cases, we use $%
\Delta x$ to denote the common length of the sub-intervals $\Delta x_{k}$.

\vspace{0.1in} 
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The length of subintervals in a uniform partition of $\left[ a,b\right] $
with $n$ subintervals is $\Delta x=\dfrac{b-a}{n}$.\pagebreak

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\textbf{Example 2.} \ Suppose that $a=2$ and $b=4$. Find the uniform
partition of $\left[ a,b\right] $ of $50$ subintervals.

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\textbf{Solution:} \ The length of the entire interval is $2$. \ Then the
length of each sub-interval is $\dfrac{b-a}{n}=\dfrac{2}{50}=0.04$. \ Then$%
\vspace{0.06in}$

\ \ \ \ \ \ $a=x_{0}=2,x_{1}=2.04,x_{2}=2.08,...x_{49}=3.96,x_{50}=4=b$ \ or 
$x_{k}=2+k\cdot 0.04$ for $k=0,1,2,..,50$

This partition can also be expressed as $\left\{ x_{k}=2+0.04k\text{ for all 
}k=0,1,2,..50\right\} \vspace{0.06in}$

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Suppose that $f$ is a function bounded on $\left[ a,b\right] $ and $P$ is a
partition of $\left[ a,b\right] $ of $n$ sub-intervals. \ We will
approximate the area under the graph by approximating $f$ as constant over
each subinterval. \ This way the area under the graph is a rectangle. \
There are several reasonable ways of approximating $f$. \ 

For example, let $f\left( x\right) =\dfrac{1}{x}$ on the interval $\left[ 1,4%
\right] $. \ \ \ Let us use a uniform partition with $n=6$.

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On each of the interval $\left[ x_{k,}x_{k+1}\right] $ of the partition, we
will approximate $f$ as constant. \ The value of that constant function will
be the function value $f\left( x_{k}\right) $ of the left end-point in the
interval. \ \ 

On the first subinterval $\left[ 1,1.5\right] $, we use $f\left( 1\right) $
as the constant value. \ On the second subinterval $\left[ 1.5,2\right] ,$
we use $f\left( 1.5\right) $ as the constant value, and so on.

This way, we approximate the area by adding areas of rectangles. The
horizontal side (we'll call it width) of each rectangle is $\dfrac{3}{6}%
=\allowbreak \dfrac{1}{2}$. \ \ 
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The heights of the six rectangles are $f\left( 1\right) $, $f\left(
1.5\right) $, $f\left( 2\right) $, $f\left( 2.5\right) $, $f\left( 3\right) $%
, and $f\left( 3.5\right) $. \ So the six areas added: 
\begin{eqnarray*}
A_{\text{appr}} &=&\dfrac{1}{2}\cdot f\left( 1\right) +\dfrac{1}{2}\cdot
f\left( 1.5\right) +\dfrac{1}{2}\cdot f\left( 2\right) +\dfrac{1}{2}\cdot
f\left( 2.5\right) +\dfrac{1}{2}\cdot f\left( 3\right) +\dfrac{1}{2}\cdot
f\left( 3.5\right) \\
&=&\dfrac{1}{2}\left( f\left( 1\right) +f\left( 1.5\right) +f\left( 2\right)
+f\left( 2.5\right) +f\left( 3\right) +f\left( 3.5\right) \right) \\
&=&\dfrac{1}{2}\left( f\left( 1\right) +f\left( \dfrac{3}{2}\right) +f\left(
2\right) +f\left( \dfrac{5}{2}\right) +f\left( 3\right) +f\left( \dfrac{7}{2}%
\right) \right) \text{ \ \ \ \ \ \ \ }f\left( x\right) =\dfrac{1}{x} \\
&=&\dfrac{1}{2}\left( 1+\dfrac{2}{3}+\dfrac{1}{2}+\dfrac{2}{5}+\dfrac{1}{3}+%
\dfrac{2}{7}\right) \approx 1.\,\allowbreak 592\,86
\end{eqnarray*}

Such an approximation is called a \textbf{left Riemann sum}.

We did not find the exact area. \ Instead, we have a somewhat crude
approximation of it. \ In this case, our approximation is certainly above
the exact value, because the function is completely covered by our
rectangles on $\left[ 1,4\right] $. \ In short, our approximation \textit{%
over-estimates} the area.

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This is not always the case. \ It happens in case of $f\left( x\right) =%
\dfrac{1}{x}$ because it is a decreasing function. \ Using a left Riemann
sum for other functions may overestimate or underestimate the area under the
graph, and in some cases we cannot know which.\vspace{0.4in}%
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Similarly, we can approximate the function over each subinterval as the
right end-point.

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In this case, the height of the first rectangle will be $f\left( 1.5\right)
, $ that of the second rectangle $f\left( 2\right) $, and so on. \ The
approximate area is then%
\begin{equation*}
A_{\text{appr}}=\dsum\limits_{k=1}^{6}f\left( x_{k}\right) \Delta x
\end{equation*}%
\vspace{0.4in} 
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\begin{eqnarray*}
A_{\text{appr}} &=&\dfrac{1}{2}\cdot f\left( 1.5\right) +\dfrac{v}{2}\cdot
f\left( 2\right) +\dfrac{1}{2}\cdot f\left( 2.5\right) +\dfrac{1}{2}\cdot
f\left( 3\right) +\dfrac{1}{2}\cdot f\left( 3.5\right) +\dfrac{1}{2}\cdot
f\left( 4\right) \\
&=&\dfrac{1}{2}\left( f\left( 1.5\right) +f\left( 2\right) +f\left(
2.5\right) +f\left( 3\right) +f\left( 3.5\right) +f\left( 4\right) \right) =%
\dfrac{1}{2}\left( f\left( \dfrac{3}{2}\right) +f\left( 2\right) +f\left( 
\dfrac{5}{2}\right) +f\left( 3\right) +f\left( \dfrac{7}{2}\right) +f\left(
4\right) \right) \text{ \ \ \ \ \ \ \ }f\left( x\right) =\dfrac{1}{x} \\
&=&\dfrac{1}{2}\left( \dfrac{2}{3}+\dfrac{1}{2}+\dfrac{2}{5}+\dfrac{1}{3}+%
\dfrac{2}{7}+\dfrac{1}{4}\right) \approx 1.\,\allowbreak 217\,86
\end{eqnarray*}

This approximation uses the left end point in each sub-interval to estimate
the function value. \ A sum like this is called a \textbf{left Riemann sum}.

There are different Riemann sums. \ The midpoint Riemann sum uses the
function value taken at the midpoint of each subinterval. \ Another Riemann
sum would approximate the function as the average of the function values
taken at the two end points of the subinterval. \ 

We have only assumed that $f$ is bounded. \ But if $f$ is also continuous,
then the function takes an absolute maximum and minimum in each subinterval.
\ A Riemann sum that uses the minimum in each subinterval obviously
underestimates the area under the graph. A Riemann sum that uses the maximum
in each subinterval obviously overestimates the area under the graph. \ We
will later see the usefulness of Riemann sums in cases we do know whether
the estmation is greater than the area or smaller.

But how do we get better results than crude approximations? \ One way to
improve the approximate values is to refine the partition. \ 

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\textbf{Definition}: \ $\ $A partition $P=\left\{
x_{0,}x_{1},x_{2},...,x_{n}\right\} $ is \textbf{refined} if we keep all
numbers from $P$ and add more.

\vspace{0.1in} 
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As we refine partitions, the estimation for the area can only improve. \
Under-estimations increase, over-estimations decrease. \ The picture shows a
left Riemann sum on a uniform partition and the same Riemann sum on a finer
partition. \ The blue regions are the improvement in the approximation of
the area. 
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Suppose that $P_{1}$, $P_{2}$,...... is \ a sequence of partition, each a
refinement of the previous one. \ Let us also suppose that as we refine the
partitions, we divide each sub-interval, i.e. the length of the longest sub
interval approaches zero. \ For each partition $P_{k}$, let $U_{k}$ be an
underestimation of the area under the graph, and $O_{k}$ an overestimation
of the same area with the same partition $P_{k}$.

Consider now the sequences $O_{1}$, $O_{2}$,... and $U_{1}$, $U_{2}$, ... .
\ Every over-estimation of the area is naturally an upper \ bound for any
under-estimation of the area under the graph. \ 
\begin{equation*}
O_{n}\geq U_{m}\text{ \ \ \ for \ all }n\text{, }m\text{ natural numbers.}
\end{equation*}%
Since refining partitions can only improve estimations, the sequence $O_{1}$%
, $O_{2}$,... \ is decreasing and the sequence $U_{1}$, $U_{2}$, .... is
increasing. \vspace{0.08in}

The sequence $O_{1}$, $O_{2}$,... \ is decreasing and is bounded below by
any $U_{k}$. \ Therefore, it must have a limit, $L$. \ The sequence $U_{1}$, 
$U_{2}$,... \ is increasing and is bounded above by any $O_{k}$. \
Therefore, it must have a limit, $M$, where $L\geq M$. \ If $L>M$, we say
that the area under the graph doesn't exist. \ If $L=M$, we define this
common limit to be the area under the graph. \ If this area exists, we say
that $f$ is \textbf{integrable} on $\left[ a,b\right] $.\vspace{0.08in}

The following theorems will not be proven, but we should state them here.%
\vspace{0.08in}

Theorem: \ If the area under the graph exists, it is the same, no matter
what kind of partitions and Riemann sums we use.\vspace{0.08in}

Theorem: \ If a function $f$ is continuous on $\left[ a,b\right] $, then it
is integrable there.\vspace{0.08in}

Theorem: \ If a function $f$ has only finitely many discontinuities on $%
\left[ a,b\right] $, then it is integrable there.\vspace{0.08in}

Theorem: \ If a function $f$ is increasing on $\left[ a,b\right] $, then it
is integrable there. \ \bigskip

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"0";croptop "1";cropright "1";cropbottom "0";filename
'sample.jpg';file-properties "XNPEU";}} \ \ {\Large Sample Problems}%
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\begin{enumerate}
\item Consider the function $f\left( x\right) =\dfrac{1}{x}$ on the interval 
$\left[ 1,4\right] $.

a) \ Compute the left Riemann sum for $f$ on this interval using a regular
partition with $n=6$ subintervals.\newline
b) \ Compute the right Riemann sum for $f$ on this interval using a regular
partition with $n=6$ subintervals.

\item Consider the function $f\left( x\right) =x^{2}$ on the interval $\left[
0,6\right] .$

a) \ Compute the left Riemann sum for $f$ on this interval with $n=6$
subintervals.\newline
b) \ Compute the right Riemann sum for $f$ on this interval with $n=6$
subintervals.\newline
c) \ Compute the left Riemann sum for $f$ on this interval with $n=12$
subintervals.\newline
d) \ Compute the right Riemann sum for $f$ on this interval with $n=12$
subintervals.\newline
e) \ Compute the left Riemann sum for $f$ on this interval with $n=100$
subintervals.\newline
f) \ Compute the right Riemann sum for $f$ on this interval with $n=100$
subintervals.\newline
g) \ Compute the left Riemann sum for $f$ on this interval with $n$
subintervals.\newline
h) \ Compute the limit of the left Riemann sum for $f$ on this interval with 
$n$ intervals, as $n$ approaches infinity.\newline
i) \ Compute the right Riemann sum for $f$ on this interval with $n$
subintervals.\newline
j) \ Compute the limit of the right Riemann sum for $f$ on this interval
with $n$ intervals, as $n$ approaches infinity.\bigskip \bigskip
\end{enumerate}

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'work.jpg';file-properties "XNPEU";}} \ \ \ {\Large Practice Problems}%
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\begin{enumerate}
\item Consider the function $f\left( x\right) =\sqrt{x}$ on the interval $%
\left[ 0,4\right] .$

a) \ Compute the left Riemann sum for $f$ on this interval with $n=4$
subintervals.\newline
b) \ Compute the right Riemann sum for $f$ on this interval with $n=4$
subintervals.\newline
c) \ Compute the left Riemann sum for $f$ on this interval with $n=10$
subintervals.\newline
d) \ Compute the right Riemann sum for $f$ on this interval with $n=10$
subintervals.

\item Consider the function $f\left( x\right) =\ln \left( x+1\right) $ on
the interval $\left[ 0,10\right] .$

a) \ Compute the left Riemann sum for $f$ on this interval with $n=10$
subintervals.\newline
b) \ Compute the right Riemann sum for $f$ on this interval with $n=10$
subintervals.

\item Consider the function $f\left( x\right) =x^{3}$ on the interval $\left[
0,1\right] .$

a) \ Compute the left Riemann sum for $f$ on this interval with $n=4$
subintervals.\newline
b) \ Compute the right Riemann sum for $f$ on this interval with $n=4$
subintervals.\newline
c) \ Compute the left Riemann sum for $f$ on this interval with $n=10$
subintervals.\newline
d) \ Compute the right Riemann sum for $f$ on this interval with $n=10$
subintervals.\newline
e) \ Compute the left Riemann sum for $f$ on this interval with $n=100$
subintervals.\newline
f) \ Compute the right Riemann sum for $f$ on this interval with $n=100$
subintervals.\newline
\pagebreak
\end{enumerate}

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{\large Sample Problems}%
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\begin{enumerate}
\item a) \ $\dfrac{223}{140}\approx 1.\,592\,857$ \ \ \ \ \ \ \ \ \ \ b) \ $%
\dfrac{341}{280}\approx 1.\,217\,857$\vspace{0.06in}

\item a) \ $55$ \ \ \ \ \ b) \ $91$ \ \ \ \ c) \ $\dfrac{253}{4}=63.\,25$ \
\ \ \ \ \ \ d) \ $\dfrac{325}{4}=81.\,25$ \ \ \ \ \ \ e) \ $\dfrac{177\,309}{%
2500}=70.\,924$ \vspace{0.06in}

f) \ $\dfrac{182\,709}{2500}=73.\,084$ \ \ \ \ \ \ g) \ $\dfrac{36\left(
n-1\right) \left( 2n-1\right) }{n^{2}}=\dfrac{72n^{2}-108n+36}{n^{2}}$ \ \ \
\ \ h) \ $72$ \vspace{0.06in}

i) \ $\dfrac{36\left( 2n^{2}+3n+1\right) }{n^{2}}=\dfrac{72n^{2}+108n+36}{%
n^{2}}$ \ \ \ \ \ j) $\ 72$\vspace{0.06in}\bigskip
\end{enumerate}

{\large Practice Problems}%
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%EndExpansion

\begin{enumerate}
\item a) \ $1+\sqrt{2}+\sqrt{3}\approx 4.\,146\,264\,37$ \ \ \ \ \ \ \ b) \ $%
3+\sqrt{2}+\sqrt{3}\approx 6.\,146\,264\,37$\bigskip

c) \ $\dfrac{2\sqrt{10}}{25}\left( 6+\sqrt{2}+\sqrt{3}+\sqrt{5}+\sqrt{6}+%
\sqrt{7}+\sqrt{8}\right) \approx 4.\,884\,075$\bigskip

d) \ $\dfrac{2\sqrt{10}}{25}\left( 6+\sqrt{2}+\sqrt{3}+\sqrt{5}+\sqrt{6}+%
\sqrt{7}+\sqrt{8}+\sqrt{10}\right) \approx 5.\,684\,075$\bigskip

\item a) \ $\ln \left( 10!\right) =\ln 3628\,800\approx 15.\,104\,412\,57$ \
\ \ \ \ b) \ $\ln \left( 11!\right) =\ln 39\,916\,800\approx
17.\,502\,307\,85$\bigskip

\item a) \ $\dfrac{9}{64}=0.140\,625\,\ \ \ \ \ \ \ \ \ \ \ \ $b) \ $\dfrac{%
25}{64}=0.390\,625\,\ \ \ \ \ \ \ \ \ \ $c) \ $\dfrac{81}{400}=0.202\,5$ \ \
\ \ \ \ \ \ d) \ $\dfrac{121}{400}=0.302\,5$\bigskip

\item e) \ $\dfrac{9801}{40\,000}=0.245\,025\,\ \ \ \ \ $f) \ $\dfrac{10\,201%
}{40\,000}=0.255\,025\,$\pagebreak
\end{enumerate}

\begin{center}
{\Large Sample Problems \FRAME{itbpF}{0.8527in}{0.4999in}{0.211in}{}{}{%
pencil.bmp}{\special{language "Scientific Word";type
"GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width
0.8527in;height 0.4999in;depth 0.211in;original-width
1.9735in;original-height 1.1467in;cropleft "0";croptop "1";cropright
"1";cropbottom "0";filename 'pencil.bmp';file-properties "XNPEU";}} Solutions%
}%
%TCIMACRO{\TeXButton{ resume}{\setcounter{enumi}{0}\RESUME}}%
%BeginExpansion
\setcounter{enumi}{0}\RESUME%
%EndExpansion
\end{center}

\begin{enumerate}
\item Consider the function $f\left( x\right) =\dfrac{1}{x}$ on the interval 
$\left[ 1,4\right] $.

a) \ Compute the left Riemann sum for $f$ on this interval using a regular
partition with $n=6$ subintervals.

Solution: \ The interval $\left[ 1,4\right] $ is $3$ units long. \ The
regular partition will contain intervals of length $\dfrac{3}{6}=\dfrac{1}{2}
$. \ The partition consists of $\left\{ 1,~1\dfrac{1}{2}~,2,~2\dfrac{1}{2}%
~,3,~3\dfrac{1}{2},~4\right\} $ \ Notice that these are seven numbers. \ We
usually start labeling with zero. \ In this case, these seven numbers are $%
\left\{ x_{0},x_{1},x_{2},x_{3},x_{4},x_{5},x_{6}\right\} $. \ On each
interval, we approximate the area under the graph by a rectangle as tall as
the function value of the left endpoint of the interval. \ For example, on
the first interval, we approximate the area under the graph using a
rectangle with height $\dfrac{1}{1}=1$. \ On the second interval, the height
of the rectangle is $\dfrac{1}{~1\dfrac{1}{2}~}=\dfrac{2}{3}$.\FRAME{dtbpFX}{%
2.8945in}{1.6648in}{0pt}{}{}{left1.bmp}{\special{language "Scientific
Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file
"F";width 2.8945in;height 1.6648in;depth 0pt;original-width
9.5in;original-height 5.4267in;cropleft "0";croptop "1";cropright
"1";cropbottom "0";filename 'left1.bmp';file-properties "XNPEU";}}The first
rectangle has width $\dfrac{1}{2}$ \ and height $1$. \ The area is $A_{1}=%
\dfrac{1}{2}\cdot 1=\dfrac{1}{2}$.

The second rectangle has width $\dfrac{1}{2}$ \ and height $\dfrac{1}{\left( 
\dfrac{3}{2}\right) }=\dfrac{2}{3}$. The area is \ $A_{2}=\dfrac{1}{2}\cdot 
\dfrac{2}{3}=\dfrac{1}{3}$.

Let us notice that all rectangles have the same width of $\dfrac{1}{2}$. \
This is an advantage of a regular partition.

The third rectangle has height $\dfrac{1}{2}$. \ Its area is $A_{3}=\dfrac{1%
}{2}\cdot \dfrac{1}{2}=\dfrac{1}{4}$.

The fourth rectangle has height $\dfrac{1}{\left( \dfrac{5}{2}\right) }=%
\dfrac{2}{5}$. \ Its area is $A_{4}=\dfrac{1}{2}\cdot \dfrac{2}{5}=\dfrac{1}{%
5}$

The fifth rectangle has height $\dfrac{1}{3}$. \ Its area is $A_{4}=\dfrac{1%
}{2}\cdot \dfrac{1}{3}=\dfrac{1}{6}$.

The sixth rectangle has height $\dfrac{1}{\left( \dfrac{7}{2}\right) }=%
\dfrac{2}{7}$. \ Its area is $A_{4}=\dfrac{1}{2}\cdot \dfrac{2}{7}=\dfrac{1}{%
7}$. \ \ In short, the left-hand approximation is%
\begin{eqnarray*}
L_{f,n=6} &=&\dfrac{1}{2}\cdot \dfrac{1}{1}+\dfrac{1}{2}\cdot \dfrac{1}{1.5}+%
\dfrac{1}{2}\cdot \dfrac{1}{2}+\dfrac{1}{2}\cdot \dfrac{1}{2.5}+\dfrac{1}{2}%
\cdot \dfrac{1}{3}+\dfrac{1}{2}\cdot \dfrac{1}{3.5}=\dfrac{1}{2}\left( 
\dfrac{1}{1}+\dfrac{1}{1.5}+\dfrac{1}{2}+\dfrac{1}{2.5}+\dfrac{1}{3}+\dfrac{1%
}{3.5}\right) \\
&=&\dfrac{1}{2}\left( 1+\dfrac{2}{3}+\dfrac{1}{2}+\dfrac{2}{5}+\dfrac{1}{3}+%
\dfrac{2}{7}\right) =\dfrac{223}{140}\approx 1.\,592\,857
\end{eqnarray*}

We can clearly see from the picture that this approximation is an
overestimation of the area.

b) \ Compute the right Riemann sum for $f$ on this interval using a regular
partition with $n=6$ subintervals.\FRAME{dtbpFX}{3.1782in}{1.6648in}{0pt}{}{%
}{right1.bmp}{\special{language "Scientific Word";type
"GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width
3.1782in;height 1.6648in;depth 0pt;original-width 9.5in;original-height
4.9372in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename
'right1.bmp';file-properties "XNPEU";}}%
\begin{eqnarray*}
R &=&\dfrac{1}{2}\cdot \dfrac{1}{1.5}+\dfrac{1}{2}\cdot \dfrac{1}{2}+\dfrac{1%
}{2}\cdot \dfrac{1}{2.5}+\dfrac{1}{2}\cdot \dfrac{1}{3}+\dfrac{1}{2}\cdot 
\dfrac{1}{3.5}+\dfrac{1}{2}\cdot \dfrac{1}{4}=\dfrac{1}{2}\left( \dfrac{1}{%
1.5}+\dfrac{1}{2}+\dfrac{1}{2.5}+\dfrac{1}{3}+\dfrac{1}{3.5}+\dfrac{1}{4}%
\right) \\
&=&\dfrac{1}{2}\left( \dfrac{2}{3}+\dfrac{1}{2}+\dfrac{2}{5}+\dfrac{1}{3}+%
\dfrac{2}{7}+\dfrac{1}{4}\right) =\dfrac{341}{280}\approx 1.\,217\,857
\end{eqnarray*}

We can clearly see from the picture that this approximation is an
underestimation of the area. \ Thus we now know that the area under the
graph is between those two values:%
\begin{equation*}
1.\,217\,857<A<1.\,592\,857
\end{equation*}%
Note: \ We sometimes use summation notation when writing such expressions. \
Using summation notation, these Riemann sums are%
\begin{eqnarray*}
L &=&\dsum\limits_{k=0}^{5}\dfrac{1}{2}\cdot \dfrac{1}{1+k\left( \dfrac{1}{2}%
\right) }=\dfrac{1}{2}\dsum\limits_{k=0}^{5}\dfrac{1}{1+k\left( \dfrac{1}{2}%
\right) }=\dfrac{1}{2}\dsum\limits_{k=0}^{5}\dfrac{1}{\dfrac{2+k}{2}}=\dfrac{%
1}{2}\dsum\limits_{k=0}^{5}\dfrac{2}{2+k}=\dsum\limits_{k=0}^{5}\dfrac{1}{2+k%
}\text{ } \\
\text{and }R &=&\dsum\limits_{k=1}^{6}\dfrac{1}{2}\cdot \dfrac{1}{1+k\left( 
\dfrac{1}{2}\right) }=\dsum\limits_{k=1}^{6}\dfrac{1}{2+k}\text{ }
\end{eqnarray*}%
\pagebreak

\item For this problem, we will need the following theorem: \ for all
natural numbers $n,$ 
\begin{equation*}
1^{2}+2^{2}+3^{2}+...+n^{2}=\dfrac{n\left( n+1\right) \left( 2n+1\right) }{6}
\end{equation*}%
Consider the function $f\left( x\right) =x^{2}$ on the interval $\left[ 0,6%
\right] .$

a) \ Compute the left Riemann sum for $f$ on this interval with $n=6$
subintervals.

Solution: \ Each subinterval is of length $1,$ and so the partition is $%
\left\{ 0,1,2,3,4,5,6\right\} $. \ \FRAME{dtbpFX}{2.6013in}{1.6648in}{0pt}{}{%
}{left2.bmp}{\special{language "Scientific Word";type
"GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width
2.6013in;height 1.6648in;depth 0pt;original-width 11.1353in;original-height
7.0837in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename
'left2.bmp';file-properties "XNPEU";}}The left-hand sum is%
\begin{equation*}
L_{f,n=6}=1\cdot 0^{2}+1\cdot 1^{2}+1\cdot 2^{2}+1\cdot 3^{2}+1\cdot
4^{2}+1\cdot 5^{2}=1+4+9+16+25=55
\end{equation*}%
Using summation notation, 
\begin{equation*}
L_{f,n=6}=\dsum\limits_{k=0}^{5}1\cdot k^{2}=\dsum\limits_{k=0}^{5}k^{2}=55
\end{equation*}%
We can see on the picture that this Riemann sum underestimates the area.

b) \ Compute the right Riemann sum for $f$ on this interval with $n=6$
subintervals.\FRAME{dtbpFX}{2.6662in}{1.6648in}{0pt}{}{}{rigth2.bmp}{\special%
{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio
TRUE;display "USEDEF";valid_file "F";width 2.6662in;height 1.6648in;depth
0pt;original-width 11.7813in;original-height 7.3129in;cropleft "0";croptop
"1";cropright "1";cropbottom "0";filename 'rigth2.bmp';file-properties
"XNPEU";}}%
\begin{equation*}
R_{f,n=6}=1\cdot 1^{2}+1\cdot 2^{2}+1\cdot 3^{2}+1\cdot 4^{2}+1\cdot
5^{2}+1\cdot 6^{2}=1+4+9+16+25+36=91
\end{equation*}%
Using summation notation, 
\begin{equation*}
R_{f,n=6}=\dsum\limits_{k=1}^{6}1\cdot k^{2}=\dsum\limits_{k=1}^{6}k^{2}=91
\end{equation*}%
We can see on the picture that this Riemann sum underestimates the area. \
Thus, we have that 
\begin{equation*}
55<A<91
\end{equation*}%
c) \ Compute the left Riemann sum for $f$ on this interval with $n=12$
subintervals.

Solution: \ Each subinterval will have length $\dfrac{6}{12}=\dfrac{1}{2}$.
\ The partition is \newline
$\left\{ 0,~0.5,~1,~1.5~,2~,2.5,~3,~3.5,~4,~4.5,~5,~5.5,~6\right\} $. \ The
left Riemann sum is 
\begin{eqnarray*}
L_{f,n=12} &=&\dfrac{1}{2}\cdot 0^{2}+\dfrac{1}{2}\cdot 0.5^{2}+\dfrac{1}{2}%
\cdot 1^{2}+\dfrac{1}{2}\cdot 1.5^{2}+\dfrac{1}{2}\cdot 2^{2}+\dfrac{1}{2}%
\cdot 2.5^{2}+\dfrac{1}{2}\cdot 3^{2}+\dfrac{1}{2}\cdot 3.5^{2}+\dfrac{1}{2}%
\cdot 4^{2} \\
&&+\dfrac{1}{2}\cdot 4.5^{2}+\dfrac{1}{2}\cdot 5^{2}+\dfrac{1}{2}\cdot
5.5^{2}
\end{eqnarray*}%
Although this looks like a lot of computation, it can be made quite simple
using a bit of algebra and the theorem stated above. \ We first factor out $%
\dfrac{1}{2}$ and write the rest as fractions, with a common denominator of $%
2$.%
\begin{eqnarray*}
L_{f,n=12} &=&\dfrac{1}{2}\left(
0^{2}+0.5^{2}+1^{2}+1.5^{2}+...+5.5^{2}\right) =\dfrac{1}{2}\left( \left( 
\dfrac{1}{2}\right) ^{2}+\left( \dfrac{2}{2}\right) ^{2}+\left( \dfrac{3}{2}%
\right) ^{2}+\left( \dfrac{4}{2}\right) ^{2}+...+\left( \dfrac{11}{2}\right)
^{2}\right) \\
&=&\dfrac{1}{2}\left( \dfrac{1^{2}}{4}+\dfrac{2^{2}}{4}+\dfrac{3^{2}}{4}+%
\dfrac{4^{2}}{4}+...+\dfrac{11^{2}}{4}\right) \text{ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ factor out }\dfrac{1}{4} \\
&=&\dfrac{1}{2}\cdot \dfrac{1}{4}\left( 1^{2}+2^{2}+3^{2}+...+11^{2}\right) 
\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ use theorem with }n=11
\\
&=&\dfrac{1}{8}\cdot \dfrac{11\cdot 12\cdot 23}{6}=\dfrac{253}{4}=63.\,25
\end{eqnarray*}%
The same computation, using summation notation:%
\begin{equation*}
L_{f,n=12}=\dsum\limits_{k=0}^{11}\dfrac{1}{2}\cdot \left( \dfrac{1}{2}%
k\right) ^{2}=\dfrac{1}{2}\dsum\limits_{k=0}^{11}\dfrac{k^{2}}{4}=\dfrac{1}{8%
}\dsum\limits_{k=0}^{11}k^{2}=\dfrac{1}{8}\dfrac{11\cdot 12\cdot 23}{6}=%
\dfrac{253}{4}=63.\,25
\end{equation*}

d) \ Compute the right Riemann sum for $f$ on this interval with $n=12$
subintervals.

Solution: \ The difference between the left and right Riemann sums is just
the first and the last rectangle.%
\begin{eqnarray*}
R_{f,n=12} &=&\dfrac{1}{2}\left( 0.5^{2}+1^{2}+...+5.5^{2}+6^{2}\right) =%
\dfrac{1}{2}\left( \left( \dfrac{1}{2}\right) ^{2}+\left( \dfrac{2}{2}%
\right) ^{2}+\left( \dfrac{3}{2}\right) ^{2}+\left( \dfrac{4}{2}\right)
^{2}+...+\left( \dfrac{12}{2}\right) ^{2}\right) \\
&=&\dfrac{1}{2}\left( \dfrac{1^{2}}{4}+\dfrac{2^{2}}{4}+\dfrac{3^{2}}{4}+%
\dfrac{4^{2}}{4}+...+\dfrac{12^{2}}{4}\right) \text{ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ factor out }\dfrac{1}{4} \\
&=&\dfrac{1}{2}\cdot \dfrac{1}{4}\left( 1^{2}+2^{2}+3^{2}+...+12^{2}\right) 
\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ use theorem with }n=12
\\
&=&\dfrac{1}{8}\cdot \dfrac{12\cdot 13\cdot 25}{6}=\dfrac{325}{4}=81.\,25
\end{eqnarray*}%
\newline
Using summation notation,%
\begin{equation*}
R_{f,n=12}=\dsum\limits_{k=1}^{12}\dfrac{1}{2}\cdot \left( \dfrac{1}{2}%
k\right) ^{2}=\dfrac{1}{2}\dsum\limits_{k=1}^{12}\dfrac{k^{2}}{4}=\dfrac{1}{8%
}\dsum\limits_{k=1}^{12}k^{2}=\dfrac{1}{8}\dfrac{12\cdot 13\cdot 25}{6}=%
\dfrac{325}{4}=81.\,25
\end{equation*}%
Because this function is increasing, all left sums underestimate the area
and all right sums overestimate the area under the graph. Thus%
\begin{equation*}
63.\,25<A<81.\,25
\end{equation*}%
e) \ Compute the left Riemann sum for $f$ on this interval with $n=100$
subintervals.\newline
Solution: \ Each subinterval is $\dfrac{6}{100}$ units long. \ The partition
is \newline
$\left\{ x_{0}=0,~x_{1}=\dfrac{6}{100},~x_{2}=\dfrac{12}{100},~....,x_{k}=%
\dfrac{6k}{100},...,x_{100}=\dfrac{600}{100}=6\right\} $\ \newline
The left Riemann sum is%
\begin{eqnarray*}
L_{f,n=100} &=&\dfrac{6}{100}\cdot 0^{2}+\dfrac{6}{100}\cdot \left( \dfrac{6%
}{100}\right) ^{2}+\dfrac{6}{100}\cdot \left( \dfrac{12}{100}\right)
^{2}+....+\dfrac{6}{100}\cdot \left( \dfrac{6\cdot 99}{100}\right) ^{2} \\
&=&\dfrac{6}{100}\left( 0^{2}+\left( \dfrac{6\cdot 1}{100}\right)
^{2}+\left( \dfrac{6\cdot 2}{100}\right) ^{2}+....+\left( \dfrac{6\cdot 99}{%
100}\right) ^{2}\right) \\
&=&\dfrac{6}{100}\cdot \left( \dfrac{6}{100}\right) ^{2}\left(
0^{2}+1^{2}+2^{2}+....+99^{2}\right) =\left( \dfrac{6}{100}\right) ^{3}%
\dfrac{99\cdot 100\cdot 199}{6} \\
&=&\left( \dfrac{6}{100}\right) ^{2}\left( 99\cdot 199\right) =\dfrac{%
177\,309}{2500}=70.\,924
\end{eqnarray*}%
Using summation notation,%
\begin{eqnarray*}
L_{f,n=100} &=&\dsum\limits_{k=0}^{99}\dfrac{6}{100}\cdot \left( \dfrac{6}{%
100}k\right) ^{2}=\dfrac{6}{100}\dsum\limits_{k=0}^{99}\left( \dfrac{6}{100}%
\right) ^{2}k^{2}=\left( \dfrac{6}{100}\right)
^{3}\dsum\limits_{k=0}^{99}k^{2}=\left( \dfrac{6}{100}\right) ^{3}\dfrac{%
99\cdot 100\cdot 199}{6} \\
&=&\left( \dfrac{6}{100}\right) ^{2}\left( 99\cdot 199\right) =\dfrac{%
177\,309}{2500}=70.\,924
\end{eqnarray*}

f) \ Compute the right Riemann sum for $f$ on this interval with $n=100$
subintervals.

The right Riemann sum is%
\begin{eqnarray*}
R_{f,n=100} &=&\dfrac{6}{100}\cdot \left( \dfrac{6}{100}\right) ^{2}+\dfrac{6%
}{100}\cdot \left( \dfrac{12}{100}\right) ^{2}+....+\dfrac{6}{100}\cdot
\left( \dfrac{6\cdot 100}{100}\right) ^{2} \\
&=&\dfrac{6}{100}\left( \left( \dfrac{6\cdot 1}{100}\right) ^{2}+\left( 
\dfrac{6\cdot 2}{100}\right) ^{2}+....+\left( \dfrac{6\cdot 100}{100}\right)
^{2}\right) \\
&=&\dfrac{6}{100}\cdot \left( \dfrac{6}{100}\right) ^{2}\left(
0^{2}+1^{2}+2^{2}+....+100^{2}\right) =\left( \dfrac{6}{100}\right) ^{3}%
\dfrac{100\cdot 101\cdot 201}{6} \\
&=&\left( \dfrac{6}{100}\right) ^{2}\left( 101\cdot 201\right) =\dfrac{%
182\,709}{2500}=73.\,\allowbreak 083\,6
\end{eqnarray*}%
Using summation notation,%
\begin{eqnarray*}
R_{f,n=100} &=&\dsum\limits_{k=1}^{100}\dfrac{6}{100}\cdot \left( \dfrac{6}{%
100}k\right) ^{2}=\dfrac{6}{100}\dsum\limits_{k=1}^{100}\left( \dfrac{6}{100}%
\right) ^{2}k^{2}=\left( \dfrac{6}{100}\right)
^{3}\dsum\limits_{k=1}^{100}k^{2}=\left( \dfrac{6}{100}\right) ^{3}\dfrac{%
100\cdot 101\cdot 201}{6} \\
&=&\left( \dfrac{6}{100}\right) ^{2}\left( 100\cdot 201\right) =\dfrac{%
182\,709}{2500}=73.\,084
\end{eqnarray*}%
Thus%
\begin{equation*}
70.\,924<A<73.\,084
\end{equation*}%
\pagebreak

g) \ Compute the left Riemann sum for $f$ on this interval with $n$
subintervals.

Solution: \ Each subinterval is $\dfrac{6}{n}$ units long. \ The numbers in
the partition are\newline
$\left\{ x_{0}=0,~x_{1}=\dfrac{6}{n},~x_{2}=2\left( \dfrac{6}{n}\right)
,~x_{3}=3\left( \dfrac{6}{n}\right) ,~x_{4}=4\left( \dfrac{6}{n}\right)
,....,~x_{n}=n\left( \dfrac{6}{n}\right) =6\right\} $

The left Riemann sum is 
\begin{eqnarray*}
L_{f,n} &=&\dfrac{6}{n}\cdot 0^{2}+\dfrac{6}{n}\left( \dfrac{6}{n}\right)
^{2}+\dfrac{6}{n}\left( 2\cdot \dfrac{6}{n}\right) ^{2}+\dfrac{6}{n}\left(
3\cdot \dfrac{6}{n}\right) ^{2}+....+\dfrac{6}{n}\left( \left( n-1\right)
\cdot \dfrac{6}{n}\right) ^{2} \\
&=&\dfrac{6}{n}\left( 0^{2}+\left( \dfrac{6}{n}\right) ^{2}+\left( 2\cdot 
\dfrac{6}{n}\right) ^{2}+\left( 3\cdot \dfrac{6}{n}\right) ^{2}+....+\left(
\left( n-1\right) \cdot \dfrac{6}{n}\right) ^{2}\right) \\
&=&\dfrac{6}{n}\left( 1^{2}\cdot \left( \dfrac{6}{n}\right) ^{2}+2^{2}\cdot
\left( \dfrac{6}{n}\right) ^{2}+3^{2}\cdot \left( \dfrac{6}{n}\right)
^{2}+....+\left( n-1\right) ^{2}\cdot \left( \dfrac{6}{n}\right) ^{2}\right)
\\
&=&\dfrac{6}{n}\left( \dfrac{6}{n}\right) ^{2}\left(
1^{2}+2^{2}+3^{2}+....+\left( n-1\right) ^{2}\right) =\left( \dfrac{6}{n}%
\right) ^{3}\dfrac{\left( n-1\right) \left( \left( n-1\right) +1\right)
\left( 2\left( n-1\right) +1\right) }{6} \\
&=&\left( \dfrac{6}{n}\right) ^{3}\dfrac{\left( n-1\right) n\left(
2n-1\right) }{6}=\left( \dfrac{6}{n}\right) ^{2}\left( n-1\right) \left(
2n-1\right) =\dfrac{36\left( n-1\right) \left( 2n-1\right) }{n^{2}} \\
&=&\dfrac{36\left( 2n^{2}-3n+1\right) }{n^{2}}=\dfrac{72n^{2}-108n+36}{n^{2}}
\end{eqnarray*}%
Using summation notation,%
\begin{eqnarray*}
L_{f,n} &=&\dsum\limits_{k=0}^{n-1}\dfrac{6}{n}\cdot \left( \dfrac{6}{n}%
k\right) ^{2}=\dfrac{6}{n}\dsum\limits_{k=0}^{n-1}\left( \dfrac{6}{n}\right)
^{2}k^{2}=\left( \dfrac{6}{n}\right)
^{3}\dsum\limits_{k=0}^{n-1}k^{2}=\left( \dfrac{6}{n}\right) ^{3}\dfrac{%
\left( n-1\right) \left( \left( n-1\right) +1\right) \left( 2\left(
n-1\right) +1\right) }{6} \\
&=&\left( \dfrac{6}{n}\right) ^{3}\dfrac{\left( n-1\right) n\left(
2n-1\right) }{6}=\left( \dfrac{6}{n}\right) ^{2}\left( n-1\right) \left(
2n-1\right) =\dfrac{36\left( n-1\right) \left( 2n-1\right) }{n^{2}}=\dfrac{%
72n^{2}-108n+36}{n^{2}}
\end{eqnarray*}

h) \ Compute the limit of the left Riemann sum for $f$ on this interval with 
$n$ intervals, as $n$ approaches infinity.

Solution: \ 
\begin{equation*}
\lim\limits_{n\rightarrow \infty }\dfrac{72n^{2}-108n+36}{n^{2}}%
=\lim\limits_{n\rightarrow \infty }\left( 72-\dfrac{108}{n}+\dfrac{36}{n^{2}}%
\right) =72
\end{equation*}%
i) \ Compute the right Riemann sum for $f$ on this interval with $n$
subintervals.

Solution: \ The right Riemann sum is%
\begin{eqnarray*}
R_{f,n} &=&\dfrac{6}{n}\left( \dfrac{6}{n}\right) ^{2}+\dfrac{6}{n}\left(
2\cdot \dfrac{6}{n}\right) ^{2}+\dfrac{6}{n}\left( 3\cdot \dfrac{6}{n}%
\right) ^{2}+....+\dfrac{6}{n}\left( n\cdot \dfrac{6}{n}\right) ^{2} \\
&=&\dfrac{6}{n}\left( \left( \dfrac{6}{n}\right) ^{2}+\left( 2\cdot \dfrac{6%
}{n}\right) ^{2}+\left( 3\cdot \dfrac{6}{n}\right) ^{2}+....+\left( n\cdot 
\dfrac{6}{n}\right) ^{2}\right) \\
&=&\dfrac{6}{n}\left( 1^{2}\cdot \left( \dfrac{6}{n}\right) ^{2}+2^{2}\cdot
\left( \dfrac{6}{n}\right) ^{2}+3^{2}\cdot \left( \dfrac{6}{n}\right)
^{2}+....+n^{2}\cdot \left( \dfrac{6}{n}\right) ^{2}\right)
\end{eqnarray*}%
\begin{eqnarray*}
R_{f,n} &=&\dfrac{6}{n}\left( \dfrac{6}{n}\right) ^{2}\left(
1^{2}+2^{2}+3^{2}+....+n^{2}\right) =\left( \dfrac{6}{n}\right) ^{3}\dfrac{%
n\left( n+1\right) \left( 2n+1\right) }{6} \\
&=&\left( \dfrac{6}{n}\right) ^{2}\left( n+1\right) \left( 2n+1\right) =%
\dfrac{36\left( n+1\right) \left( 2n+1\right) }{n^{2}}=\dfrac{36\left(
2n^{2}+3n+1\right) }{n^{2}}=\dfrac{72n^{2}+108n+36}{n^{2}}
\end{eqnarray*}%
Using summation notation,%
\begin{eqnarray*}
R_{f,n} &=&\dsum\limits_{k=1}^{n}\dfrac{6}{n}\cdot \left( \dfrac{6}{n}%
k\right) ^{2}=\dfrac{6}{n}\dsum\limits_{k=1}^{n}\left( \dfrac{6}{n}\right)
^{2}k^{2}=\left( \dfrac{6}{n}\right) ^{3}\dsum\limits_{k=1}^{n}k^{2}=\left( 
\dfrac{6}{n}\right) ^{3}\dfrac{n\left( n+1\right) \left( 2n+1\right) }{6} \\
&=&\left( \dfrac{6}{n}\right) ^{2}\left( n+1\right) \left( 2n+1\right) =%
\dfrac{36\left( n+1\right) \left( 2n+1\right) }{n^{2}}=\dfrac{72n^{2}+108n+36%
}{n^{2}}
\end{eqnarray*}

j) \ Compute the limit of the right Riemann sum for $f$ on this interval
with $n$ intervals, as $n$ approaches infinity.

Solution: \ 
\begin{equation*}
\lim\limits_{n\rightarrow \infty }\dfrac{72n^{2}+108n+36}{n^{2}}%
=\lim\limits_{n\rightarrow \infty }\left( 72+\dfrac{108}{n}+\dfrac{36}{n^{2}}%
\right) =72
\end{equation*}
\end{enumerate}

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