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\centerline{MATH 136 -- Calculus 2}
\centerline{Solutions for Practice on Approximating Areas}
\centerline{January 22, 2020}
\vskip 10pt
\noindent
{\it Background}
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We have discussed the idea of approximating the area between a graph
$y = f(x) \ge 0$ and the $x$-axis for $a \le b\le x$ by using sums
of areas of rectangles and worked out a first example.  To practice
using these ideas you will work through another similar example using
a different function, interval, and subdivision.  Here's the description
of the region we want in formulas:  The region between $y = x^2 - 4x + 6$
and the $x$-axis, for $1 \le x \le 2$.  
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\noindent
{\it Questions}
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\begin{enumerate}
\item Sketch the region (you may want to complete the square in $x^2 - 4x + 6$
to do this).  Check your work by looking at the figure on the back of this page.
\item[] {\it Answer:} Completing the square, $x^2 - 4x + 6 = (x-2)^2 + 2$.
So the graph is a parabola opening up, with the vertex at the point $(2,2)$.  See graphs on back,
which show the segment of this parabola for $x$ in $[1,2]$.
\item To subdivide the $x$-interval $[1,2]$ into $N = 5$ equal smaller intervals, what 
$\Delta x$ should we use?  What are the endpoints $x_0,x_1,x_2,x_3,x_4,x_5$
of the intervals?
\item[] {\it Answer:}  $\Delta x = \frac{2 - 1}{5} = \frac{1}{5} = 0.2$.  So the end points are
$x_i = 1 + i\Delta x = 1 + (.2)i$, or 
$$x_0 = 1, x_1 = 1.2, x_2 = 1.4, x_3 = 1.6, x_4 = 1.8, x_5 = 2$$ 
\item Make rectangles using the value of the function at the \emph{left endpoint}
of each smaller interval to get the heights. 
Sketch these rectangles together with the graph.
\item[] {\it Answer:} See Figure 1 on back.
\item Write the sum giving the areas
of these rectangles using summation notation, and compute the numerical value.
\item[] {\it Answer:}  The sum is 
$$\sum_{i=0}^4 (x_i^2 - 4x_i + 6)\cdot \Delta x = \sum_{i=0}^4 ((1 + .2i)^2 - 4(1 + .2i) + 6)(.2)$$
The numerical value is $2.44$.
\item Now make rectangles using the value of the function at the \emph{right endpoint}
of each smaller interval to get the heights.  Sketch these rectangles together with the graph.
\item[] {\it Answer:}  See Figure 2 on back.
\item  Write the sum giving the areas
of these rectangles using summation notation, and compute the numerical value.
\item[] {\it Answer:}  The sum is 
$$\sum_{i=1}^5 (x_i^2 - 4x_i + 6)\cdot \Delta x = \sum_{i=1}^5 ((1 + .2i)^2 - 4(1 + .2i) + 6)(.2)$$
The numerical value is $2.24$.
\item What can you say about the area of the region from your computations?  
\item[] {\it Answer:}  The function $f(x) = x^2 - 4x + 6$ is decreasing on the whole interval
$[1,2]$.  So the left-endpoint rectangles completely cover the area under the graph, while
the right-endpoint rectangles ``miss'' some of the area under the graph.  This says
$$2.24 < {\rm Area} < 2.44$$
\end{enumerate}

\begin{figure}
\begin{center}
\includegraphics[width=.5\linewidth]{L5.png}
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\caption{The left endpoint rectangles}
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\begin{figure}
\begin{center}
\includegraphics[width=.5\linewidth]{R5.png}
\end{center}
\caption{The right endpoint rectangles}
\end{figure}



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