1.3 The Area Problem

In this section we do two things: on one hand, we solve the area problem in cases for which calculus is not required. Namely, area under the graph of linear functions and their absolute values, parabolas, and semicircles; on the other hand, we compute several rectangular approximations to the area under the graph of elementary functions. Along the way we will accomplish the following:

• review of absolute value functions and their graphs;
• review functions whose graphs are semicircles;
• compute the area under the graph of semicircles and absolut value functions;
• review trigonometric functions and their graphs;
• review inverse trigonometric functions and their graphs;
• review exponential and logarithmic functions and their graphs;
• become falimiar with rectangular approximations in basic cases.

1.3.1 Preliminaries.

 

Notation

 

The conjunctions “and” and “or” play a central role when writing clear/precise arguments. In mathematics they have a very precise meaning. They are called logic connectives, because they serve to connect statements.

The conjunction or is used to connect two statements \(S_1\) and \(S_2\), and has the meaning that the new statement \(S_1\) or \(S_2\) is true when at least one of the statements \(S_1\) or \(S_2\) is true. In what follows we will denote this by
\[\left[ \begin{array}{l}{S_1}\\{S_2}\end{array} \right. \;\; \equiv \;\; S_1\;\text{ or }\; S_2.\]

The conjunction and is used to connect two statements \(S_1\) and \(S_2\), and has the meaning that the new statement \(S_1\) and \(S_2\) is true when both of the statements \(S_1\) and \(S_2\) are true. In what follows we will denote this by
\[\left\{
\begin{array}{l}{S_1}\\{S_2}\end{array} \right. \;\; \equiv \;\; S_1\;\text{ and }\; S_2.\]

Recall that two statements \({S_1},\;{S_2}\) are called equivalent, and we write \(S_1 \Longleftrightarrow S_2\), if \(S_1\) implies \(S_2\), and \(S_2\) implies \(S_1\), that is, if \(S_1 \Longrightarrow S_2\) and \(S_2 \Longrightarrow S_1\).

 

Absolute Value and Piecewise Functions

 

Recall that
\[|\,x\,| = \left [\renewcommand{\arraystretch}{1.2}\begin{array}{ll}
x&\text{if } \; x \geq 0\\ -x&\text{if } \; x < 0 \end{array}\right ..\] Example 1. Rewrite the function \(f(x) = |\, 2x-3\,| + 2x-1\) as a piecewise function and sketch its graph.

Solution. We have
\[f(x)=|\,2x+3\,|+2x-1 = \left [\renewcommand{\arraystretch}{1.3}\begin{array}{rl}
(2x+3) + 2x-1&\text{if } \; 2x+3 \geq 0\\ -(2x+3)+2x-1&\text{if } \; 2x + 3 < 0 \end{array}\right .= \left [\renewcommand{\arraystretch}{2}\begin{array}{ll} 4x+2& \displaystyle \text{if } \; x \geq -\frac{3}{2}\\ -4&\displaystyle \text{if } \; x < -\frac{3}{2} \end{array}\right ..\] The graph of \(f\) is shown in Figure 1.9.

 

Semicircles

 

Consider the function \(\displaystyle f(x)=\sqrt{r^2-x^2}\), where \(r>0\) is constant. The domain of \(f\), \(D_f\), is obtained as follows:
\[x\in D_f \;\; \Longleftrightarrow \;\; r^2-x^2\geq 0 \;\; \Longleftrightarrow \;\; x^2 < r^2 \;\; \Longleftrightarrow \;\; -r \leq x \leq r \;\; \Longleftrightarrow \;\; D_f=[-r,r].\] The graph of \(f\) is the semicircle with center at the origin and radius \(r\) above the \(x\)-axis. This can be seen as follows:
\[y = \sqrt{r^2-x^2} \;\; \Longleftrightarrow \;\; \left \{\renewcommand{\arraystretch}{1.2}\begin{array}{l}
y^2=r^2-x^2\\y \geq 0
\end{array}\right .
\;\; \Longleftrightarrow \;\; \left \{\renewcommand{\arraystretch}{1.2}\begin{array}{l}
x^2+ y^2=r^2\\y \geq 0
\end{array}\right . \hspace{-.15in},\]
that is, the graph is the portion of the circle \(x^2+ y^2=r^2\) above the \(x\)-axis.

 

Inverse Trigonometric Functions

 

Basic Properties of Arccos. Recall that arccosine is the inverse function of cosine.

The domain and range of \(y=\arccos x\) are, respectively,
\[D_{\arccos}=[-1,1] \hspace{.5in} R_{\arccos}=[0,\pi].\]
Furthermore,
\[\cos(\arccos x)=x \;\; \text{for all} \; x\in D_{\arccos} \hspace{.5in} \arccos(\cos x)=x \;\; \text{for all} \; x \in [0,\pi].\]

 
Basic Properties of Arcsin. Recall that arcsine is the inverse function of sine.

The domain and range of \(y=\arcsin x\) are, respectively,
\[D_{\arccos}=[-1,1] \hspace{.5in} R_{\arccos}=\left [-\frac{\pi}{2},\frac{\pi}{2}\right ].\]
Furthermore,
\[\sin(\arcsin x)=x \;\; \text{for all} \; x\in D_{\arcsin} \hspace{.5in} \arcsin(\sin x)=x \;\; \text{for all} \; x \in \left [-\frac{\pi}{2},\frac{\pi}{2}\right ].\]

 
Basic Properties of Arctan. Recall that arctangent is the inverse function of sine.

The domain and range of \(y=\arctan x\) are, respectively,
\[D_{\arctan}=(-\infty, \infty \hspace{.5in} R_{\arctan}=\left (-\frac{\pi}{2},\frac{\pi}{2}\right ).\]
Furthermore,
\[\tan(\arctan x)=x \;\; \text{for all} \; x\in D_{\arctan} \hspace{.5in} \arctan(\tan x)=x \;\; \text{for all} \; x \in \left (-\frac{\pi}{2},\frac{\pi}{2}\right ).\]

 

1.3.2 The Area Problem

 

Convention. The phrase

“find the area under the graph of the function \(y=f(x)\), \(x\in [a,b]\)”

will implicitly mean that \(f(x)\geq 0\) for all \(x\in [a,b]\), and that the area—\(A(\mathcal{R})\)—is the area of the region \(\mathcal{R}\) enclosed by the graph of \(f\) and the \(x\)-axis.

Example 1. Find the area under the graph of the function \(f(x)= 2x+1\), \(x\in [1,3]\).

Solution. Figure 1.11 (a) shows that the region \(\mathcal{R}\) is a trapezoid. Since the area of a trapezoid is given by the formula
\[A =\frac{b_1+b_2}{2}h,\]
where \(b_1\) and \(b_2\) are the bases of the trapezoid and \(h\) is the height, we need to determine these lengths. It is clear from the graph that
\[b_1=f(1) = 3, \;\; b_2=f(3) = 7, \;\; h = 3-1 = 2. \;\; \Longrightarrow \;\; A= \frac{3+7}{2}\cdot 2 = 10\; (\text{units}^2).\]

Example 2. Find the area under the graph of the function \(f(x)= |\,x+2\,|-x+1\), \(x\in [-4,4]\).

Solution. First, we need to rewrite \(f\) as a piecewise function. Since \( x+2 \) changes sign at \(x=-2\),
\[
f(x)=|\, x+2\,|- x+1
= \left [ \renewcommand{\arraystretch}{1.25}
\begin{array}{cl}
-( x+2)- x+1 & \text{if }\; x<-2\\ x+2 - x+1 & \text{if }\; -2\leq x \end{array} \right. = \left [ \renewcommand{\arraystretch}{1.5} \begin{array}{cl} -2x-1 & \text{if }\; x < -2\\ 3 & \text{if }\; -2\leq x \end{array} \right.\hspace{-.1in}.\] It follows that the graph of \(f\) is as in Figure 1.11 (b). Thus, in order to compute the area under the graph, we need to find the area of the trapezoid and the rectangular region. \[A= \frac{f(-4)+f(-2)}{2}(-2-(-4)) + (4-(-2))\cdot 3 = \frac{7+3}{2}\cdot2 + 18 = 28\;(\text{units}^2).\]

 

Area Enclosed By the Graphs of Two Functions

 

Example 3. Find the area of the region \(\mathcal{R}\) enclosed by the graphs of \(\displaystyle f(x)=\left |\,\frac{3}{2} x-2\,\right | +1\) and \(\displaystyle g(x)=\frac{1}{2} x + 3\).

Solution. We rewrite \(f\) as a piecewise function.
\[
f(x)=\left |\,\frac{3}{2} x-2\,\right | +1
= \left [ \renewcommand{\arraystretch}{2.2}
\begin{array}{rl}
\displaystyle -\left ( \frac{3}{2} x-2\right ) +1 & \displaystyle \text{if }\; x < \frac{4}{3}\\ \displaystyle \left ( \frac{3}{2} x-2\right ) +1 & \displaystyle \text{if }\; x \geq \frac{4}{3}\\ \end{array} \right. = \left [ \renewcommand{\arraystretch}{2.2} \begin{array}{rl} \displaystyle - \frac{3}{2} x +3 & \displaystyle \text{if }\; x < \frac{4}{3}\\ \displaystyle \frac{3}{2} x-1 & \displaystyle \text{if }\; x \geq \frac{4}{3}\\ \end{array} \right.\hspace{-.05in}.\] It follows that the graphs of \(f\) and \(g\) are as shown in Figure 1.12. Hence, the region \(\mathcal{R}\) enclosed by their graphs is the triangle \(ABC\). To compute \(A(\mathcal{R})\), we consider \(AB\) as the base of the triangle. Then the height is the line segment \(CQ\), where \(Q\) is the point of intersection of \(AB\) and the line perpendicular to \(AB\) through \(C\).

To find the points of intersection of the graphs, \(A\) and \(B\), we solve
\begin{align*}
f(x)= g(x) \Longleftrightarrow \left |\,\frac{3}{2} x-2\,\right | +1 = \frac{1}{2}x+3 \Longleftrightarrow \left [ \renewcommand{\arraystretch}{2.2}
\begin{array}{rl}
\displaystyle – \frac{3}{2} x +3 = \frac{1}{2}x+3 & \displaystyle x < \frac{4}{3}\\ \displaystyle \frac{3}{2} x-1 = \frac{1}{2}x+3 & \displaystyle x \geq \frac{4}{3}\\ \end{array} \right. \Longleftrightarrow x\in \left \{0, 4\right \}. \end{align*} Thus, the vertices of \(\triangle ABC\) are given by \[A= \left (0, f(0)\right ) = (0,3), \;\;\;\;\; B= \left (4, f(4)\right ) = (4,5),\;\;\;\;\; C= \left (\frac{4}{3}, f\left (\frac{4}{3}\right )\right ) = \left (\frac{4}{3}, 1\right ).\] We now need to find \(Q\). The graph of \(g\) is the line \(\displaystyle \mathcal{L}:\, y = \frac{1}{2}x+3\). Thus, the line perpendicular to \(\mathcal{L}\) is the line with equation--recall that \(\displaystyle m_{\perp}= -\frac{1}{m}\), for \(m\neq 0\) \[\mathcal{L}_{\perp}: y = -2\left (x-\frac{4}{3}\right ) + 1.\] To find \(Q = \mathcal{L} \, \cap \mathcal{L}_{\perp}\), we solve \[\frac{1}{2}x+3 = -2\left (x-\frac{4}{3}\right ) + 1 \;\; \Longleftrightarrow \;\; \frac{5}{2}x = \frac{2}{3}\;\; \Longleftrightarrow \;\; x = \frac{4}{15}. \] Thus \[Q= \left (\frac{4}{15}, g\left (\frac{4}{15}\right )\right ) = \left (\frac{4}{15}, \frac{47}{15}\right ). \] We are now ready to compute \(A(\mathcal{R})\). For this, we use the distance formula: \begin{align*} \text{base } & = |\,AB\,| =\sqrt{(4-0)^2+(5-3)^2} = \sqrt{20}=2\sqrt{5}, \vphantom{\frac{A}{A_{\frac{a}{a}}}}\\ \text{height } &= |\,CQ\,| = \sqrt{\left (\frac{4}{15}-\frac{4}{3}\right )^2+\left (\frac{47}{15}-1\right )^2} = \frac{16}{3\sqrt{5}}, \vphantom{\frac{A}{A_{\frac{A}{A}}}}\\ A(\mathcal{R}) &= \frac{1}{2}\cdot 2\sqrt{5} \cdot \frac{16}{3\sqrt{5}} = \frac{16}{3}. \end{align*}

 

Area of a Parabolic Sector and Archimedes’ Theorem

 

Example 4. Find the area of the region \(\mathcal{R}\) enclosed by the graphs of \(\displaystyle f(x)=4x-x^2 \) and \(\displaystyle g(x)=x\).

Figure 1.13 shows the region enclosed by the graphs of the functions. To compute its area, we are going to use a result first proven by Archimedes (3rd century BC):

 
Archimede’s Theorem. Let \(y =a x^2+bx + c\), \(a\neq 0\), let \(A\) and \(B\) be two distinct points on the parabola, and let \(AB\) be the secant line to the parabola connecting the points. Let \(C\) be the point on the parabola whose tangent line is parallel to \(AB\). Then, the area \(A(\mathcal{R})\) of the parabolic sector enclosed by the parabola and the secant line is given by
\[A(\mathcal{R}) = \frac{4}{3}A(\triangle ABC).\]

Solution. Here is an outline of the calculations to obtain \(A(\triangle ABC)\) (see Example 3 above for a more detailed example):

• points of intersection of the line and the parabola:
  \[4x-x^2 = x\;\; \Longleftrightarrow \;\;x(x-3)=0 \;\;\Longleftrightarrow \;\; x\in \left \{0, 3\right \}.\;\; \Longrightarrow \left \{(0,0), (3,3)\right \}.\]
• let \(AB\) be the base of the triangle, then \(b = |\,AB\,| = \sqrt{3^2+3^2} = 3\sqrt{2}\).
• to find the coordinates of the point \(C\), first find the value of \(b\) for which the line \(y=mx + b\) is tangent to the parabola, where \(m =m_{AB} =1\)—parallel lines have equal slopes. (See Exaample 2 in 1.2.4 for an example with more details.)
  \[4x-x^2 = x+b \;\;\Longleftrightarrow\;\; x^2-3x+b = 0\;\;\Longleftrightarrow\;\; x=\frac{3\pm\sqrt{9-4\cdot 1\cdot b}}{2}.\]
  For the line to be tangent, there must be a unique solution \(x\). Thus
  \[9-4\cdot 1\cdot b = 0 \;\;\Longleftrightarrow\;\; b=\frac{9}{4}. \;\; \Longrightarrow\;\; x= \frac{3}{2} \;\; \Longrightarrow\;\; C\left (\frac{3}{2}, f\left (\frac{3}{2}\right )\right )=C\left (\frac{3}{2}, \frac{15}{4}\right ).\]
• to find the coordinates of \(Q\), solve simultaneously
  \[y = x\;\;\;\text{ and }\;\;\; y-\frac{15}{4}= -\left (x-\frac{3}{2}\right ) \;\; \Longleftrightarrow \;\; (x, y ) = \left (\frac{21}{8}, \frac{21}{8}\right ) = Q.\]
• The height of the triangle is given by \(\displaystyle h= |\,CQ\,| = \frac{9\sqrt{2}}{8}\).
• \(\displaystyle A(\triangle ABC) = \frac{1}{2}bh = \frac{1}{2}\cdot 3\sqrt{2}\cdot \frac{9\sqrt{2}}{8} = \frac{27}{8}\; (\text{units}^2) \).

It follows that, using Archimede’s Theorem, the area of the parabolic sector \(\mathcal{R}\) enclosed by \(y=4x-x^2\) and \(y=x\) is given by
\[A(\mathcal{R}) = \frac{4}{3}A(\triangle ABC) = \frac{4}{3}\cdot\frac{27}{8}= \frac{9}{2}\;(\text{units}^2).\]

Area under a Semicircle and Trigonometry

 

Example 5. Find the area under the graph of the function \(\displaystyle f(x)= \sqrt{4-x^2}\), \(x\in [1,2\,]\).

Solution. Figure 1.14(a) shows the region \(\mathcal{R}\) whose area is to be computed. The steps we follow to compute \(A(\mathcal{R})\) are as follows:

• compute the area of the shaded circular sector \(A(\mathcal{S})\) shown in Figure 1.14(b). The area of such a sector is given by the formula
  \[A = \frac{1}{2}r^2\theta,\]
  where \(r\) is the radius of the circle and \(\theta\) the angle subtended by the sector in radias. Thus, we have
  \[ A_S= \frac{1}{2}\cdot 2^2\,\theta.\]
• To determine \(\theta\), we use the right triangle \(OAB\) as in Figure (c). Since the hypotenuse is 2, and the base is 1, it follows that the height is \(\sqrt{3}\) (verify this via the Pythagorean theorem). It follows that
  \[\sin \theta =\frac{\sqrt{3}}{2}\;\;\; \text{ and } \;\;\; \cos \theta = \frac{1}{2} \;\; \Longrightarrow \;\; \theta = \frac{\pi}{3} \;\; \Longrightarrow \;\; A_S= \frac{1}{2}\cdot 2^2\cdot\frac{\pi}{3} = \frac{2\pi}{3}.\]
• We now compute the area—\(A\left (\triangle OAB\right )\)—of \(\triangle OAB\) as in Figure 1.14(c):
  \[A\left (\triangle OAB\right ) = \frac{1}{2}\cdot 1\cdot \sqrt{3} = \frac{\sqrt{3}}{2}.\]
• Then, the area we are looking for is given by
  \[A\left (\mathcal{R}\right ) = A(\mathcal{S})-A\left (\triangle OAB\right )= \frac{2\pi}{3} – \frac{\sqrt{3}}{2}.\]

The above computations lead to the following formula (see Figure 1.15(b)):
\[\text{if } \;f(x) = \sqrt{r^2-x^2}, \; 0 < x_1 \leq x \leq r, \;\; \Longrightarrow \]
\begin{equation}\label{ch01-03-eq01}
A\left (\mathcal{R}\right ) = \left [\renewcommand{\arraystretch}{2.5}\begin{array}{l}
\displaystyle \frac{1}{2}\, r^2\, \arccos\left (\frac{x_1}{r}\right ) – \frac{1}{2}\, x_1 \,\sqrt{r^2-x_1^2}\\
\displaystyle \frac{1}{2}\, r^2\, \arcsin\left (\frac{\displaystyle \sqrt{r^2-x_1^2}}{r}\right ) – \frac{1}{2}\, x_1 \,\sqrt{r^2-x_1^2}\\
\displaystyle \frac{1}{2}\, r^2\, \arctan\left (\frac{\displaystyle \sqrt{r^2-x_1^2}}{x_1}\right ) – \frac{1}{2}\, x_1 \,\sqrt{r^2-x_1^2}
\end{array}\right. .\tag{1.8}
\end{equation}

Note that if \(\theta\) is as in Figure 1.15(b), then \(\displaystyle \theta \in \left [ 0, \frac{\pi}{2}\right )\) and
\[\cos \theta = \frac{x_1}{r}, \;\;\; \sin \theta = \frac{\displaystyle \sqrt{r^2-x_1^2}}{r}, \;\;\; \tan \theta = \frac{\displaystyle \sqrt{r^2-x_1^2}}{x_1}\;\; \Longrightarrow \]
\[\Longrightarrow \;\; \theta = \arccos\left (\frac{x_1}{r}\right ) \;\; \text{ or }\;\; \theta = \arcsin\left (\frac{\displaystyle \sqrt{r^2-x_1^2}}{r}\right )\;\;\text{ or }\;\; \theta = \arctan\left (\frac{\displaystyle \sqrt{r^2-x_1^2}}{x_1}\right ).\]

Example 6. Find the area under the graph of the function \(\displaystyle f(x)= \sqrt{4-x^2}\), \(x\in [1,\sqrt{3}\,]\).

Solution. Figure 1.16(a) shows the region \(\mathcal{R}\) whose area is to be computed. The steps we follow to compute \(A(\mathcal{R})\) are as follows:

• compute the area \(A_1\) of the shaded region shown in Figure 1.16(b) via (\ref{ch01-03-eq01}):
  \[A_1= \frac{1}{2}\cdot 2^2\,\arccos\left (\frac{1}{2}\right ) – \frac{1}{2}\cdot1\cdot\sqrt{3} = \frac{2\pi}{3}-\frac{\sqrt{3}}{2}.\]
• compute the area \(A_{\sqrt{3}}\) of the shaded region shown in Figure 1.16(c) via (\ref{ch01-03-eq01}):
  \[A_{\sqrt{3}}= \frac{1}{2}\cdot 2^2\,\arccos\left (\frac{\sqrt{3}}{2}\right ) – \frac{1}{2}\cdot\sqrt{3}\cdot1. = \frac{\pi}{3} -\frac{\sqrt{3}}{2}.\]
• Then, the area we are looking for is given by
  \[A\left (\mathcal{R}\right ) = A_{1}-A_{\sqrt{3}}=\left (\frac{2\pi}{3}-\frac{\sqrt{3}}{2} \right)-\left (\frac{\pi}{3}-\frac{\sqrt{3}}{2}\right ) =\frac{\pi}{3}.\]

The following example shows the differences between using arcsine, arccosine, or arctangent to determine an angle. These differences arise because of their domains. In addition, we consider different methods to compute the area.

Example 7. Find the area under the graph of the function \(\displaystyle f(x)= \sqrt{4-x^2}\), \(x\in \left [-\sqrt{2},2\,\right ]\).

Solution 1. Figure 1.17(a) shows the region whose are is to be computed. We add the area of the circular sector \(\mathcal{S}\) that subtends the angle \(\theta\) and the area of the triangle \(OAB\) as shown in Figure 1.17(b):
\[A(\mathcal{S}) + A(\triangle OAB) = \frac{1}{2}\cdot2^2\cdot \arccos\left (-\frac{\sqrt{2}}{2}\right ) + \frac{1}{2}\cdot \sqrt{2}\cdot f\left (-\sqrt{2}\right ) =
2\cdot \frac{3\pi}{4} + \frac{1}{2}\cdot \sqrt{2}\cdot \sqrt{2} =\frac{3\pi}{2}+ 1. \]

 
Remark. Since the terminal side of the angle \(\theta\) is in the second quadrant, it follows that
\[\theta=\arccos\left (-\frac{\sqrt{2}}{2}\right ) = \pi + \arcsin \left (-\frac{\sqrt{2}}{2}\right ) = \pi + \arctan\left (-\frac{\sqrt{2}}{2}\right ).\]

Solution 2. Figure 1.17(c) shows that to computed the desired area, all we need to do is to add the area of one quarter of a circle, the circular sector \(\mathcal{S}\) that subtends the angle \(\alpha\) and the area of the triangle \(OAB\):
\begin{align*}
\frac{1}{4}\cdot \pi\cdot2^2 + A(\mathcal{S}) + A(\triangle OAB) &= \pi + \frac{1}{2}\cdot2^2\cdot \arcsin\left (\frac{\sqrt{2}}{2}\right ) + \frac{1}{2}\cdot \sqrt{2}\cdot f\left (-\sqrt{2}\right ) =
\vphantom{\frac{A}{A_A}}\\
&= \pi +
2\cdot \frac{\pi}{4} + \frac{1}{2}\cdot \sqrt{2}\cdot \sqrt{2}= \pi + \frac{\pi}{2}+ 1 =\frac{3\pi}{2}+ 1.
\end{align*}

 

1.3.4 Rectangular Approximations

 

As stated in 1.1.3, the method to solve the area problem in general consists on using rectangular approximations. Here are some examples of how this method works. Along the way we will accomplish the following:

• review the graphs of trigonometric, inverse trigonometric, exponential, and logarithmic functions;
• evaluate and simplify, using their basic properties, the above functions;
• become familiar with the approximation process to area under the graph of a function via rectangular approximations.

Example 1. Let \(f(x)=-x^2+2x+4\), \(x\in [-1,2]\).

(a) Compute the sum of the areas of the rectangles shown in the rectangular approximation in Figure 1.10(b).

(b) Use Archimedes’ theorem to find the exact area of under the graph of \(f\).

Solution. (a)

(1) To compute the area of the rectangles, note that the rectangles have the same width. The standard notation for the width is \(\Delta x\), thus, we write \(\Delta x = 1\). It follows that we need to determine the heights of the rectangles at the points:
\[x_0=-1,\;\; x_1=-1+1=0,\;\; x_2=-1+2(1)=1,\;\; x_3=-1+3(1)=2.\]

(2) Height and area of each rectangular region.

Note that the height of each rectangle is determined by the value of \(f\) at each one of the above points.
\[\renewcommand{\arraystretch}{2.}
\begin{array}{c|l|l}
\text{Rectangle} & \hspace{.6in}\text{Height} & \hspace{.5in}\text{Area}\\ \hline
R_1 &
h_1=f\left(x_1\right)=f\left(0\right)=4&
A_1=f\left(x_1\right)\Delta x=4\cdot 1\\
R_2 &
h_2=f\left(x_2\right)=f(1)=5&
A_2=f\left(x_2\right)\Delta x=5\cdot 1\\
R_3 &
h_3=f\left(x_3\right)=f\left(2\right)=4&
A_3=f\left(x_3\right)\Delta x=4\cdot 1
\end{array}
\]

(3) Sum of the areas of the rectangular regions: \(R(f,3)=4\cdot 1+5\cdot 1+4\cdot 1=13\).

(b) Here is an outline of the calculations. You are asked to verify them in Exercise 11. If needed, review Example 4 in 1.3.2. In the notation of Figure 1.18 (c), the coordinates of the points \(A\), \(B\), \(C\), \(Q\) are
\[A(-1, f(-1)) = (-1,-1),\;\; B(2, f(2)) = (1, 4);\;\; C\left (\frac{1}{2}, \frac{19}{4}\right ); \;\; Q\left (\frac{13}{8}, \frac{29}{8}\right ).\]
Furthermore,
\[|\,AB\,| = 3\sqrt{2}, \;\; |\,CQ|, = \frac{9\sqrt{2}}{8}, \;\; \Longrightarrow \;\; A\left (\triangle ABC\right ) = \frac{1}{2}\cdot 3\sqrt{2} \cdot \frac{9\sqrt{2}}{8} = \frac{27}{8}.\]
It follows that the area of the parabolic sector is \(\displaystyle A_{\text{Parabolic Sector}} = \frac{4}{3}\cdot \frac{27}{8} = \frac{9}{2}\).

The area of the trapezoid below the parabolic sector is given by \(\displaystyle A_{\text{Trapezoid}} = \frac{1+4}{2}\cdot3 = \frac{15}{2}\).

Hence, the area under the graph of the parabola is given by
\[A = A_{\text{Parabolic Sector}}+ A_{\text{Trapezoid}} = \frac{9}{2}+ \frac{15}{2}= 12.\]
The rectangular approximation in (a) overestimates the area by \(1\).

Example 2. Let \(f(x)=\cos x + 1\), \(\displaystyle x\in \left[-\frac{\pi}{2},\pi \right]\). Compute the sum of the areas of the rectangles shown in Figure 1.19 (b).

Solution.

(1) The rectangles have the same width, and this is given by
\[\Delta x=\frac{\pi -(-\pi /2)}{6}=\frac{\pi }{4}.\]
Note that there are six rectangles, but one of them has zero height. That is, it has been reduced to a line.

We need to find the heights of the rectangles at each one of the following points:
\[x_1= -\frac{\pi }{4}, \;\; x_2 =0, \;\; x_3= \frac{\pi }{4}, \;\; x_4= \frac{\pi }{2},\;\; x_5= \frac{3\pi }{4},\;\; x_6=b= \pi.
\]

(2) Height and area of each rectangular region.

\[\renewcommand{\arraystretch}{2.2}
\begin{array}{c|l|l}
\text{Rectangle} & \hspace{.8in}\text{Height} & \hspace{.75in}\text{Area}\\ \hline
R_1 & \displaystyle
h_1=f\left(x_1\right)=f\left(-\frac{\pi }{4}\right)=1+\frac{\sqrt{2}}{2}& \displaystyle
A_1=f\left(x_1\right)\Delta x=\left(1+\frac{\sqrt{2}}{2}\right)\cdot \frac{\pi}{4}\\
R_2 & \displaystyle
h_2=f\left(x_2\right)=f(0)=2& \displaystyle
A_2=f\left(x_2\right)\Delta x=3\cdot \frac{\pi}{4}\\
R_3 & \displaystyle
h_3=f\left(x_3\right)=f\left(\frac{\pi }{4}\right)=1+\frac{\sqrt{2}}{2}& \displaystyle
A_3=f\left(x_3\right)\Delta x=\left(1+\frac{\sqrt{2}}{2}\right)\cdot \frac{\pi}{4}\\
R_4 & \displaystyle
h_4=f\left(x_4\right)=f\left(\frac{\pi }{2}\right)=1& \displaystyle
A_4=f\left(x_4\right)\Delta x=1\cdot \frac{\pi}{4}\\
R_5 & \displaystyle
h_5=f\left(x_5\right)=f\left(\frac{3\pi }{4}\right)=1-\frac{\sqrt{2}}{2}& \displaystyle
A_5=f\left(x_5\right)\Delta x=\left(1-\frac{\sqrt{2}}{2}\right)\cdot \frac{\pi}{4}\\
R_6 & \displaystyle
h_6=f\left(x_6\right)=f(\pi)= -1+1=0& \displaystyle
A_6=f\left(x_6\right)\Delta x=0\cdot \frac{\pi}{4}
\end{array}
\]

(3) Sum of the areas of the rectangular regions.
\begin{align*}
R(f,6)&=\left(1+\frac{\sqrt{2}}{2}\right)\left(\frac{\pi }{4}\right)+(2)\left(\frac{\pi }{4}\right)+\left(1+\frac{\sqrt{2}}{2}\right)\left(\frac{\pi
}{4}\right)+(1)\left(\frac{\pi }{4}\right)+ \\
&\hspace{.5in} +\left(1-\frac{\sqrt{2}}{2}\right)\left(\frac{\pi }{4}\right)+(0)\left(\frac{\pi }{4}\right)=\left(\frac{12+\sqrt{2}}{2}\right)\frac{\pi }{4}.
\end{align*}

Example 3. Let \(\displaystyle f(x)=\arcsin x + \frac{\pi}{2}\), \( x\in \left[-1,1 \right]\). Compute the sum of the areas of the rectangles shown in Figure 1.20 (b)(b). Assume all rectangles have the same width.

Solution.

(1) Given that the interval \([-1,1]\) has length \(2 = 1-(-1)\) and there are four rectangles, then the width of each rectangle is
\[\Delta x=\frac{1-(-1))}{4}=\frac{1 }{2}.\]
It follows that we need to determine the heights of the rectangles at the following points:
\[x_1=-\frac{1}{2}, \;\; x_2=0, \;\; x_3=\frac{1}{2}, \;\; x_4 = 1.\]

(2) Height and area of each rectangular region.

\[
\renewcommand{\arraystretch}{2.2}
\begin{array}{c|l|l}
\text{Rectangle} & \hspace{.8in}\text{Height} & \hspace{.75in}\text{Area}\\ \hline
R_1 & \displaystyle
h_1=\arcsin\left(-\frac{1}{2}\right)+\frac{\pi}{2}=-\frac{\pi}{6}+\frac{\pi}{2}=\frac{\pi}{3}& \displaystyle
A_1=f\left(x_1\right)\Delta x=\frac{\pi}{3}\cdot \frac{1}{2}=\frac{\pi}{6}\\
R_2 & \displaystyle
h_2=\arcsin\left(0\right)+\frac{\pi}{2}=0+\frac{\pi}{2}=\frac{\pi}{2}& \displaystyle
A_2=f\left(x_2\right)\Delta x=\frac{\pi}{2}\cdot \frac{1}{2}=\frac{\pi}{4}\\
R_3 & \displaystyle
h_3=\arcsin\left(\frac{1}{2}\right)+\frac{\pi}{2}=\frac{\pi}{6}+\frac{\pi}{2}=\frac{2\pi}{3}& \displaystyle
A_3=f\left(x_3\right)\Delta x=\frac{2\pi}{3}\cdot \frac{1}{2}=\frac{\pi}{3}\\
R_4 & \displaystyle
h_4=\arcsin\left(1\right)+\frac{\pi}{2}=\frac{\pi}{1}+\frac{\pi}{2}=\pi& \displaystyle
A_4=f\left(x_4\right)\Delta x=\pi\cdot \frac{1}{2}=\frac{\pi}{2}
\end{array}
\]

(3) Sum of the areas of the rectangular regions.
\[R(f,4)=\frac{\pi}{6} + \frac{\pi}{4} + \frac{\pi}{3} +\frac{\pi}{2} = \frac{2\pi+3\pi + 4\pi +6\pi}{12} =\frac{15\pi}{12} =\frac{5\pi}{4}.\]

Example 4. Let \(\displaystyle f(x)=e^x\), \( x\in \left[-2,1 \right]\). Compute the sum of the areas of the rectangles shown in Figure 1.21 (b). Assume all rectangles have the same width.

Solution.

(1) The width of the rectangles is \(\displaystyle \Delta x = \frac{2-(-2)}{4} = 1\). It follows that we need to determine the heights of the rectangles at the points:
\[x_1=-1,\;\; x_2= 0,\;\; x_3= 1,\;\; x_4=2.\]

(2) Height and area of each rectangular region.

Note that the height of each rectangle is determined by the value of \(f\) at each one of the above points.
\[\renewcommand{\arraystretch}{2.2}
\begin{array}{c|l|l}
\text{Rectangle} & \hspace{.8in}\text{Height} & \hspace{.75in}\text{Area}\\ \hline
R_1 & \displaystyle
h_1=f\left(x_1\right)=f\left(2e\right)=e^{-1}& \displaystyle
A_1=f\left(x_1\right)\Delta x=e^{-1}\cdot 1\\
R_2 & \displaystyle
h_2=f\left(x_2\right)=f(0)=e^0=1& \displaystyle
A_2=f\left(x_2\right)\Delta x=1\cdot 1\\
R_3 & \displaystyle
h_3=f\left(x_3\right)=f\left(1\right)=e& \displaystyle
A_3=f\left(x_3\right)\Delta x=e\cdot 1\\
R_4 & \displaystyle
h_4=f\left(x_4\right)=f\left(2\right)=e^2& \displaystyle
A_4=f\left(x_4\right)\Delta x=e^2\cdot 1
\end{array}
\]

(3) Sum of the areas of the rectangular regions:
\[\frac{1}{e}+1 + e + e^2 = \frac{1}{e}\left (e+e^2+e^3\right ).\]

Example 5. Let \(\displaystyle f(x)= \ln x\), \( x\in \left[e,6e \right]\). Compute the sum of the areas of the rectangles shown in Figure 1.22 (b). Assume all rectangles have the same width.

Solution.

(1) The width of the rectangles is
\[\Delta x = \frac{6e-e}{5} = e.\]
It follows that we need to determine the heights of the rectangles at the points:
\[x_1=2e,\;\; x_2= 3e,\;\; x_3= 4e,\;\; x_4=5e, \;\; x_5=6e.\]

(2) Height and area of each rectangular region.

Note that the height of each rectangle is determined by the value of \(f\) at each one of the above points.
\[\renewcommand{\arraystretch}{1.5}
\begin{array}{c|l|l}
\text{Rectangle} & \hspace{.8in}\text{Height} & \hspace{.75in}\text{Area}\\ \hline
R_1 & \displaystyle
h_1=f\left(x_1\right)= \ln (2e) = \ln 2 + 1& \displaystyle
A_1=f\left(x_1\right)\Delta x=\left(\ln 2 + 1 \right)\cdot e\\
R_2 & \displaystyle
h_2=f\left(x_2\right)=\ln (3e) = \ln 3 + 1& \displaystyle
A_2=f\left(x_2\right)\Delta x=\left(\ln 3 + 1 \right)\cdot e\\
R_3 & \displaystyle
h_3=f\left(x_3\right)=\ln (4e) = \ln 4 + 1=e& \displaystyle
A_3=f\left(x_3\right)\Delta x=\left(2\ln 2 + 1 \right)\cdot e\\
R_4 & \displaystyle
h_4=f\left(x_4\right)=\ln (5e) = \ln 5 + 1& \displaystyle
A_4=f\left(x_4\right)\Delta x=\left(\ln 5 + 1 \right)\cdot e\\
R_5 & \displaystyle
h_5=f\left(x_5\right)=\ln (2e) = \ln 6 + 1& \displaystyle
A_5=f\left(x_5\right)\Delta x=\left(\ln 2 +\ln 3 + 1 \right)\cdot e
\end{array}
\]

(3) Sum of the areas of the rectangular regions:
\[\left(\ln 2 + 1 \right)\cdot e + \left(\ln 3 + 1 \right)\cdot e + \left(\ln 4 + 1 \right)\cdot e + \left(\ln 5 + 1 \right)\cdot e + \left(\ln 6 + 1 \right)\cdot e =\]
\[= \left(\ln \left (2\cdot 3\cdot 4\cdot5\cdot6\right )+5 \right)\cdot e = \left (\ln 720 + 5\right )\cdot e,\]
or
\[\left(\ln 2 + 1 \right)\cdot e + \left(\ln 3 + 1 \right)\cdot e + \left(2\ln 2 + 1 \right)\cdot e + \left(\ln 5 + 1 \right)\cdot e + \left(\ln 2 +\ln 3 + 1 \right)\cdot e =\]
\[= \left(4\ln 2 +2\ln 3 +\ln 5 + 5 \right)\cdot e
.\]