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As explained in Chapter 3, GC is built to create graphs. It is designed to interpret anything you type as either a definition of a mathematical term (e.g., a function) or a command to create a graph. Essentially, any wellformed statement that you type, other than a definition, will cause GC to generate a graph.
We describe regions in the plane by stating conditions that points in the plane must satisfy to be considered in that region. For example, the mathematical statement $\left\{(x,y)\mid x^2+y^2\le 1\right\}$, read as "The set of all points with coordinates x and y such that ...", describes the set of points in Figure 8.3.01 (apologies for the weird numbering). GC highlighted all the points in the Cartesian plane that have coordinates x and y such that $x^2+y^2\le 1$.Notice that in Figure 8.3.01 we typed only the condition part of the mathematical statement. This is because GC always interprets any nondefinitional statement involving x or y as a command to "highlight all the points (x,y) such that ...".
Statement  Meaning  Result in GC 
$x=3$  Highlight all the points in the plane with coordinates $(x,y)$ such that $x=3$  
$y=x^3$  Highlight all the points in the plane with coordinates $(x,y)$ such that $y=x^3$.  
$x=y^3$  Highlight all the points in the plane with coordinates $(x,y)$ such that $x=y^3$.  
$1\le x\le 2$  Highlight all the points in the plane with coordinates $(x,y)$ such that $1\le x \le 2$.  
$1\le x \le 2, 0.5\le y \le 0$  Highlight all the points in the plane with coordinates $(x,y)$ such that $1\le x \le 2$ AND $0.5\le y \le 0$.  
$f(x)=x^2$ $g(x)=\sqrt{x}$ $y=f(x)$ $y=g(x)$ $f(x)<y<g(x)$ 
Define the functions $f$ and $g$. Highlight all the points in the plane with coordinates $(x,y)$ such that $y=f(x)$. Highlight all the points in the plane with coordinates $(x,y)$ such that $y=g(x)$. Highlight all the points in the plane with coordinates $(x,y)$ such that $f(x)\lt y$ and $y\lt g(x)$. 

$f(x)=x^2$ $g(x)=x^3$ $x=f(y)$ $x=g(y)$ $f(y)<x<g(y)$ 
Define the functions $f$ and $g$. Highlight all the points in the plane with coordinates $(x,y)$ such that $x=f(y)$. Highlight all the points in the plane with coordinates $(x,y)$ such that $x=g(y)$. Highlight all the points in the plane with coordinates $(x,y)$ such that $f(y)\lt x$ and $x\lt g(y)$. 
A problem with thinking of integrals as areas of regions bounded by curves is that the common notion of area is that it is always positive. However, a region below the xaxis would have a "negative" area (Figure 8.3.2).
In this textbook we have always spoken of integrals as accumulations of some quantity at some rate of change. We avoided speaking of integrals as areas. But this does not mean that the two ideas (integrals as accumulations or as areas) are incompatible. In this section we will explain how they are indeed compatible. But we must look closely at the idea of area and its rate of change.
Examine the rectangle in Figure 8.3.3. It shows a region bounded by a rectangle that is 5 cm high and 10 cm wide. We can think of the region bounded by the rectangle as filling from left to right as the region's width varies from 0 cm to 10 cm. Since the area varies with width, we can ask about the area's rate of change with respect to changes in its width.
What is the rate of change of the enclosed region's area with respect to changes in its width? Let $w$ be the enclosed region's width as it varies. Then the value of $w$ varies from 0 to 10.
The highlighted region varies as $w$ varies, and therefore the highlighted region's area is a function of $w$, or $A(w)=5w$. To ask about the highlighted region's rate of change of area with respect to width is to ask about $r_A$, the exact rate of change function for $A$.
The rate of change of $A$ at a moment of $w$ over an interval of length
$h$ is
$$(*)\qquad \begin{align} r(w)&=\frac{A(w+h)A(w)}{h}\\[1ex]
&=\frac{5(w+h)5w}{h}\\[1ex]
&=\frac{5w+5h5w}{h}\\[1ex]
&=5 \quad \mathrm{cm^2/cm} \end{align}$$
The rate of change of the rectangle's enclosed area with respect to $w$ as $w$ varies from $w=0$ to $w=10$ is 5 $\mathrm{cm^2/cm}$. That is, the numeric value of the rectangle's height is the numeric value of the area's rate of change.
Notice that we said that the numeric value of the rectangle's height is the numeric value of the area's rate of change. The rectangle's height is not the area's rate of change. The rectangle's height is 5 cm. The area's rate of change is 5 $\mathrm{cm^2/cm}$!
There was nothing special about a height of 5 cm in equation (*) above. A height of $H$ would have given us that the region's area changes with respect to width at a rate of $H \, \mathrm{ cm^2/cm}$. Indeed, a height of $f(w)$ would have given us that the region's area changes with respect to width at a rate of $f(w) \, \mathrm{ cm^2/cm}$.
To understand that a region bounded by a function's graph can have positive or negative area we must first be clear that function's values have directions.
Figure 8.3.3a illustrates two ways of thinking about function's values.
The first way to think about a function's values is ambiguous about how to conceive of the function's value on the graph. It is unclear whether:
The second way clarifies the matter:
Reflection 8.3.1. How do you anticipate that using an arrow (vector) in Figure 8.3.3a will clarify the idea of signed (negative or positive) area? What understanding does an arrow provide that a bare line segment (as in Part 1 of Figure 8.3.3a) hides?
It would be understandable for you to wonder what this discussion about function's values has to do with areas of regions bounded by function's graphs. We hope that Figure 8.3.4 will help you make the connection.
In Figure 8.3.4, we intend that you see the region bounded by $y=0$, $x=3$, $x=3$, and $y=2\sin(x)$ being swept out as $x$ varies from 3 to 3. Upon zooming in, however, you'll see that the region is actually made of (potentially) infinitesimal rectangles, and that it is the rectangles being filled that fills the bounded region.
Each rectangle has a height of f(x), and is filled
over an interval of length dx. Each rectangle is therefore being
filled at a rate of f(x) $\mathrm{units^2/unit}$ throughout
intervals of length dx. 
Figure 8.3.4. You can think of a region bounded by the graph of a
function as composed of rectangles of infinitesimal width. The
area of each rectangle changes at the rate of f(x) $units^2$ of area per
unit of width.
Reflection 8.3.2. Explain how a negative
value of a function produces a negative change in accumulated signed
area bounded by its graph. Explain the same for positive values of a
function producing positive changes in accumulated signed area bounded
by the function's graph.
We can now relate integrals as areas of bounded regions and integrals as net accumulation from rate of change.
Let f be a function whose graph is given within the Cartesian coordinate system.
Let A be the function such that $A(x)$ is the net signed area bounded by the graph of f over the interval $[a,x]$.
Each value of $f$ is the exact rate of change
function for the area function $A$, or $f(x)=r_A(x)$. Therefore, $A(x)$,
the net signed area of the region bounded by the graph of $f$ over the
interval $[a,x]$, is the net accumulation of signed area that accumulates
at the rate of change of $f(t)$ over intervals of length $dt$ as $t$
varies from $t=a$ to $t=x$. We say this symbolically as:
$$\begin{align}
A(x)&=\int_a^xr_A(t)dt\\[1ex]
&=\int_a^x f(t)dt\end{align}$$
A region in Cartesian space bounded by the graph of a function is just a set of points. As a region, it is neither positive nor negative. But area is a measure of a region. You will see that area of a region can be positive or negative depending on how the region is swept out.
A region bounded by a graph is created by a function's value sweeping
along the independent axis. Function's values can be positive or negative.
The animation in Figure 8.3.5 illustrates this idea.
How to think of positive and negative area
The animation in Figure 8.3.5 gave away the computation of the signed area of a bounded region. You compute total signed area by accumulating bits of signed area that each accumulate at a rate of change of change. We know very well by now that we use an integral to do this.
Define a function A whose values $A(x)$ give the net signed area of the region bounded by $y=0$ and $g(t)=e^{\sin\left(t^{\cos t}\right)}1.5$ from $t=0$ to $t=x$ for all values of $x>0$. Display $A's$ graph in GC's right pane by entering $y'=A(x')$.
For Part 1, we must describe the regions bounded below and above the xaxis separately. Mathematically, the region above the xaxis and bounded by the graph of g is $\left\{(x,y)  x\ge 0, 0\le y\le g(x)\right\}$. The region below the xaxis and bounded by the graph of g is described mathematically as $\left\{(x,y)  x\ge 0, g(x)\le y\le 0\right\}$.
Figure 8.3.6 shows these commands and the resulting graphs in GC. In GC we need only state the conditions of the mathematical statements because GC interprets all nondefinitional commands as saying, "Graph all points $(x,y)$ such that ...".
For Part 2, we need only define the integral function as shown in Figure 8.3.7.
Reflection 8.3.3. Examine each statement in the left margin of Figure 8.3.7. What does each statement do in regard to generating the graphs and highlighted regions in the two graphing panes?
Reflection 8.3.4. Examine the two graphs in Figure 8.3.7. Explain how the left graph and the right graph show the same thing in regard to accumulation of displacement in relation to number of seconds elapsed.
It is important to notice that the quantity represented by
signed area in the left graph of Figure 8.3.7 does not
have $\mathrm{meter^2}$ as its unit. Rather, it has meter
as its unit. Each incremental change in signed area is created
by something moving at a velocity in meters/sec for a number of
seconds! This is one of the many ways that thinking about integrals as areas ends up being misleading. Area is measured in square units, whereas signed area (coming from an integral) has a unit that derives from the quantities involved. You will not experience this confusion when you understand integrals as accumulations. 
Integrals give net signed area of regions bounded by functions' graphs. Negative areas and positive areas cancel each other to varying extents. The absolute value of a function makes all values positive, as shown in Figure 8.3.8. The absolute value function makes all regions have positive area.
Reflection 8.3.5. The last three statements in GC produce colored regions in the displayed graphs. What regions does each statement highlight, and why does the statement highlight these regions?
Unsigned area is always positive, while signed area can be either negative or positive. Therefore an integral function that gives signed area can be increasing or decreasing over different intervals, whereas an integral function that gives unsigned area is always increasing (Figure 8.3.9).
Figure 8.3.9. Graph of $y=A(x)$ (left), the signed area function,
compared to graph of $y=\leftA(x)\right$ (right).
Connect the graphs on the left sides of Figures 8.3.8 and 8.3.9 and the
graphs on the right side of the same figures.
Figure 8.3.10 shows that the idea of signed area of regions bounded between functions' graphs is not as straightforward as the idea of the signed area of a region bounded by a functions graph and $y=0$ over an interval of $x$. The question is when to consider a region as having positive area and when to consider it as having negative area.
Reflection 8.3.6. Reset the animation in Figure 8.3.11 to its beginning after watching it once. Imagine the arrow that shows $g(x)f(x)$ in the left pane varying as the value of x varies. Then envision what will appear in the right panehow the arrow on the left will show the value of $g(x)f(x)$ in the right pane as the value of x varies. Run the animation again to confirm what you envisioned.
Reflection 8.3.7. Reset the animation in Figure 8.3.12 to its beginning after watching it once. Imagine the arrow that shows $f(x)g(x)$ in the left pane varying as the value of x varies. Then envision what will appear in the right panehow the arrow on the left will show the value of $f(x)g(x)$ in the right pane as the value of x varies. Run the animation again to confirm what you envisioned.
Reflection 8.3.8. Reset the animation in Figure 8.3.13 to its beginning after watching it once. Imagine the arrow that shows $g(x)f(x)$ in the left pane varying as the value of x varies and the signed area that it generates. Then envision what will appear in the right panehow the arrow on the left and the signed area it generates will show the value of $g(x)f(x)$ and the signed area it generates in the right pane as the value of x varies. Run the animation again to confirm what you envisioned.
Reflection 8.3.9. Reset the animation in Figure 8.3.14 to its beginning after watching it once. Imagine the arrow that shows $f(x)g(x)$ in the left pane varying as the value of x varies and the signed area that it generates. Then envision what will appear in the right panehow the arrow on the left and the signed area it generates will show the value of $f(x)g(x)$ and the signed area it generates in the right pane as the value of x varies. Run the animation again to confirm what you envisioned.
If the comparison you make is from $f(x)$ to $g(x)$, then a bounded region will have positive area when $g(x)>f(x)$, and will have negative area when $g(x)<f(x)$.
If the comparison you make is from $g(x)$ to $f(x)$, then a bounded region will have positive area when $f(x)>g(x)$, and will have negative area when $f(x)<fgx)$.
You must make the direction of comparison clear to yourself in order to decide on the sign of a bounded region's area.
This section assumes that you have learned about radian angle measure and functions graphed in polar coordinates. Review Chapter 2 if you are unfamiliar with radian angle measure; review Chapter 3, section 3.9 if you are unfamiliar with graphs in polar coordinates.
The common perception of area is that all regions bounded by graphs are measured in the same way  with rectangles that collectively approximate the region and the region's area is approximated by the sum of the rectangles' areas. An example will illustrate that this way of thinking is appropriate only when the functions graphs are displayed in Cartesian (rectangular) coordinates.
Figure 8.3.15 shows what seem to be identical graphs. The graph on the left is of $y=f(x)$, where $f$ is defined as $f(x)=\sqrt{9x^2},3\le x \le 3$, in Cartesian (rectangular) coordinates. The graph on the right is of $r=g(\theta)$, where $g$ is defined as $g(\theta)=3,0\le \theta \le \pi$, in polar coordinates.
While the graphs appear to be identical, Figure 8.3.16 shows that the integrals of the functions that produce them are not identical. $\int_{3}^{3}f(x)dx$ is approximately 14.14, while $\int_{0}^{\pi}g(\theta)d\theta$ is approximately 9.42.
The reason that these two integrals, which ostensibly "cover" the same region, have different values is this: While the unsigned area of the enclosed region in the left graph accumulates at a rate of $f(x)$ with respect to x, the unsigned area of the enclosed region in the right graph does not accumulate at a rate of $g(\theta)$ with respect to $\theta$. So when a region in the polar plane is bounded by the graph of $y=g(\theta)$, the function $\int_0^x g(\theta)d\theta$ does not give an accurate measure of the enclosed region's unsigned area.
Figure 8.3.17 shows a closeup of the regions being swept in Figure 8.3.15. The area of the incremental change in swept region of the graph in Cartesian coordinates is essentially $f(x)dx$. But the area of the incremental change in the swept region in polar coordinates is not $g(\theta)d\theta$.
Figure 8.3.18 illustrates that when a region
bounded by a circle of radius r is swept by an angle that
subtends an arc of measure $\theta$ radii, the unsigned area of the swept
region varies at a constant rate from 0 to $\pi r^2$ as $\theta$ varies
from 0 to $2\pi$. So changes in unsigned area of swept regions in polar
coordinates vary in proportion to changes in subtended arc length.
Let $A(\theta)$ be the unsigned area of a swept region in the circle of radius r. Since $A(\theta)$ changes at a constant rate with respect to $\theta$, $dA$, the differential in unsigned area, is proportional to $d\theta$, or $dA=k\cdot d\theta$ for some number k.
If we let $d\theta=2\pi$ (we sweep the entire circle), then $dA$ (the change in unsigned area) is $\pi r^2$. Therefore, $\pi r^2=k\cdot 2\pi$, or $k=\dfrac{1}{2}r^2$. The unsigned area of a swept region in a circle with radius r changes at the rate of $\dfrac{1}{2}r^2$ with respect to changes in angle measure. So, in a circle of radius r, $dA=\dfrac{1}{2}r^2d\theta$.
Reflection 8.3.10. Try to say in your own words why the rate of change of swept area in a circle with respect to angle measure is $\dfrac{1}{2}r^2$.
The relevance of the above to unsigned area of regions bounded by curves in polar coordinates is this (read the bullets in conjunction with viewing Figure 8.3.19):
We return to the example of Figure 8.3.16, which suggested that $g(\theta)$ is not the rate of change of unsigned area with respect to $\theta$. It is repeated below, as Figure 8.3.20but this time with an appropriate rate of change function for signed area with respect to angle measure.
The integrals agree to two decimal places. (Remember, GC approximates integrals, and it has particular difficulties with integrals of rate of change functions that change wildly over small intervals, which is true of $f(x)=\sqrt{9x^2}$ near $x=3$ and $x=3$.)
Reflection 8.3.11. Attend to the independent variables in the left and right panes of Figure 8.3.20. What is the independent variable in the left pane? What is the independent variable in the right pane? How do they vary differently? Over what intervals do they vary?
Reflection 8.3.12. What signs do the areas of regions generated in Figure 8.3.20 have? Given the definition of the area of a swept region in polar coordinates as $\frac{1}{2}\int_0^\theta g(t)^2dt$, is it possible for a swept region in polar coordinates to have negative area?
Figure 8.3.21 shows the graph of $r=f(\theta)$, where $f$ is defined as $f(x)=1+4\cos(x)$, being constructed as $\theta$ varies from 0 to $2\pi$.
Study Figure 8.3.21 closely.
First, looking from the polar pole in the direction of $\theta$, some of the graph's points are in the direction of $\theta$ while some are in the opposite direction of $\theta$. When $f(\theta)\lt 0$, GC plots the corresponding point in the opposite direction that $\theta$ points. It is as if $\theta$ controls a rotating number line. As $\theta$ varies, positive values of f are plotted from the pole in the direction of $\theta$ and negative values of f are plotted from the pole in the opposite direction of $\theta$.
Second, the polar pole is not like the origin in Cartesian coordinates. In Cartesian coordinates, the origin has coordinates $(0,0)$. In polar coordinates, the pole always has coordinates $(0,\theta)$, where $\theta$ can be any number. A point will be plotted at the polar pole whenever $f(\theta)=0$regardless of the value of $\theta$.
The second point is important when we think of defining regions in polar coordinates.
While it seems natural to think that negative values of a function should
sweep negative areas in the polar plane, we cannot capture this with the
integral $$A(\theta)=\frac{1}{2}\int_a^\theta g(t)^2dt.$$
The reason that we lose the sign of the rate of change is that $g(t)^2$ is
always positive! We lose the sign of the function's value when squaring
it.
We can still reflect the sign of the function in the rate of change of
swept area, however, by using the function "sgn": $\mathrm{sgn}(x)$ is
called the "signum" function ("signum"
is Latin for "sign"); sgn(x) produces 1 if $x\lt 0$, 0 if $x=0$,
and 1 if $x\gt 0$. It is defined officially as
$$\mathrm{sgn}(x)=\begin{cases}
1 & \text{if $x\lt 0$}\\[1ex]
0 & \text{if $x=0$}\\[1ex]
1 & \text{if $x\gt 0$}
\end{cases}$$
Given a function $g$ graphed in polar coordinates, we can recover the sign
of the rate of change of swept area with respect to angle measure by
multiplying $g(t)^2$ and $\mathrm{sgn}(g(t))$. The net signed area of a
region in polar coordinates as $\theta$ varies is therefore
$$(\text{***})\qquad A(\theta)=\frac{1}{2}\int_a^\theta
\left(\mathrm{sgn}(g(t))\cdot g(t)^2\right)dt.$$
We can test this statement against the function and graph in Figure in 8.3.21.
If you watch Figure 8.3.21 again, you will see that the inner loop is made by negative values of $f(\theta)$. The unsigned area of the inner loop is therefore counted twice as we sweep unsigned area as $\theta$ varies from 0 to $2\pi$.
If we use $\frac{1}{2}\int_a^\theta f(t)^2dt$ to compute the unsigned area of the outside loop (which contains the inside loop) and subtract the unsigned area of the inner loop, we should get the same result as using $\frac{1}{2}\int_a^\theta(\mathrm{sgn}(f(t))f(t)^2dt$ to compute net signed area. We can test this observation by computing signed area as swept according to Figure 8.3.21 in two waysthe first by subtracting unsigned area that is included twice, and the second by computing signed area.
For convenience, we will define functions $A_1$ and $A_2$ so that they
take their lower limit as input. $A_1(a,\theta)$ will give the unsigned
area enclosed by the graph from $t=a$ to $t=\theta$. $A_2(a,\theta)$ will
give the signed area enclosed by the graph from $t=a$ to $t=\theta$.
$$\begin{align}A_1(a,\theta)&=\frac{1}{2}\int_a^\theta f(t)^2dt\\[1ex]
A_2(a,\theta)&=\frac{1}{2}\int_a^\theta \left(\mathrm{sgn}(f(t))\cdot
f(t)^2\right)dt\end{align}$$
We will fill the upper half of the outer curve by letting $\theta$ vary from $\theta=0$ to the first value of $\theta$ that makes $1+4\cos(\theta)=0$. This will happen when $\cos(\theta)=0.25$, or when $\theta=\mathrm{acos}(0.25)$. Recall that "acos" is computerese for arc cosine, or the arc (angle) that produces a particular value of cosine. Books typically use $\cos^{1}$, while GC uses "acos". This is covered more fully under inverse functions in Chapter 3.
Figure 8.3.22 shows that letting $\theta$ vary from 0 to acos(0.25) sweeps the region enclosed in the top half of the outer curve (which includes the inner curve). The unsigned area of the region enclosed by the outer curve is therefore twice the area of the highlighted region, or $2A_1(0,\mathrm{acos}(0.25))$.
Figure 8.3.23 shows the bottom half of the region bounded by the inner curve, or the part of the graph that is generated when $\theta$ varies from $\theta=\mathrm{acos}(0.25)$ to $\theta=\pi$. The unsigned area of the inner region is therefore twice the area of the highlighted region, or $2A_1(\mathrm{acos}(0.25),\pi)$.
The signed area of the region enclosed by $r=f(\theta)$ as $\theta$ varies from 0 to $2\pi$ should therefore be the difference of the unsigned regions bounded by the outer and inner loops, or $2A_1(0,\mathrm{acos}(0.25))2A_1(\mathrm{acos}(0.25),\pi)$.
Figure 8.3.24 (copied from a GC window) confirms that the two methods produce the same result. The difference in unsigned areas is the same as the net signed area as $\theta$ varies from 0 to $2\pi$.
In this entire section we have shown that the idea of integral as signed area of a region bounded by one or more graphs is entirely consistent with the general theme of this bookthat an integral is an accumulation of changes in a measured quantity that varies at a rate of change with respect to another quantity.
Moreover, we demonstrated that the general idea of integral as accumulation is coherent with signed areas of regions bounded by graphs regardless of the coordinate system.
As is always the case with integrals as accumulations, the key to integrals as signed areas in any coordinate system is to determine the appropriate rate of change function whose values give the rate of change of signed area at every moment of its independent variable.
a.  
b.  
c.  
d.  
e. 
The function $f$ gives Stefan's displacement, in miles, to the North of his house $x$ hours since he started walking. The function $r_f$ defined as $r_f(x)=\sin(5x)\dfrac{x}{2.5}$ is the exact rate of change function for $f$ with respect to $x$.
For each of a.  d., explain the meaning in this situation of the given statement. Your answer should explain both the sign and the magnitude of the expression's value if one is given.Explain the general meaning of $r_f(x)$ in regard to Stefan's displacement.
Explain the meaning of $r_f(0.4)=0.749$ in regard to Stefan's displacement.
Explain the meaning of $r_f(1.8)=0.251$ in regard to Stefan's displacement.
Explain the meaning of $f(0.4)=0.251$ in regard to Stefan's displacement.
Label the vertical axis in the right coordinate system
with an appropriate unit.
The left graph shows regions colored in green and and regions colored in gray. What is the meaning (including units) of the signed areas of these regions in regard to Stefan's displacement?
The left graph shows a region colored in black from x=2.2 to x=2.5. What is the meaning (including units) of the signed area of this region in regard to Stefan's displacement?
Use GC to find the signed area of the black region. What does this number mean in regard to Stefan's displacement?
Represent your answer to part (e) by drawing something on the graph of $y=f(x)$. What you draw should reflect both sign and magnitude. Explain why you drew it as you did.
What does $f(x)g(x)$ represent in regard to this situation?
What does $g(x)f(x)$ represent in regard to this situation?
$f(x)=\mathrm{sin}(3x)$ and $g(x)=0.5\mathrm{cos}(3.5x)$. Use GC to calculate $\int_{2}^{1} (f(x)g(x))dx$.
What is the meaning (including sign and magnitude) of your answer to part (c) in regard to this situation?
Use GC to calculate $\int_{2}^{1} (g(x)f(x))dx$.
What is the meaning (including sign and magnitude) of your answer to part (e) in regard to this situation?
Download this file.
Click the "n" slider. Now determine which regions have positive area and which regions have negative area.
Print 2 copies of the graph. On the first printout, highlight regions that have positive area in green (or with \\\\\). In the second printout, highlight regions that have negative area in gray (or with //////).
Define the functions $f$ and $A_1$ as in Figure 8.3.24. Enter $A_1(0,2\pi)$. You will get 28.274 (to 3 decimal places). This is the unsigned area of a swept regionperhaps with parts swept more than once.
How does the definition of $A_2$, the function that gives net signed area, fix this problem?
In GC:
Enter a command that will compute the unsigned area of the region swept from $t=0$ to $t=\pi/2$. GC should report 1.4697 (rounded to 4 decimal places).
Explain why the numbers you obtained in Parts (a) and (b) are different.
Enter an expression involving only $A_1$ that produces the same result that you got in Part (b).