Calculus Of
One Real Variable By Pheng Kim Ving 
12.2 
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1. Areas Between A Curve And The xAxis 
A plane is a 2dimensional space. A plane region is, well, a
region on a plane, as opposed to, for example, a region in a
3dimensional space. We'll calculate the area A of a plane region bounded by the curve that's
the graph of a function f
continuous on [a, b] where a < b,
the xaxis, and
the vertical lines x
= a and x = b. See Figs. 1.1 and 1.2. For the sake
of simplicity we'll take the freedom to refer to such an area as area between f and [a, b].
The area of the region
bounded by a curve that's the graph of a function f and the xaxis,
without the specification of the vertical lines or from
what xvalue to
what xvalue, is
the area of the region bounded by the curve, the xaxis, and the vertical lines at the
smallest and largest xintercepts
of the curve. For the sake of simplicity we'll take the freedom to refer to
such an area as
area between f
and the xaxis.
Fig. 1.1

Fig. 1.2

Fig. 1.3

Example 1.1
Solution
Fig. 1.4 Area For Example 1.1. 
EOS
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2. Areas Between 2 Curves 
Let's compute the area A
of the region bounded by 2 curves that are the graphs of the functions f and g and the vertical
lines x = a and x = b,
where a < b and f and g
are continuous on [a,
b]. See Figs. 2.1, 2.2, and
2.3. For the sake of
simplicity we'll take the liberty to refer to such an area as area between f and g on [a,
b]. The area of the region
bounded by 2 curves that are the graphs of the functions f and g,
without the specification of the vertical lines or from
what xvalue to
what xvalue, is
the area of the region bounded by the 2 curves and the vertical lines at their
leftmost
and rightmost points of intersection. For the sake of simplicity we'll take the
liberty to refer to such an area as area
between f and g.
Fig. 2.1

Fig. 2.2

Fig. 2.3

The area A between continuous functions y = f(x) and y = g(x) on [a, b] is:
Area Between A Curve And The xAxis
The area of a plane region bounded by the curve that's the
graph of a function f,
the xaxis, the
vertical line x = a, and
the vertical line x = b is the area of the region
between 2 curves that are the graphs of the functions y = f(x) and y = 0
and between the vertical lines x
= a and x = b. The area between a function and [a, b] is a special case of the area
between 2 functions on [a,
b].
The differential dx
is an increment in x.
The product ( f(x) g(x))
dx is the area of a thin
vertical rectangle whose
width is dx and
height is f(x) g(x).
Now that dA = ( f(x) g(x)) dx, we see that area A can be regarded as the
infinite sum or definite integral of the differential of area dA. For this reason the
differential of area dA
is called the area
element. The area A
can be regarded as the infinite sum or definite integral of the area elements dA.
Fig. 2.4
Total area A is infinite sum or definite integral of area elements

The area A
can be regarded as the infinite sum or definite integral of the differential of
area dA. For
this reason, the
differential of area dA
is called the area element. The area A
between 2 curves on [a,
b] can be regarded as the
infinite sum or definite integral of the area element dA from x
= a to x = b.
Fig. 2.5
Total area A is infinite sum or definite integral of area

Note that the height of the thin vertical rectangle over any
subinterval is equal to the height y
of the upper curve minus
that of the lower curve. In abbreviation, the area element over any
subinterval is:
dA = ( y_{upper} y_{lower}) dx,
where both y_{upper} and y_{lower} are expressed in terms of x.
Example 2.1
Compute the area between the graphs of y = 2x and y = x^{2}.
Solution
Fig. 2.6
Area For Example 2.1. Area Element:
Colored Gray.
Note: Area element is enlarged and so
isn't exactly a rectangle , but it's easiest to

EOS
Example 2.2
Find the area between the graphs of y = 2x and y = x^{3}.
Solution
Fig. 2.7
Area For Example 2.2.

EOS
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3. Areas Of Unbounded Regions 
Fig. 3.1
Area of unbounded region is limit of
area of bounded region:

Calculate the area between y = (e^{x} + 1)/e^{x} and y = 1 and to the right of x = 0.
Fig. 3.2 Area For Example 3.1. 
Then:
EOS
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4. Integration Along The yAxis 
The equation y
= ln x
+ 3x defines a
function y of x, whose domain is a set of xvalues and so is on the xaxis. The
equation x = y^{2} 12
defines a function x
of y, whose
domain is a set of yvalues
and so is on the yaxis.
Suppose x
= f( y) and x = g(
y) are 2 functions x of y continuous on the yinterval [c,
d] where c < d.
See Fig. 4.1. To
find the area A of
the plane region lying between the curves x
= f( y) and x = g(
y), above the horizontal line y = c, and
Fig. 4.1
Integration Along The yAxis:

below the horizontal line y
= d, we can
integrate along the yaxis
over [c, d]. The area element dA is the area of a thin
horizontal rectangle stretching
from the left curve to the right curve. The rectangle's width is dy. Thus its length must be
expressed in terms of y
too. For each y in
[c, d], this length is  f( y)
g( y). Consequently dA =  f( y) g(
y) dy.
Then:
Remark that the length of the thin horizontal rectangle over
any ysubinterval
is equal to the xposition
of the right curve
minus that of the left curve. Hence, in abbreviation, the area element over any
ysubinterval is:
dA = ( x_{right} x_{left}) dy,
where both x_{right} and x_{left} are expressed in terms of y.
Example 4.1
Compute the area of the region which lies to the right of
the parabola x = y^{2} 12 and to the left of the
line y = x using
integration along the yaxis.
Solution
Fig. 4.2 Area For Example 4.1. 
The graphs of x
= y^{2} 12 and y = x are sketched in Fig. 4.2. We'll integrate along
the yaxis. As y = x we have x
= y.
We solve the equations x
= y^{2} 12 and x = y simultaneously for y to find the ycoordinates
of the points of intersection
of the graphs:
y^{2} 12 = y,
y^{2} y 12 = 0,
( y + 3)( y 4) = 0,
y = 3 or 4.
Let A be the area of interest. The area element is dA = ( y ( y^{2} 12)) dy = ( y^{2} + y + 12) dy. So:
EOS
When using integration along the yaxis, in calculations both functions must be in
the form x = f( y). However in the
graphs the label of a graph should be in both forms y = f(x) and x = f^{ }^{1}(
y) (recall: f^{ }^{1} = inverse of f ) if its original
equation as given in the statement of the problem is in the form y = f(x):
the form y = f(x) is the given original, and the
form x = f^{ }^{1}(
y) is helpful in guiding the
calculations. Recall that when using integration along the xaxis, both functions
must be in the form y
= f(x).
Along The xAxis Or the yAxis?
The area in the above example can also be found by
integrating along the xaxis.
Let's see what happens if we attempt to
do so:
If we evaluate these component integrals, we will get the
same result as found in the example. Now, what happens is
that attempting to integrate along the xaxis
leads to these more complicated component integrals. For regions bounded
by graphs of functions x
of y, ie
functions of the form x
= f( y), it's generally easier to
integrate along the yaxis
than
along the xaxis.
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5. Steps In Finding The Area Of A Plane Region 
Here's a suggestive set of steps in finding the area of a
plane region. In each of the following 4 steps, the symbol [  ] is
used as follows: [AB] means that A is the case for the integration
along the xaxis and B
is the case for the integration
along the yaxis.
i. Sketch the curves. Find the [x y]coordinates of their points of
intersection. Know where the region is. Notice the [x
y]subintervals where the curves change their relative positions
if any.
ii. Draw a thin [vertical  horizontal] rectangle or strip of width [dx dy]. It's enough to do this over just 1 subinterval.
iii. For each subinterval, explicitly write down the
area element, dA =
[(height) dx(length) dy]. Make sure that
[height  length] is in terms of [x y] (in accordance with [dx dy]). This step can be
skipped.
iv. Integrate the expression for the area element
along the [x y]axis. If
necessary, split the integral into a sum of
integrals on the subintervals.
Integration along the xaxis
is also named integration in the xdirection
or using dxincrement. Integration along
the
yaxis is also named integration in
the ydirection or using
dyincrement.
Problems & Solutions 
Solution
2. Compute the area between y = x^{2} and y = 2x + 3.
Solution
Solution
4. Calculate the area between y = e^{x}/(e^{x} + 1) and y = 1 and to the right of x = 0.
Let u = 1 + e^{}^{x}.
Then du = e^{}^{x}
dx. Thus:
5. Compute the area of the plane region bounded by the curve x = y^{2} 2 and the
line y = x using integration along the
yaxis.
Solution
We'll integrate along the yaxis. As y
= x we have x = y. Solving the equations x = y^{2} 2 and x = y
simultaneously for y
to find the ycoordinates
of the points of intersection of their graphs we get:
y^{2} 2 = y,
y^{2} + y 2 = 0,
( y 1)( y + 2) = 0,
y = 1 or 2.
Let A be the area of interest. The area element is dA = ( y ( y^{2} 2)) dy = ( y^{2} y + 2) dy. So:
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