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Calculus Of
One Real Variable By Pheng Kim Ving |
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12.2 |
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1. Areas Between A Curve And The x-Axis |
A plane is a 2-dimensional space. A plane region is, well, a
region on a plane, as opposed to, for example, a region in a
3-dimensional 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 x-axis, 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 x-axis,
without the specification of the vertical lines or from
what x-value to
what x-value, is
the area of the region bounded by the curve, the x-axis, and the vertical lines at the
smallest and largest x-intercepts
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 x-axis.
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Fig. 1.1
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Fig. 1.2
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Fig. 1.3
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Example 1.1
Solution
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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 x-value to
what x-value, 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.
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Fig. 2.1
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Fig. 2.2
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Fig. 2.3
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The area A between continuous functions y = f(x) and y = g(x) on [a, b] is:
Area Between A Curve And The x-Axis
The area of a plane region bounded by the curve that's the
graph of a function f,
the x-axis, 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].
Differential Of Area Area Element
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.
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Fig. 2.4
Total area A is infinite sum or definite integral of area elements
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Area As Infinite Sum Or Definite Integral Of Area Element
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.
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Fig. 2.5
Total area A is infinite sum or definite integral of area
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Note that the height of the thin vertical rectangle over any
sub-interval is equal to the height y
of the upper curve minus
that of the lower curve. In abbreviation, the area element over any
sub-interval is:
dA = ( yupper ylower) dx,
where both yupper and ylower are expressed in terms of x.
Example 2.1
Compute the area between the graphs of y = 2x and y = x2.
Solution
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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
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EOS
Example 2.2
Find the area between the graphs of y = 2x and y = x3.
Solution
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Fig. 2.7
Area For Example 2.2.
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EOS
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3. Areas Of Unbounded Regions |
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Fig. 3.1
Area of unbounded region is limit of
area of bounded region:
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Example 3.1
Calculate the area between y = (ex + 1)/ex and y = 1 and to the right of x = 0.
Solution
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Fig. 3.2 Area For Example 3.1. |
Then:
EOS
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4. Integration Along The y-Axis |
The equation y
= ln x
+ 3x defines a
function y of x, whose domain is a set of x-values and so is on the x-axis. The
equation x = y2 12
defines a function x
of y, whose
domain is a set of y-values
and so is on the y-axis.
Suppose x
= f( y) and x = g(
y) are 2 functions x of y continuous on the y-interval [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
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Fig. 4.1
Integration Along The y-Axis:
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below the horizontal line y
= d, we can
integrate along the y-axis
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 y-sub-interval
is equal to the x-position
of the right curve
minus that of the left curve. Hence, in abbreviation, the area element over any
y-sub-interval is:
dA = ( xright xleft) dy,
where both xright and xleft are expressed in terms of y.
Example 4.1
Compute the area of the region which lies to the right of
the parabola x = y2 12 and to the left of the
line y = x using
integration along the y-axis.
Solution
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Fig. 4.2 Area For Example 4.1. |
The graphs of x
= y2 12 and y = x are sketched in Fig. 4.2. We'll integrate along
the y-axis. As y = x we have x
= y.
We solve the equations x
= y2 12 and x = y simultaneously for y to find the y-coordinates
of the points of intersection
of the graphs:
y2 12 = y,
y2 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 ( y2 12)) dy = ( y2 + y + 12) dy. So:
EOS
When using integration along the y-axis, 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 x-axis, both functions
must be in the form y
= f(x).
Along The x-Axis Or the y-Axis?
The area in the above example can also be found by
integrating along the x-axis.
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 x-axis
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 y-axis
than
along the x-axis.
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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: [A|B] means that A is the case for the integration
along the x-axis and B
is the case for the integration
along the y-axis.
i. Sketch the curves. Find the [x| y]-coordinates of their points of
intersection. Know where the region is. Notice the [x|
y]-sub-intervals 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 sub-interval.
iii. For each sub-interval, 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 sub-intervals.
Other Names
Integration along the x-axis
is also named integration in the x-direction
or using dx-increment. Integration along
the
y-axis is also named integration in
the y-direction or using
dy-increment.
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Problems & Solutions |
Solution
2. Compute the area between y = x2 and y = 2x + 3.
Solution
Solution
4. Calculate the area between y = ex/(ex + 1) and y = 1 and to the right of x = 0.
Solution
Let u = 1 + ex.
Then du = ex
dx. Thus:
5. Compute the area of the plane region bounded by the curve x = y2 2 and the
line y = x using integration along the
y-axis.
Solution
We'll integrate along the y-axis. As y
= x we have x = y. Solving the equations x = y2 2 and x = y
simultaneously for y
to find the y-coordinates
of the points of intersection of their graphs we get:
y2 2 = y,
y2 + 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 ( y2 2)) dy = ( y2 y + 2) dy. So:
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