Calculus Of One Real Variable – By Pheng Kim Ving Chapter 12: Applications Of The Integral – Section 12.2: Areas Of Plane Regions 12.2 Areas Of Plane Regions

 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”. #  #  #  Example 1.1 Solution # Fig. 1.4

Area For Example 1.1. EOS

 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”. #  #  #  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. # Total area A is infinite sum or definite integral of area elements dA = ( f(x) – g(x)) dx: ## 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. # 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 = ( yupperylower) 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 # Note: Area element is enlarged and so isn't exactly a “rectangle”, but it's easiest to draw and more suggestive of the fact that it's a part of total area. So we can call it “strip” instead of “rectangle”. EOS

Example 2.2

Find the area between the graphs of y = 2x and y = x3.

Solution # Area For Example 2.2. EOS

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

Calculate the area between y = (ex + 1)/ex and y = 1 and to the right of x = 0.

### Solution # Fig. 3.2

Area For Example 3.1. Then: EOS

 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 # Integration Along The y-Axis: 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 = ( xrightxleft) 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 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,
y2y – 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.

 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.

 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 = (– y2y + 2) dy. So: 