6.4 Areas in the xy-Plane
Integral properties we will be using
This section shows how to use the
definite integral to compute the area of a
region that lies between the graphs of two
or more functions.
Consider a region bounded both above
and below by graphs of functions.
[area of bounded region] =
We want to find a simple expression for
the area bounded above by y = f(x) and
below by y = g(x) from x = a to x = b.
[area under f(x)] – [area under g(x)]
Note that this is the region under f(x) with
the region under g(x) taken away.
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Area Between Two Curves
If y = f(x) lies above y = g(x) from x = a to
x = b, then the area of the region between
f(x) and g(x) from x = a to x = b is
Find the area of the region between
y = 2x2 – 4x + 6 and y = -x2 + 2x + 1
from x = 1 to x = 2.
Our formula for the area is then
If we sketch the
two graphs, we
can discover the
region of interest.
We note that
2x2 – 4x + 6
lies above
–x2 + 2x + 1
from
x = 1 to x = 2.
The solution to such problems is not always
quite so straightforward.
We first sketch the graphs to discover the region of interest.
Find the area of the region between
y = x2 and y = x2 – 4x + 4
from
Notice that the two
graphs cross.
x = 0 to x = 3.
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By setting x2 = x2 – 4x + 4, we find that
the graphs cross at x = 1.
We observe that y = x2 – 4x + 4 is on top
from x = 0 to x = 1, and y = x2 is on top
from x = 1 to x = 3.
We cannot directly apply our rule to
calculate the area.
How do we find our answer?
For x = 0 to x = 1, y = x2 – 4x + 4 is on top
When we have a situation where graphs
cross, we determine the area of a desired
region by breaking the region into
separate parts.
For our situation, we will divide our
problem into two parts.
We will find the area from x = 0 to x = 1,
and then find the area from x = 1 to x = 3.
For x = 1 to x = 3, y = x2 is on top
So, the total area we want is the sum of
the two areas:
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So far, the cases we have examined used
nonnegative functions.
What happens when we consider
functions that are nonnegative?
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Suppose we have two graphs of
functions f(x) and g(x) that are not
always positive.
Select a constant c so that the graphs of f(x) + c
and g(x) + c lie completely above the x –axis from
x = a to x = b.
We want to find the area between f(x)
and g(x) from x = a to x = b.
Note that the region between
f(x) + c and g(x) + c has the
same area as the original
region.
If we use our rule as applied to nonnegative
functions, we have
Set up the integral that gives the area
between the curves
y = x2 – 2x and y = -ex
from x = -1 to x = 2.
We see that our rule is valid for any
functions f(x) and g(x) as long as the graph
of f(x) lies above g(x) from x = a to x = b.
Sketching the
graphs, we
discover that
y = x2 – 2x lies
above –ex for the
entire interval.
Our rule for determining the area
between two curves can be applied
directly.
The area is given by
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Set up the integral that gives the area
bounded by the two curves
Sketching the
graphs, we will
note that they
cross.
y = 2x2 and y = x3 – 3x.
We find the points where the two graphs cross
by setting x3 – 3x = 2x2 and solving for x.
The area from x = -1 to x = 0 is given by
We get the equation x3 – 2x2 – 3x = 0
If we factor the left side, we get
x(x – 3)(x + 1) = 0
So, the solutions are x = -1, x = 0, and x = 3.
These are the points at which the graphs
cross.
Recall that from x = 0 to x = 3, y = 2x2 lies
above y = x3 – 3x.
The area from x = 0 to x = 3 is given by
We note that the from x = -1 to x = 0,
y = x3 – 3x lies above y = 2x2
So, the area between the curves on the
interval is
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