6.4 Areas in the xy-Plane
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.
Integral properties we will be using
Consider a region bounded both above
and below by graphs of functions.
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.
Note that this is the region under f(x) with
the region under g(x) taken away.
[area of bounded region] =
[area under f(x)] – [area under g(x)]
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.
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.
Our formula for the area is then
The solution to such problems is not always
quite so straightforward.
Find the area of the region between
y = x2 and y = x2 – 4x + 4
from
x = 0 to x = 3.
We first sketch the graphs to discover the region of interest.
Notice that the two
graphs cross.
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?
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 = 0 to x = 1, y = x2 – 4x + 4 is on top
For x = 1 to x = 3, y = x2 is on top
So, the total area we want is the sum of
the two areas:
2
+8
----10
So far, the cases we have examined used
nonnegative functions.
What happens when we consider
functions that are nonnegative?
Suppose we have two graphs of
functions f(x) and g(x) that are not
always positive.
We want to find the area between f(x)
and g(x) from x = a to x = b.
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.
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
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.
Set up the integral that gives the area
between the curves
y = x2 – 2x and y = -ex
from x = -1 to x = 2.
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
Set up the integral that gives the area
bounded by the two curves
y = 2x2 and y = x3 – 3x.
Sketching the
graphs, we will
note that they
cross.
We find the points where the two graphs cross
by setting x3 – 3x = 2x2 and solving for x.
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.
We note that the from x = -1 to x = 0,
y = x3 – 3x lies above y = 2x2
The area from x = -1 to x = 0 is given by
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
So, the area between the curves on the
interval is