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6
APPLICATIONS OF
THE INTEGRAL
6.1 Area Between Two Curves
Preliminary Questions
1.
What is the area interpretation of
Z
b
a
(
f
(
x
)
−
g
(
x
)
)
dx
if
f
(
x
)
≥
g
(
x
)
?
SOLUTION
Because
f
(
x
)
≥
g
(
x
)
,
R
b
a
(
f
(
x
)
−
g
(
x
))
represents the area of the region bounded between the graphs
of
y
=
f
(
x
)
and
y
=
g
(
x
)
, bounded on the left by the vertical line
x
=
a
and on the right by the vertical line
x
=
b
.
2.
Is
Z
b
a
(
f
(
x
)
−
g
(
x
)
)
still equal to the area between the graphs of
f
and
g
if
f
(
x
)
≥
0but
g
(
x
)
≤
0?
Yes. Since
f
(
x
)
≥
0and
g
(
x
)
≤
0, it follows that
f
(
x
)
−
g
(
x
)
≥
0.
3.
Suppose that
f
(
x
)
≥
g
(
x
)
on
[
0
,
3
]
and
g
(
x
)
≥
f
(
x
)
on
[
3
,
5
]
. Express the area between the graphs over
[
0
,
5
]
as a
sum of integrals.
Remember that to calculate an area between two curves, one must subtract the equation for the lower curve
from the equation for the upper curve. Over the interval
[
0
,
3
]
,
y
=
f
(
x
)
is the upper curve. On the other hand, over the
interval
[
3
,
5
]
,
y
=
g
(
x
)
is the upper curve. The area between the graphs over the interval
[
0
,
5
]
is therefore given by
Z
3
0
(
f
(
x
)
−
g
(
x
))
+
Z
5
3
(
g
(
x
)
−
f
(
x
))
.
4.
Suppose that the graph of
x
=
f
(
y
)
lies to the left of the
y
axis. Is
Z
b
a
f
(
y
)
dy
positive or negative?
If the graph of
x
=
f
(
y
)
lies to the left of the
y
axis, then for each value of
y
, the corresponding value of
x
is less than zero. Hence, the value of
R
b
a
f
(
y
)
is negative.
Exercises
1.
Find the area of the region between
y
=
3
x
2
+
12 and
y
=
4
x
+
4 over
[−
3
,
3
]
(Figure 8).
50
25
y
x
y
=
3
x
2
+
12
y
=
4
x
+
4
3
−
1
−
2
−
31
2
FIGURE 8
As the graph of
y
=
3
x
2
+
12 lies above the graph of
y
=
4
x
+
4 over the interval
[−
3
,
3
]
, the area
between the graphs is
Z
3
−
3
³
(
3
x
2
+
12
)
−
(
4
x
+
4
)
´
=
Z
3
−
3
(
3
x
2
−
4
x
+
8
)
=
³
x
3
−
2
x
2
+
8
x
´ ¯
¯
¯
3
−
3
=
102
.
2.
Compute the area of the region in Figure 9(A), which lies between
y
=
2
−
x
2
and
y
=−
2 over
[−
2
,
2
]
.
y
x
2
−
2
−
2
−
2
−
2
y
x
1
y
=
2
−
x
2
y
=
−
2
y
=
2
−
x
2
y
=
x
(A)
(B)
FIGURE 9
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CHAPTER 6
APPLICATIONS OF THE INTEGRAL
SOLUTION
As the graph of
y
=
2
−
x
2
lies above the graph of
y
=−
2 over the interval
[−
2
,
2
]
, the area between the
graphs is
Z
2
−
2
³
(
2
−
x
2
)
−
(
−
2
)
´
dx
=
Z
2
−
2
(
4
−
x
2
)
=
µ
4
x
−
1
3
x
3
±¯
¯
¯
¯
2
−
2
=
32
3
.
3.
Let
f
(
x
)
=
x
and
g
(
x
)
=
2
−
x
2
[Figure 9(B)].
(a)
Find the points of intersection of the graphs.
(b)
Find the area enclosed by the graphs of
f
and
g
.
(a)
Setting
f
(
x
)
=
g
(
x
)
yields 2
−
x
2
=
x
, which simpli±es to
0
=
x
2
+
x
−
2
=
(
x
+
2
)(
x
−
1
).
Thus, the graphs of
y
=
f
(
x
)
and
y
=
g
(
x
)
intersect at
x
2and
x
=
1.
(b)
As the graph of
y
=
x
lies below the graph of
y
=
2
−
x
2
over the interval
[−
2
,
1
]
, the area between the graphs is
Z
1
−
2
³
(
2
−
x
2
)
−
x
´
=
µ
2
x
−
1
3
x
3
−
1
2
x
2
±¯
¯
¯
¯
1
−
2
=
9
2
.
4.
Let
f
(
x
)
=
8
x
−
10 and
g
(
x
)
=
x
2
−
4
x
+
10.
(a)
Find the points of intersection of the graphs.
(b)
Compute the area of the region
below
the graph of
f
and
above
the graph of
g
.
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 Fall '09

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