Math 220
Calculus I
Section 6.4 Areas in the x-y Plane
Example F
6.4 Example A-1
Find the area of the geometric figure pictured below.
2
3
6.4 Example A-1
Find the area of the geometric figure pictured below.
The figure consists of a rectangle and a semicircle.
2
3
6.4 Example A-1
Find the area of the geometric figure pictured below.
The figure consists of a rectangle and a semicircle. (Note that the width of the rectangle, 2,
is the same as the radius of the semi-circle.)
2
3
6.4 Example A-1
Find the area of the geometric figure pictured below.
The figure consists of a rectangle and a semicircle. (Note that the width of the rectangle, 2,
is the same as the radius of the semi-circle.)
The total area is the sum of the two areas.
2
3
6.4 Example A-1
Find the area of the geometric figure pictured below.
The figure consists of a rectangle and a semicircle. (Note that the width of the rectangle, 2,
is the same as the radius of the semi-circle.)
The total area is the sum of the two areas.
 
1
A  23   2 2
2
2
3
6.4 Example A-1
Find the area of the geometric figure pictured below.
The figure consists of a rectangle and a semicircle. (Note that the width of the rectangle, 2,
is the same as the radius of the semi-circle.)
The total area is the sum of the two areas.
1
A  23   2 2 
2
A  6  2

2
3
6.4 Theory
In Lectures 6.2 and 6.3 we were able to equate the area
under a curve on an interval a ≤ x ≤ b, with the definite
integral from a to b: b
 f x  dx
a
Combining this concept with geometric ideas such as those
used in Example A supports the assertion of some of the
properties of integrals already used in Lecture 6.1.
6.4 Theory
In Lectures 6.2 and 6.3 we were able to equate the area
under a curve on an interval a ≤ x ≤ b, with the definite
integral from a to b: b
 f x  dx
a
Combining this concept with geometric ideas such as those
used in Example A supports the assertion of some of the
properties of integrals already used in Lecture 6.1.
 a  f x   g x  dx   a f x 
b
b
dx   g x  dx
b
a
6.4 Theory
In Lectures 6.2 and 6.3 we were able to equate the area
under a curve on an interval a ≤ x ≤ b, with the definite
integral from a to b: b
 f x  dx
a
Combining this concept with geometric ideas such as those
used in Example A supports the assertion of some of the
properties of integrals already used in Lecture 6.1.
 a  f x   g x  dx   a f x 
dx   g x  dx
 a  f x   g x  dx   a f x 
dx   g x  dx
b
b
b
b
b
a
b
a
6.4 Theory
In Lectures 6.2 and 6.3 we were able to equate the area
under a curve on an interval a ≤ x ≤ b, with the definite
integral from a to b: b
 f x  dx
a
Combining this concept with geometric ideas such as those
used in Example A supports the assertion of some of the
properties of integrals already used in Lecture 6.1.
 a  f x   g x  dx   a f x 
dx   g x  dx
 a  f x   g x  dx   a f x 
dx   g x  dx
b
b
b
b
b
a
 a k  f x  dx  k   a f x 
b
b
b
a
dx
6.4 Example B
1
Find the area between the curves f(x) = x + 1 and g x   on
x
the interval 1 ≤ x ≤ 3.
6.4 Example B
1
Find the area between the curves f(x) = x + 1 and g x   on
x
the interval 1 ≤ x ≤ 3.
As long as f(x) > g(x) for all values of
x in the interval, we can use the
second property above and subtract
the areas.
6.4 Example B
1
Find the area between the curves f(x) = x + 1 and g x   on
x
the interval 1 ≤ x ≤ 3.
As long as f(x) > g(x) for all values of
x in the interval, we can use the
second property above and subtract
the areas.
3
1
1 x  1  x dx
“above”
“below”
6.4 Example B
1
Find the area between the curves f(x) = x + 1 and g x   on
x
the interval 1 ≤ x ≤ 3.
As long as f(x) > g(x) for all values of
x in the interval, we can use the
second property above and subtract
the areas.
3
1
1 2
3
1 x  1  x dx  2 x  x  ln x 1
6.4 Example B
1
Find the area between the curves f(x) = x + 1 and g x   on
x
the interval 1 ≤ x ≤ 3.
As long as f(x) > g(x) for all values of
x in the interval, we can use the
second property above and subtract
the areas.
3
1
1 2
3


x

1

dx

x

x

ln
x
1
1
x
2
1 2
 1 2

  3  3  ln 3   1  1  ln 1
2
 2

6.4 Example B
1
Find the area between the curves f(x) = x + 1 and g x   on
x
the interval 1 ≤ x ≤ 3.
As long as f(x) > g(x) for all values of
x in the interval, we can use the
second property above and subtract
the areas.
3
1
1 2
3
1 x  1  x dx  2 x  x  ln x 1
1 2
 1 2

  3  3  ln 3   1  1  ln 1
2
 2

9
1

 3  ln 3 
1  0
2
2
6.4 Example B
1
Find the area between the curves f(x) = x + 1 and g x   on
x
the interval 1 ≤ x ≤ 3.
As long as f(x) > g(x) for all values of
x in the interval, we can use the
second property above and subtract
the areas.
3
3
1
1 2
1 x  1  x dx  2 x  x  ln x 1
1 2
 1 2

  3  3  ln 3   1  1  ln 1
2
 2

9
1

 3  ln 3 
1  0
2
2
 6  ln 3
6.4 Example C
1
Find the area between the curves f(x) = x – 1 and g  x    2
x
on the interval 1 ≤ x ≤ 3.
6.4 Example C
1
Find the area between the curves f(x) = x – 1 and g  x    2
x
on the interval 1 ≤ x ≤ 3.
Looking at the graph, the shape
remains the same because the
curves from Example B were simply
shifted down by 2. The area should
be the same.
6.4 Example C
1
Find the area between the curves f(x) = x – 1 and g  x    2
x
on the interval 1 ≤ x ≤ 3.
Looking at the graph, the shape
remains the same because the
curves from Example B were simply
shifted down by 2. The area should
be the same. However, in this case
some of the area lies below the xaxis, and would thus be a negative
area (as noted in Lecture 6.3).
6.4 Example C
1
Find the area between the curves f(x) = x – 1 and g  x    2
x
on the interval 1 ≤ x ≤ 3.
Looking at the graph, the shape
remains the same because the
curves from Example B were simply
shifted down by 2. The area should
be the same. However, in this case
some of the area lies below the xaxis, and would thus be a negative
area (as noted in Lecture 6.3).
Does this affect the calculation of area between the two
curves?
6.4 Example C
1
Find the area between the curves f(x) = x – 1 and g  x    2
x
on the interval 1 ≤ x ≤ 3.
1



x

1


2

 dx
1
x

3
“above”
“below”
6.4 Example C
1
Find the area between the curves f(x) = x – 1 and g  x    2
x
on the interval 1 ≤ x ≤ 3.
1 2
3
1

1 x  1   x  2  dx  2 x  x  ln x  2 x 1
3
6.4 Example C
1
Find the area between the curves f(x) = x – 1 and g  x    2
x
on the interval 1 ≤ x ≤ 3.
1 2
3
1



x

1


2
dx

x

x

ln
x

2
x


1
1
2
x

1 2
3
 x  x  ln x
1
2
3
6.4 Example C
1
Find the area between the curves f(x) = x – 1 and g  x    2
x
on the interval 1 ≤ x ≤ 3.
1 2
3
1

1 x  1   x  2  dx  2 x  x  ln x  2 x 1
1 2
3
 x  x  ln x
1
2
3
1 2
 1 2





  3  3  ln 3   1  1  ln 1
2
 2

6.4 Example C
1
Find the area between the curves f(x) = x – 1 and g  x    2
x
on the interval 1 ≤ x ≤ 3.
1 2
3
1



x

1


2
dx

x

x

ln
x

2
x


1
1
2
x

3
1 2
3
 x  x  ln x
1
2
1 2
 1 2

  3  3  ln 3   1  1  ln 1
2
 2

9
1
  3  ln 3   1  0
2
2
6.4 Example C
1
Find the area between the curves f(x) = x – 1 and g  x    2
x
on the interval 1 ≤ x ≤ 3.
1 2
3
1



x

1


2
dx

x

x

ln
x

2
x


1
1
2
x

3
1 2
3
 x  x  ln x
1
2
1 2
 1 2

  3  3  ln 3   1  1  ln 1
2
 2

9
1
  3  ln 3   1  0  6  ln 3
2
2
6.4 Example C
1
Find the area between the curves f(x) = x – 1 and g  x    2
x
on the interval 1 ≤ x ≤ 3.
So the integral calculation gives us
the same result as in Example B:
Area = 6 – ln3.
It doesn’t matter whether the two
curves lie above or below the x-axis,
only that we subtract the
“higher/above” curve minus the
“lower/below”.
6.4 Example D
Find the area between the curves y = x2 and y = x2 – 4x + 12
on the interval 0 ≤ x ≤ 4.
15
0
0
5
6.4 Example D
Find the area between the curves y = x2 and y = x2 – 4x + 12
on the interval 0 ≤ x ≤ 4.
15
Looking at the graph, y = x2 – 4x + 12
lies above y = x2 for a portion of the
interval, but is below for the rest of the
interval.
0
0
5
6.4 Example D
Find the area between the curves y = x2 and y = x2 – 4x + 12
on the interval 0 ≤ x ≤ 4.
15
Looking at the graph, y = x2 – 4x + 12
lies above y = x2 for a portion of the
interval, but is below for the rest of the
interval.
We need to determine where the two
intersect, then set up two integrals:
one for each portion.
0
0
5
6.4 Example D
Find the area between the curves y = x2 and y = x2 – 4x + 12
on the interval 0 ≤ x ≤ 4.
15
We need to determine where the two
intersect, then set up two integrals:
one for each portion.
x 2  x 2  4 x  12
0
0
5
6.4 Example D
Find the area between the curves y = x2 and y = x2 – 4x + 12
on the interval 0 ≤ x ≤ 4.
15
We need to determine where the two
intersect, then set up two integrals:
one for each portion.
x 2  x 2  4 x  12
4 x  12
0
0
5
6.4 Example D
Find the area between the curves y = x2 and y = x2 – 4x + 12
on the interval 0 ≤ x ≤ 4.
15
We need to determine where the two
intersect, then set up two integrals:
one for each portion.
x 2  x 2  4 x  12
4 x  12
x3
x=3
0
0
5
6.4 Example D
Find the area between the curves y = x2 and y = x2 – 4x + 12
on the interval 0 ≤ x ≤ 4.

3
0
“above”

“below”

2
2
x

4
x
12

x

 dx

15

  3 x 2  x 2  4 x 12 dx
4
x=3
“above”
“below”
0
0
5
6.4 Example D
Find the area between the curves y = x2 and y = x2 – 4x + 12
on the interval 0 ≤ x ≤ 4.
 x
3
2
0

15

 4 x 12 x 2 dx

  3 x  x  4 x 12 dx
4
2

2
x=3
 4 x 12 dx   4 x 12 dx
3
4
0
3
0
0
5
6.4 Example D
Find the area between the curves y = x2 and y = x2 – 4x + 12
on the interval 0 ≤ x ≤ 4.

3
0

15

2
2
x

4
x
12

x
dx




  3 x  x  4 x 12 dx
4
2

2
x=3
 4 x 12 dx   4 x 12 dx
3
4
0
3
 2x 12x   2x 12x 
2
3
0
2
4
3
Combining like terms is not an option—the
two polynomials have different boundaries
to evaluate.
0
0
5
6.4 Example D
Find the area between the curves y = x2 and y = x2 – 4x + 12
on the interval 0 ≤ x ≤ 4.
15
 x
3
2
0

 4 x 12 x 2 dx

x=3

  3 x 2  x 2  4 x 12 dx
4

 4 x 12 dx   4 x 12 dx
3
4
0
3
0
0
5
 2x 2 12x 30  2x 2 12x 43





 232 123  20 2 120  24 2 124  232 123
6.4 Example D
Find the area between the curves y = x2 and y = x2 – 4x + 12
on the interval 0 ≤ x ≤ 4.
15
 x
3
2
0

 4 x 12 x 2 dx

x=3

  3 x 2  x 2  4 x 12 dx
4

 4 x 12 dx   4 x 12 dx
3
4
0
3
0
 2x 12x   2x 12x 
2

3
0
2
0
4
3


5


 232 123  20 2 120  24 2 124  232 123
= –18
+ 36
+ 0
–
0
+ 32 – 48 – 18
+ 36
6.4 Example D
Find the area between the curves y = x2 and y = x2 – 4x + 12
on the interval 0 ≤ x ≤ 4.
 x
3
2
0
15

 4 x 12 x 2 dx


  3 x 2  x 2  4 x 12 dx
4

x=3
 4 x 12 dx   4 x 12 dx
3
4
0
3
 2x 12x   2x 12x 
2
3
0

2
0
4
3
0


5


 232 123  20 2 120  24 2 124  232 123
= –18
= 20
+ 36
+ 0
–
0
+ 32 – 48 – 18
+ 36
6.4 Example E
Find the area between the curves y = 2x and y = x2 – 4x + 8.
10
0
0
5
6.4 Example E
Find the area between the curves y = 2x and y = x2 – 4x + 8.
In this case, although no interval is
specified, by looking at the graph we
can see that there are two points of
intersection.
10
0
0
5
6.4 Example E
Find the area between the curves y = 2x and y = x2 – 4x + 8.
In this case, although no interval is
specified, by looking at the graph we
can see that there are two points of
intersection.
Between these two points the area is
bounded, and lies between the two
curves.
10
0
0
5
6.4 Example E
Find the area between the curves y = 2x and y = x2 – 4x + 8.
In this case, although no interval is
specified, by looking at the graph we
can see that there are two points of
intersection.
Between these two points the area is
bounded, and lies between the two
curves.
We need to solve for the intersections
to find the boundaries (or limits) of
integration.
10
0
0
5
6.4 Example E
Find the area between the curves y = 2x and y = x2 – 4x + 8.
10
We need to solve for the intersections
to find the boundaries (or limits) of
integration.
2x  x2  4x  8
0
0
5
6.4 Example E
Find the area between the curves y = 2x and y = x2 – 4x + 8.
10
We need to solve for the intersections
to find the boundaries (or limits) of
integration.
2x  x2  4x  8
2
0  x  6x  8
0
0
5
6.4 Example E
Find the area between the curves y = 2x and y = x2 – 4x + 8.
10
We need to solve for the intersections
to find the boundaries (or limits) of
integration.
2x  x2  4x  8
2
0  x  6x  8
0  x  2x  4
0
0
5
6.4 Example E
Find the area between the curves y = 2x and y = x2 – 4x + 8.
10
We need to solve for the intersections
to find the boundaries (or limits) of
integration.
x=4
x=2
2x  x 2  4x  8
2
0  x  6x  8
0  x  2 x  4 
x  2, 4
0
0
5
6.4 Example E
Find the area between the curves y = 2x and y = x2 – 4x + 8.
10
x=4

4
2
2x  x
2

 4 x  8 dx
x=2
“above”
“below”
0
0
5
6.4 Example E
Find the area between the curves y = 2x and y = x2 – 4x + 8.
10
x=4
x=2
0

4
2


2x  x  4 x  8 dx 
2

0
4
2
2
x
  6x  8 dx
5
6.4 Example E
Find the area between the curves y = 2x and y = x2 – 4x + 8.
10

4
2
2x  x
2

 4 x  8 dx 
 x
4
2
2
 6x  8 dx
x=4
 1 3
4
2
  x  3x  8x 2
 3

x=2
0
0
5
6.4 Example E
Find the area between the curves y = 2x and y = x2 – 4x + 8.
10
x=4

4
2
2x  x
2

 4 x  8 dx 
 x
4
2
2
 6x  8 dx
0
 1 3
4
2
  x  3x  8x 2
0
 3

 1 3
  1 3

2
2
  4   34   84   2  32  82
 3
  3

x=2
5
6.4 Example E
Find the area between the curves y = 2x and y = x2 – 4x + 8.
10
x=4

4
2


2x  x  4 x  8 dx 
2

4
2
2
x
  6x  8 dx
x=2
 1 3
4
2
  x  3x  8x 2
0
 3

0
 1 3
  1 3

2
2
  4   34   84   2  32  82
 3
  3

64
8
 
 48  32

 12  16
3
3
5
6.4 Example E
Find the area between the curves y = 2x and y = x2 – 4x + 8.
10
x=4

4
2
2x  x
2

 4 x  8 dx 
 x
4
2
2
 6x  8 dx
 1 3
4
2
0
  x  3x  8x 2
 3

0
 1 3
  1 3

2
2
  4  34   84  2  32  82
 3
  3

64
8
 
 48  32

 12  16
3
3
4

3
x=2
5
6.4 Example F
Find the area bounded by the curves y = x2, y = 6x and
y = 8 – 2x on the interval 0 ≤ x ≤ 2.
10
0
0
5
6.4 Example F
Find the area bounded by the curves y = x2, y = 6x and
y = 8 – 2x on the interval 0 ≤ x ≤ 2.
10
How many integrals will be needed?
0
0
5
6.4 Example F
Find the area bounded by the curves y = x2, y = 6x and
y = 8 – 2x on the interval 0 ≤ x ≤ 2.
10
How many integrals will be needed?
Note that there are three points of
intersection.
0
0
5
6.4 Example F
Find the area bounded by the curves y = x2, y = 6x and
y = 8 – 2x on the interval 0 ≤ x ≤ 2.
10
How many integrals will be needed?
Note that there are three points of
intersection.
A well-placed vertical indicated that two
integrals will be sufficient—there are two
sub-intervals, each of which has its own
“function above” and “function below”.
0
0
5
6.4 Example F
Find the area bounded by the curves y = x2, y = 6x and
y = 8 – 2x on the interval 0 ≤ x ≤ 2.
10
How many integrals will be needed?
Note that there are three points of
intersection.
A well-placed vertical indicated that two
integrals will be sufficient—there are two
sub-intervals, each of which has its own
“function above” and “function below”.
Finding the intersections will give us the
correct boundaries of integration for
each integral.
0
0
5
6.4 Example F
Find the area bounded by the curves y = x2, y = 6x and
y = 8 – 2x on the interval 0 ≤ x ≤ 2.
10
Looking at the graph, one of the corners
of the bounded area is clearly (0, 0).
x=0
0
0
5
6.4 Example F
Find the area bounded by the curves y = x2, y = 6x and
y = 8 – 2x on the interval 0 ≤ x ≤ 2.
10
Looking at the graph, one of the corners
of the bounded area is clearly (0, 0).
We’ll need to solve for the other two.
x=0
0
0
5
6.4 Example F
Find the area bounded by the curves y = x2, y = 6x and
y = 8 – 2x on the interval 0 ≤ x ≤ 2.
10
The “top” corner is the intersection of
y = 6x and y = 8 – 2x.
x=0
0
0
5
6.4 Example F
Find the area bounded by the curves y = x2, y = 6x and
y = 8 – 2x on the interval 0 ≤ x ≤ 2.
10
The “top” corner is the intersection of
y = 6x and y = 8 – 2x.
6x  8  2x
x=0
0
0
5
6.4 Example F
Find the area bounded by the curves y = x2, y = 6x and
y = 8 – 2x on the interval 0 ≤ x ≤ 2.
10
The “top” corner is the intersection of
y = 6x and y = 8 – 2x.
6x  8  2x
8x  8
x=0
0
0
5
6.4 Example F
Find the area bounded by the curves y = x2, y = 6x and
y = 8 – 2x on the interval 0 ≤ x ≤ 2.
10
The “top” corner is the intersection of
y = 6x and y = 8 – 2x.
6x  8  2x
8x  8
x 1
x=1
x=0
0
0
5
6.4 Example F
Find the area bounded by the curves y = x2, y = 6x and
y = 8 – 2x on the interval 0 ≤ x ≤ 2.
10
The “right” corner is the intersection of
y = x2 and y = 8 – 2x.
x=1
x=0
0
0
5
6.4 Example F
Find the area bounded by the curves y = x2, y = 6x and
y = 8 – 2x on the interval 0 ≤ x ≤ 2.
10
The “right” corner is the intersection of
y = x2 and y = 8 – 2x.
x2  8  2x
x=1
x=0
0
0
5
6.4 Example F
Find the area bounded by the curves y = x2, y = 6x and
y = 8 – 2x on the interval 0 ≤ x ≤ 2.
10
The “right” corner is the intersection of
y = x2 and y = 8 – 2x.
x2  8  2x
2
x  2x  8  0
x=1
x=0
0
0
5
6.4 Example F
Find the area bounded by the curves y = x2, y = 6x and
y = 8 – 2x on the interval 0 ≤ x ≤ 2.
10
The “right” corner is the intersection of
y = x2 and y = 8 – 2x.
x2  8  2x
2
x  2x  8  0
x  4x  2  0
x=1
x=0
0
0
5
6.4 Example F
Find the area bounded by the curves y = x2, y = 6x and
y = 8 – 2x on the interval 0 ≤ x ≤ 2.
10
The “right” corner is the intersection of
y = x2 and y = 8 – 2x.
x 2  8  2x
2
x  2x  8  0
x  4x  2  0
x  4, 2
x=1
x=0
0
0
5
6.4 Example F
Find the area bounded by the curves y = x2, y = 6x and
y = 8 – 2x on the interval 0 ≤ x ≤ 2.
10
The “right” corner is the intersection of
y = x2 and y = 8 – 2x.
x 2  8  2x
2
x  2x  8  0
x  4x  2  0
x  4, 2
x=1
x=2
x=0
0
0
Since –4 is outside our interval of concern, we’ll only be using
x = 2.
5
6.4 Example F
Find the area bounded by the curves y = x2, y = 6x and
y = 8 – 2x on the interval 0 ≤ x ≤ 2.
10
We’ll need to be careful in setting up the
two needed integrals.
x=1
x=2
x=0
0
0
5
6.4 Example F
Find the area bounded by the curves y = x2, y = 6x and
y = 8 – 2x on the interval 0 ≤ x ≤ 2.
10
We’ll need to be careful in setting up the
two needed integrals.
For 0 ≤ x ≤ 1, y = 6x is “above” and y = x2
is “below”.
y = 6x
x=1
x=2
x=0
y = x2
0
0
5
6.4 Example F
Find the area bounded by the curves y = x2, y = 6x and
y = 8 – 2x on the interval 0 ≤ x ≤ 2.
10
We’ll need to be careful in setting up the
two needed integrals.
For 0 ≤ x ≤ 1, y = 6x is “above” and y = x2
is “below”.
For 1 ≤ x ≤ 2, y = 8 – 2x is “above” and
y = x2 is “below”.
y = 6x
x=1
x=2
y = 8 – 2x
x=0
y = x2
0
0
5
6.4 Example F
Find the area bounded by the curves y = x2, y = 6x and
y = 8 – 2x on the interval 0 ≤ x ≤ 2.
10
 6x  x  dx   8  2x  x  dx
1
2
0
2
2
y = 6x
1
“above”
“below”
x=1
“above”
x=2
y = 8 – 2x
“below”
x=0
y = x2
0
0
5
6.4 Example F
Find the area bounded by the curves y = x2, y = 6x and
y = 8 – 2x on the interval 0 ≤ x ≤ 2.
10
 6x  x  dx   8  2x  x  dx
1
0
2
2
2
y = 6x
1
x=1
x=2
y = 8 – 2x
x=0
y = x2
0
0
5
6.4 Example F
Find the area bounded by the curves y = x2, y = 6x and
y = 8 – 2x on the interval 0 ≤ x ≤ 2.
10
1
0
6x  x dx   8  2x  x dx
2
2
2
y = 6x
1
x=1
 2 1 3 1 
2 1 3 2
 3x  x   8 x  x  x 
3 0 
3 1

x=2
y = 8 – 2x
x=0
Combining like terms is not an option—the
two polynomials have different boundaries
to evaluate.
y = x2
0
0
5
6.4 Example F
Find the area bounded by the curves y = x2, y = 6x and
y = 8 – 2x on the interval 0 ≤ x ≤ 2.
10
 6x  x  dx   8  2x  x  dx
1
0
2
2
2
1
 2 1 3 1 
1 3 2
2
 3x  x   8x  x  x 

3 0 
3 1
y = 6x
x=1
x=2
y = 8 – 2x
x=0
 2 1 3   2 1 3 
y = x2
 31  1  30  0 
0

 

3
3
0

1 3  
1 3 
2
2
 82  2  2  81  1  1 

 

3
3
5
6.4 Example F
Find the area bounded by the curves y = x2, y = 6x and
y = 8 – 2x on the interval 0 ≤ x ≤ 2.
10
 6x  x  dx   8  2x  x  dx
1
0
2
2
2
1
 2 1 3 1 
1 3 2
2
 3x  x   8x  x  x 

3 0 
3 1
y = 6x
x=1
x=2
y = 8 – 2x
x=0
 2 1 3   2 1 3 
y = x2
 31  1  30  0 
0

 

3
3
0

1 3  
1 3 
2
2
 82  2  2  81  1  1 

 

3
3
1
8
1
 3   0  0 16  4   8 1
3
3
3
5
6.4 Example F
Find the area bounded by the curves y = x2, y = 6x and
y = 8 – 2x on the interval 0 ≤ x ≤ 2.
10
 6x  x  dx   8  2x  x  dx
1
0
2
2
2
1
 2 1 3 1 
1 3 2
2
 3x  x   8x  x  x 

3 0 
3 1
y = 6x
x=1
x=2
y = 8 – 2x
x=0
 2 1 3   2 1 3 
y = x2
 31  1  30  0 
0

 

3
3
0

1 3  
1 3 
2
2
 82  2  2  81  1  1 

 

3
3
1
8
1
 3   0  0 16  4   8 1
3
3
3
16

3
5
6.4 Final Notes
The key to all of these is looking at the graph to determine
which function is above the other, finding intersections as
needed, then setting up and evaluating the appropriate
integrals.
6.4 Final Notes
The key to all of these is looking at the graph to determine
which function is above the other, finding intersections as
needed, then setting up and evaluating the appropriate
integrals.
In the next section we’ll delve into applications which use all
of the integration concepts developed so far. If a function
represents a rate of change, we’ll use an integral (= area
under the curve) to determine an amount.
6.4 Final Notes
The key to all of these is looking at the graph to determine
which function is above the other, finding intersections as
needed, then setting up and evaluating the appropriate
integrals.
In the next section we’ll delve into applications which use all
of the integration concepts developed so far. If a function
represents a rate of change, we’ll use an integral (= area
under the curve) to determine an amount.
For example, from a velocity function, which is the rate of
change of position with respect to time, we can integrate to
determine the distance traveled.
6.4 Final Notes
The key to all of these is looking at the graph to determine
which function is above the other, finding intersections as
needed, then setting up and evaluating the appropriate
integrals.
In the next section we’ll delve into applications which use all
of the integration concepts developed so far. If a function
represents a rate of change, we’ll use an integral (= area
under the curve) to determine an amount.
For example, from a velocity function, which is the rate of
change of position with respect to time, we can integrate to
determine the distance traveled.
Another example: Knowing how costs are changing with
respect to the number of units produced (Marginal Cost) we
can integrate to find the amount of Cost.