§6.1 Areas between curves
Dr K
Dr K
§6.1 Areas between curves
The Riemann sum
In Chapter 5 we defined and calculated areas of regions that
lie under the graphs of functions. Here we use integrals to find
areas of regions that lie between the graphs of two functions.
Dr K
§6.1 Areas between curves
The Riemann sum
In Chapter 5 we defined and calculated areas of regions that
lie under the graphs of functions. Here we use integrals to find
areas of regions that lie between the graphs of two functions.
Consider the region S that lies between two curves
y = f (x) and y = g (x) and between the vertical lines
x = a and x = b, where f and g are continuous functions
and f (x) ≥ g (x) for all x in [a, b].
Dr K
§6.1 Areas between curves
Just as we did for areas under curves in Section 5.1, we divide
S into n strips of equal width and then we approximate the ith
strip by a rectangle with base ∆x and height f (xi∗ ) − g (xi∗ ).
Dr K
§6.1 Areas between curves
Just as we did for areas under curves in Section 5.1, we divide
S into n strips of equal width and then we approximate the ith
strip by a rectangle with base ∆x and height f (xi∗ ) − g (xi∗ ).
Dr K
§6.1 Areas between curves
This approximation appears to become better and better as
n → ∞. Therefore we define the area A of the region S as
the limiting value of the sum of the areas of these
approximating rectangles.
Dr K
§6.1 Areas between curves
This approximation appears to become better and better as
n → ∞. Therefore we define the area A of the region S as
the limiting value of the sum of the areas of these
approximating rectangles.
A = lim
n→∞
n
X
[f (xi∗ ) − g (xi∗ )]∆x
i=1
Dr K
§6.1 Areas between curves
(1)
We recognize the limit in (1) as the definite integral of f − g .
Therefore, we have the following formula for area.
Dr K
§6.1 Areas between curves
We recognize the limit in (1) as the definite integral of f − g .
Therefore, we have the following formula for area.
The area A of the region bounded by the curves
y = f (x), y = g (x), and the lines x = a, x = b, where f
and g are continuous and f (x) ≥ g (x) for all x in [a, b], is
Z b
[f (x) − g (x)] dx
A=
a
Dr K
§6.1 Areas between curves
(2)
We recognize the limit in (1) as the definite integral of f − g .
Therefore, we have the following formula for area.
The area A of the region bounded by the curves
y = f (x), y = g (x), and the lines x = a, x = b, where f
and g are continuous and f (x) ≥ g (x) for all x in [a, b], is
Z b
[f (x) − g (x)] dx
A=
(2)
a
Notice that in the special case where g (x) = 0, S is the region
under the graph of f .
Dr K
§6.1 Areas between curves
In the case where both f and g are positive, you can see from
Figure below why Equation (2) is true:
Dr K
§6.1 Areas between curves
In the case where both f and g are positive, you can see from
Figure below why Equation (2) is true:
A = [area under y = f (x)] − [area under y = g (x)]
Dr K
§6.1 Areas between curves
In the case where both f and g are positive, you can see from
Figure below why Equation (2) is true:
A = [area under y = f (x)] − [area under y = g (x)]
Z b
Z b
Z b
=
f (x) dx −
g (x) dx =
[f (x) − g (x)] dx
a
a
a
Dr K
§6.1 Areas between curves
In the case where both f and g are positive, you can see from
Figure below why Equation (2) is true:
A = [area under y = f (x)] − [area under y = g (x)]
Z b
Z b
Z b
=
f (x) dx −
g (x) dx =
[f (x) − g (x)] dx
a
a
a
Dr K
§6.1 Areas between curves
Example
Find the area of the region bounded above by y = e x ,
bounded below by y = x, and bounded on the sides by x = 0
and x = 1.
Dr K
§6.1 Areas between curves
Example
Find the area of the region bounded above by y = e x ,
bounded below by y = x, and bounded on the sides by x = 0
and x = 1.
SOLUTION
The region is shown in Figure below. The upper boundary
curve is y = e x and the lower boundary curve is y = x.
Dr K
§6.1 Areas between curves
Example
Find the area of the region bounded above by y = e x ,
bounded below by y = x, and bounded on the sides by x = 0
and x = 1.
SOLUTION
The region is shown in Figure below. The upper boundary
curve is y = e x and the lower boundary curve is y = x.
Dr K
§6.1 Areas between curves
So we use the area formula (2) with f (x) = e x , g (x) = x,
a = 0, and b = 1:
Dr K
§6.1 Areas between curves
So we use the area formula (2) with f (x) = e x , g (x) = x,
a = 0, and
1
Z 1b = 1:
1 2
x
x
A =
(e − x) dx = e − x
2
0
0
1
= e − − 1 = e − 1.5
2
Dr K
§6.1 Areas between curves
Example
Find the area of the region enclosed by the parabolas y = x 2
and y = 2x − x 2 .
Dr K
§6.1 Areas between curves
Example
Find the area of the region enclosed by the parabolas y = x 2
and y = 2x − x 2 .
SOLUTION
We first find the points of intersection of the parabolas by
solving their equations simultaneously.
Dr K
§6.1 Areas between curves
Example
Find the area of the region enclosed by the parabolas y = x 2
and y = 2x − x 2 .
SOLUTION
We first find the points of intersection of the parabolas by
solving their equations simultaneously.
This gives x 2 = 2x − x 2 , or 2x 2 − 2x = 0.
Dr K
§6.1 Areas between curves
Example
Find the area of the region enclosed by the parabolas y = x 2
and y = 2x − x 2 .
SOLUTION
We first find the points of intersection of the parabolas by
solving their equations simultaneously.
This gives x 2 = 2x − x 2 , or 2x 2 − 2x = 0.
Thus, 2x(x − 1) = 0, so x = 0 or 1.
Dr K
§6.1 Areas between curves
Example
Find the area of the region enclosed by the parabolas y = x 2
and y = 2x − x 2 .
SOLUTION
We first find the points of intersection of the parabolas by
solving their equations simultaneously.
This gives x 2 = 2x − x 2 , or 2x 2 − 2x = 0.
Thus, 2x(x − 1) = 0, so x = 0 or 1.
Dr K
§6.1 Areas between curves
The area of a typical rectangle is
(yT − yB )∆x = (2x − x 2 − x 2 )∆x
Dr K
§6.1 Areas between curves
The area of a typical rectangle is
(yT − yB )∆x = (2x − x 2 − x 2 )∆x
and the region lies between x = 0 and x = 1. So, the total
area is
Dr K
§6.1 Areas between curves
The area of a typical rectangle is
(yT − yB )∆x = (2x − x 2 − x 2 )∆x
and the region lies between x = 0 and x = 1. So, the total
area is Z
Z
1
1
(x − x 2 ) dx
(2x − 2x ) dx = 2
0 0 2
1
x
x3
1 1
1
= 2
−
=2
−
=
2
3 0
2 3
3
A =
2
Dr K
§6.1 Areas between curves
Example
Find the approximate area of the region bounded by the curves
x
and y = x 4 − x.
y=√
2
x +1
Dr K
§6.1 Areas between curves
Example
Find the approximate area of the region bounded by the curves
x
and y = x 4 − x.
y=√
2
x +1
SOLUTION
If we were to try to find the exact intersection points, we
would have to solve the equation
Dr K
§6.1 Areas between curves
Example
Find the approximate area of the region bounded by the curves
x
and y = x 4 − x.
y=√
2
x +1
SOLUTION
If we were to try to find the exact intersection points, we
would have to solve the equation
x
√
= y = x4 − x
2
x +1
Dr K
§6.1 Areas between curves
Example
Find the approximate area of the region bounded by the curves
x
and y = x 4 − x.
y=√
2
x +1
SOLUTION
If we were to try to find the exact intersection points, we
would have to solve the equation
x
√
= y = x4 − x
2
x +1
This looks like a very difficult equation to solve exactly (in
fact, it’s impossible), so instead we use a graphing device to
draw the graphs of the two curves:
Dr K
§6.1 Areas between curves
Example
Find the approximate area of the region bounded by the curves
x
and y = x 4 − x.
y=√
2
x +1
SOLUTION
If we were to try to find the exact intersection points, we
would have to solve the equation
x
√
= y = x4 − x
2
x +1
This looks like a very difficult equation to solve exactly (in
fact, it’s impossible), so instead we use a graphing device to
draw the graphs of the two curves:
Dr K
§6.1 Areas between curves
One intersection point is the origin. We zoom in toward the
other point of intersection and find that x ≈ 1.18.
Dr K
§6.1 Areas between curves
One intersection point is the origin. We zoom in toward the
other point of intersection and find that x ≈ 1.18. So an
approximation to the area between the curves is
Dr K
§6.1 Areas between curves
One intersection point is the origin. We zoom in toward the
other point of intersection and find that x ≈ 1.18. So an
approximation to the area between the curves is
Z 1.18
x
√
A≈
− (x 4 − x) dx
2
x
+
1
0
Dr K
§6.1 Areas between curves
One intersection point is the origin. We zoom in toward the
other point of intersection and find that x ≈ 1.18. So an
approximation to the area between the curves is
Z 1.18
x
√
A≈
− (x 4 − x) dx
2
x
+
1
0
To integrate the first term, we use the substitution
u = x 2 + 1. Then du = 2x dx, and when x = 1.18, we have
u ≈ 2.39; when x = 0, u = 1.
Dr K
§6.1 Areas between curves
So,
Dr K
§6.1 Areas between curves
So,
1
A ≈
2
Z 2.29
1
du
√ −
u
Z 1.18
(x 4 − x) dx
0
Dr K
§6.1 Areas between curves
So,
Z
Z 1.18
1 2.29 du
√ −
(x 4 − x) dx
A ≈
2 1 u
0
1.18
√ x5 x2
=
u −
−
5
2 0
Dr K
§6.1 Areas between curves
So,
Z
Z 1.18
1 2.29 du
√ −
(x 4 − x) dx
A ≈
2 1 u
0
1.18
√ x5 x2
=
u −
−
5
2 0
√
(1.18)5 (1.18)2
=
2.39 − 1 −
+
5
2
Dr K
§6.1 Areas between curves
So,
A ≈
=
=
≈
Z
Z 1.18
1 2.29 du
√ −
(x 4 − x) dx
2 1 u
0
1.18
√ x5 x2
u −
−
5
2 0
√
(1.18)5 (1.18)2
2.39 − 1 −
+
5
2
0.785
Dr K
§6.1 Areas between curves
If we are asked to find the area between the curves y = f (x)
and y = g (x) where
Dr K
§6.1 Areas between curves
If we are asked to find the area between the curves y = f (x)
and y = g (x) where f (x) ≥ g (x) for some values of x but
f (x) ≤ g (x) for other values of x, then
Dr K
§6.1 Areas between curves
If we are asked to find the area between the curves y = f (x)
and y = g (x) where f (x) ≥ g (x) for some values of x but
f (x) ≤ g (x) for other values of x, then we split the given
region S into several regions S1 , S2 , · · · with areas A1 , A2 , · · ·
as shown in Figure 11.
Dr K
§6.1 Areas between curves
We then define the area of the region S to be the sum of the
areas of the smaller regions S1 , S2 , · · · , that is,
A = A1 + A2 + · · · . Since
Dr K
§6.1 Areas between curves
We then define the area of the region S to be the sum of the
areas of the smaller regions S1 , S2 , · · · , that is,
A = A1 + A2 + · · · . Since
(
f (x) − g (x) when f (x) ≥ g (x)
|f (x) − g (x)| =
g (x) − f (x) when g (x) ≥ f (x)
Dr K
§6.1 Areas between curves
We then define the area of the region S to be the sum of the
areas of the smaller regions S1 , S2 , · · · , that is,
A = A1 + A2 + · · · . Since
(
f (x) − g (x) when f (x) ≥ g (x)
|f (x) − g (x)| =
g (x) − f (x) when g (x) ≥ f (x)
We have the following expression for A.
Dr K
§6.1 Areas between curves
We then define the area of the region S to be the sum of the
areas of the smaller regions S1 , S2 , · · · , that is,
A = A1 + A2 + · · · . Since
(
f (x) − g (x) when f (x) ≥ g (x)
|f (x) − g (x)| =
g (x) − f (x) when g (x) ≥ f (x)
We have the following expression for A. The area between the
curves y = f (x) and y = g (x) and between x = a and x = b
is
Dr K
§6.1 Areas between curves
We then define the area of the region S to be the sum of the
areas of the smaller regions S1 , S2 , · · · , that is,
A = A1 + A2 + · · · . Since
(
f (x) − g (x) when f (x) ≥ g (x)
|f (x) − g (x)| =
g (x) − f (x) when g (x) ≥ f (x)
We have the following expression for A. The area between the
curves y = f (x) and y = g (x) and between x = a and x = b
is
Z b
A=
|f (x) − g (x)| dx
(3)
a
Dr K
§6.1 Areas between curves
Example
Find the area of the region bounded by the curves y = sin x,
y = cos x, x = 0, and x = π2 .
Dr K
§6.1 Areas between curves
Example
Find the area of the region bounded by the curves y = sin x,
y = cos x, x = 0, and x = π2 .
SOLUTION
The points of intersection occur when sin x = cos x, that is,
when x = π/4 (since 0 ≤ x ≤ π/2). The region is sketched in
Figure .
Dr K
§6.1 Areas between curves
Example
Find the area of the region bounded by the curves y = sin x,
y = cos x, x = 0, and x = π2 .
SOLUTION
The points of intersection occur when sin x = cos x, that is,
when x = π/4 (since 0 ≤ x ≤ π/2). The region is sketched in
Figure .
Dr K
§6.1 Areas between curves
Observe that cos x ≥ sin x when 0 ≤ x ≤ π/4 but
sin x ≥ cos x when π/4 ≤ x ≤ π/2. Therefore the required
area is
Dr K
§6.1 Areas between curves
Observe that cos x ≥ sin x when 0 ≤ x ≤ π/4 but
sin x ≥ cos x when π/4 ≤ x ≤ π/2. Therefore the required
area is Z
π/2
| cos x − sin x| dx = A1 + A2
A =
0
Dr K
§6.1 Areas between curves
Observe that cos x ≥ sin x when 0 ≤ x ≤ π/4 but
sin x ≥ cos x when π/4 ≤ x ≤ π/2. Therefore the required
area is Z
π/2
A =
Z0 π/4
=
| cos x − sin x| dx = A1 + A2
Z π/2
(cos x − sin x) dx +
(sin x − cos x) dx
0
π/4
Dr K
§6.1 Areas between curves
Observe that cos x ≥ sin x when 0 ≤ x ≤ π/4 but
sin x ≥ cos x when π/4 ≤ x ≤ π/2. Therefore the required
area is Z
π/2
| cos x − sin x| dx = A1 + A2
Z π/2
=
(cos x − sin x) dx +
(sin x − cos x) dx
0
π/4
π/4 π/2
=
sin x + cos x
+ − cos x − sin x
A =
Z0 π/4
0
π/4
Dr K
§6.1 Areas between curves
Observe that cos x ≥ sin x when 0 ≤ x ≤ π/4 but
sin x ≥ cos x when π/4 ≤ x ≤ π/2. Therefore the required
area is Z
π/2
| cos x − sin x| dx = A1 + A2
Z π/2
=
(cos x − sin x) dx +
(sin x − cos x) dx
0
π/4
π/4 π/2
=
sin x + cos x
+ − cos x − sin x
0
π/4
1
1
1
1
√ + √ −0−1 + −0−1+ √ + √
=
2
2
2
2
A =
Z0 π/4
Dr K
§6.1 Areas between curves
Observe that cos x ≥ sin x when 0 ≤ x ≤ π/4 but
sin x ≥ cos x when π/4 ≤ x ≤ π/2. Therefore the required
area is Z
π/2
A =
| cos x − sin x| dx = A1 + A2
Z π/2
(cos x − sin x) dx +
(sin x − cos x) dx
0
π/4
π/4 π/2
sin x + cos x
+ − cos x − sin x
0
π/4
1
1
1
1
√ + √ −0−1 + −0−1+ √ + √
2
2
2
√2
2 2−2
Z0 π/4
=
=
=
=
Dr K
§6.1 Areas between curves
Observe that cos x ≥ sin x when 0 ≤ x ≤ π/4 but
sin x ≥ cos x when π/4 ≤ x ≤ π/2. Therefore the required
area is Z
π/2
| cos x − sin x| dx = A1 + A2
Z π/2
=
(cos x − sin x) dx +
(sin x − cos x) dx
0
π/4
π/4 π/2
=
sin x + cos x
+ − cos x − sin x
0
π/4
1
1
1
1
√ + √ −0−1 + −0−1+ √ + √
=
2
2
2
√2
= 2 2−2
In this particular example, we could have saved some work by
noticing that the region is symmetric about x = π/4 and so
A =
Z0 π/4
Dr K
§6.1 Areas between curves
Observe that cos x ≥ sin x when 0 ≤ x ≤ π/4 but
sin x ≥ cos x when π/4 ≤ x ≤ π/2. Therefore the required
area is Z
π/2
| cos x − sin x| dx = A1 + A2
Z π/2
=
(cos x − sin x) dx +
(sin x − cos x) dx
0
π/4
π/4 π/2
=
sin x + cos x
+ − cos x − sin x
0
π/4
1
1
1
1
√ + √ −0−1 + −0−1+ √ + √
=
2
2
2
√2
= 2 2−2
In this particular example, we could have saved some work by
noticing that the region is symmetric about x = π/4 and so
Z π/4
A = 2A1 = 2
(cos x − sin x) dx
A =
Z0 π/4
0
Dr K
§6.1 Areas between curves
Some regions are best treated by regarding x as a function of
y.
Dr K
§6.1 Areas between curves
Some regions are best treated by regarding x as a function of
y.
If a region is bounded by curves with equations x = f (y ),
x = g (y ), y = c, and y = d, where f and g are continuous
and f (y ) ≥ g (y ) for c ≤ y ≤ d,
Dr K
§6.1 Areas between curves
Some regions are best treated by regarding x as a function of
y.
If a region is bounded by curves with equations x = f (y ),
x = g (y ), y = c, and y = d, where f and g are continuous
and f (y ) ≥ g (y ) for c ≤ y ≤ d,
then its area is
Dr K
§6.1 Areas between curves
Some regions are best treated by regarding x as a function of
y.
If a region is bounded by curves with equations x = f (y ),
x = g (y ), y = c, and y = d, where f and g are continuous
and f (y ) ≥ g (y ) for c ≤ y ≤ d,
then its area is
Z d
[f (y ) − g (y )] dy
A=
c
Dr K
§6.1 Areas between curves
Example
Find the area enclosed by the line y = x − 1 and the parabola
y 2 = 2x + 6.
Dr K
§6.1 Areas between curves
Example
Find the area enclosed by the line y = x − 1 and the parabola
y 2 = 2x + 6.
SOLUTION
By solving the two equations, we find that the points of
intersection are (−1, −2) and (5, 4).
Dr K
§6.1 Areas between curves
Example
Find the area enclosed by the line y = x − 1 and the parabola
y 2 = 2x + 6.
SOLUTION
By solving the two equations, we find that the points of
intersection are (−1, −2) and (5, 4). We solve the equation of
the parabola for x and notice from Figure below that the left
and right boundary curves are
Dr K
§6.1 Areas between curves
Example
Find the area enclosed by the line y = x − 1 and the parabola
y 2 = 2x + 6.
SOLUTION
By solving the two equations, we find that the points of
intersection are (−1, −2) and (5, 4). We solve the equation of
the parabola for x and notice from Figure below that the left
and right boundary curves are
1
x = y2 − 3
and
x =y +1
2
Dr K
§6.1 Areas between curves
We must integrate between the appropriate y -values, y = −2
and y = 4.
Thus,
Dr K
§6.1 Areas between curves
We must integrate between the appropriate y -values, y = −2
and y = 4.
Thus,
Z 4
1
A =
(y + 1) − ( y 2 − 3) dy
2
−2
Dr K
§6.1 Areas between curves
We must integrate between the appropriate y -values, y = −2
and y = 4.
Thus,
Z 4
1
A =
(y + 1) − ( y 2 − 3) dy
2
Z−24
1 2
=
− y + y + 4 dy
2
−2
Dr K
§6.1 Areas between curves
We must integrate between the appropriate y -values, y = −2
and y = 4.
Thus,
Z 4
1
A =
(y + 1) − ( y 2 − 3) dy
2
Z−24
1 2
=
− y + y + 4 dy
−2 2
4
1 y3
y2
= −
+
+ 4y
2 3
2
−2
Dr K
§6.1 Areas between curves
We must integrate between the appropriate y -values, y = −2
and y = 4.
Thus,
Z 4
1
A =
(y + 1) − ( y 2 − 3) dy
2
Z−24
1 2
=
− y + y + 4 dy
−2 2
4
1 y3
y2
= −
+
+ 4y
2 3
2
−2
1
4
= − (64) + 8 + 16 − ( + 2 − 8) = 18
6
3
Dr K
§6.1 Areas between curves