Calculus II
Nguyen Dinh Thi
ndthi@hcmus.edu.vn
VNUHCM University of Science
Introduction
Course description: The course is designed to provide students with the fundamental ideas of applications of integration, Differential Equations and Infinite
Sequences and Series.
References:
▶ Calculus, James Stewart, Daniel K. Clegg, Saleem Watson, Cengage
Learning, 9th Edition, 2020.
Softwares:
▶ Mathematical computation softwares: Maxima, Maple, Matlab, Wolfram
Alpha, GeoGebra, Python, C/C++ ...
▶ Software for typesetting mathematics: LaTeX, or LyX.
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Calculus II
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Applications of Integration
1
Applications of Integration
Areas between Curves
Volumes
Volumes by cylindrical shell
Arc length
Average Value of a Function
Applications to Physics, Engineering, Economics, and Biology
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Areas between Curves
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].
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Areas between Curves
Just as we did for areas under curves, we divide S into n strips of equal
width and then we approximate the i th strip by a rectangle with base
∆x and height f (xi∗ ) − g (xi∗ ).
Figure 1.2
If we like, we could take all of the sample points to be right endpoints,
in which case xi∗ = xi . The Riemann sum therefore
n
X
[f (xi∗ ) − g (xi∗ )] ∆x
i=1
is an approximation to what we intuitively think of as the area of S.
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Areas between Curves
Formular 1.1
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
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Areas between Curves
Z b
Z b
f (x) dx −
Figure 1.3: A =
a
g (x) dx
a
In the case where both f and g are positive, you can see from Figure 1.3
why Formular 1.1 is true:
A = [area under y = f (x)] − [area under g = f (x)]
Z b
Z b
Z b
=
f (x) dx −
g (x) dx =
[f (x) − g (x)] dx
a
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a
a
Calculus II
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Areas between Curves
Example 1.1
Find the area of the region bounded above by y = x 2 + 1, bounded
below by y = x, and bounded on the sides by x = 0 and x = 1.
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Areas between Curves
Example 1.2
Find the area of the region enclosed by the parabolas y = x 2 and
y = 2x − x 2 .
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Areas between Curves
Figure 1.4
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 1.4. 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 + . . ..
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Areas between Curves
Since
|f (x) − g (x)| =
(
f (x) − g (x)
when f (x) ≥ g (x)
g (x) − f (x)
when f (x) ≤ g (x)
we have the following expression for A
Formular 1.2
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
a
When evaluating the integral in Formular 1.2, however, we must still
split it into integrals corresponding to A = A1 + A2 + . . .
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Areas between Curves
Example 1.3
Find the area of the region bounded by the curves y = sin x, y = cos x,
x = 0, and x = π/2.
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Areas between Curves
Figure 1.5
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
andf (y ) > g (y ) for c < y < d (see Figure 1.5),
then its area is
Z d
A=
[f (y ) − g (y )] dy .
c
If we write xR for the right boundary and xL for
the left boundary, then, as Figure 1.6 illustrates,
we have
Z
d
(xR − xL ) dy .
A=
c
Figure 1.6
D-T. Nguyen (VNU-HCMUS)
Here a typical approximating rectangle has
dimensions xR − xL and ∆y .
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Areas between Curves
Example 1.4
Find the area enclosed by the line y = x−1 and the parabola y 2 = 2x+6.
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Volumes
We start with a simple type of solid called a cylinder
▶ A cylinder is bounded by a plane region B1 , called the base, and a
congruent region B2 in a parallel plane.
▶ The cylinder consists of all points on line segments that are perpendicular to the base and join B1 to B2 .
▶ If the area of the base is A and the height of the cylinder (the
distance from B1 to B2 ) is h, then the volume V of the cylinder is
defined as
V = Ah.
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Volumes
In particular, if the base is a circle with radius r , then the cylinder is a
circular cylinder with volume V = πr 2 h.
If the base is a rectangle with length ℓ and width w , then the cylinder is
a rectangular box (also called a rectangular parallelepiped) with volume
V = ℓwh
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Volumes
▶ For a solid S that isn’t a cylinder, we start by intersecting S with
a plane and obtaining a plane region that is called a cross-section
of S.
▶ Let A(x) be the area of the cross-section of S in a plane Px perpendicular to the x−axis and passing through the point x, where
x ∈ [a, b]. The cross-sectional area A(x) will vary as x increases
from a to b.
Figure 1.7: Think of slicing S with a knife through x and computing the area
of this slice.
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Volumes
Let’s divide S into n “slabs” of equal width ∆x by using the planes
Px1 , Px2 , ... to slice the solid.
Figure 1.8: Think of slicing a loaf of bread.
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Volumes
If we choose sample points xi∗ in [xi−1 , xi ], we can approximate the i th
slab Si by a cylinder with base area A(xi∗ ) and height ∆x.
The volume of this cylinder is A(xi∗ )∆x, so an approximation to our
intuitive conception of the volume of the ith slab Si is
V (Si ) ≈ A(xi∗ )∆x.
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Volumes
Adding the volumes of these slabs, we get an approximation to the total volume
V ≈
n
X
A(xi∗ )∆x.
i=1
This approximation appears to become better and better as n → ∞. Therefore
we define the volume as the limit of these sums as n → ∞.
We recognize the limit of Riemann sums as a definite integral. So, we have
the following definition
Definition 1.1 (Volume)
Let S be a solid that lies between x = a and x = a. If the cross-sectional
area of S in the plane Px , through x and perpendicular to the x-axis, is A (x),
where A is a continuous function, then the volume of S is
V = lim
n→∞
D-T. Nguyen (VNU-HCMUS)
n
X
A (xi∗ ) ∆x =
Z b
A (x) dx
a
i=1
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Volumes
Example 1.5
4
Show that the volume of a sphere of radius r is V = πr 3 .
3
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Volumes
Example 1.5
4
Show that the volume of a sphere of radius r is V = πr 3 .
3
Solution: If we place the sphere so that its center is at the origin,
then
p the plane Px intersects the sphere in a circle whose radius is y =
r 2 − x 2.
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Volumes
So, the cross-sectional area is
A(x) = πy 2 = π r 2 − x 2 .
Using the definition of volume with a = −r and b = r , we have
Z r
V =
A(x)dx
−r
Z r
=
π r 2 − x 2 dx
−r
4
= πr 3 .
3
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Volumes by cylindrical shell
Example 1.6
Find the volume
√ of the solid obtained by rotating about the x-axis the region under
the curve y = x from 0 to 1.
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Volumes by cylindrical shell
Example 1.6
Find the volume
√ of the solid obtained by rotating about the x-axis the region under
the curve y = x from 0 to 1.
Solution:
The region is shown in the following
figure
D-T. Nguyen (VNU-HCMUS)
If we rotate about the x-axis, we get the
solid shown in the following figure
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Volumes by cylindrical shell
When we slice through the point x, we get a disk with radius
area of this cross-section is
√ 2
A(x) = π x = πx.
√
x. The
The solid lies between x = 0 and x = 1, so its volume is
Z 1
Z 1
π
V =
A(x)dx =
πxdx = .
2
0
0
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Volumes by cylindrical shell
Example 1.7
Find the volume of the solid obtained by rotating the region bounded
by y = x 3 , y = 8 and y = 0 about the y − axis.
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Volumes by cylindrical shell
Example 1.7
Find the volume of the solid obtained by rotating the region bounded
by y = x 3 , y = 8 and y = 0 about the y − axis.
Solution: The region and the resulting solid are shown in the following
figures
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Because the region is rotated about the y -axis, it makes sense to slice
the solid perpendicular to the y -axis and therefore to integrate with
respect to y . If we slice at height y , we get a circular disk with radius
√
x, where x = 3 y .
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Volumes by cylindrical shell
So the area of a cross-section through y is
A(y ) = πx 2 = πy 2/3 .
Since the solid lies between y = 0 and y = 8, its volume is
Z 8
96π
V =
A(y )dy =
.
5
0
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Volumes by cylindrical shell
Example 1.8
The region R enclosed by the curves y = x and y = x 2 is rotated about
the x-axis. Find the volume of the resulting solid.
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Volumes by cylindrical shell
Example 1.8
The region R enclosed by the curves y = x and y = x 2 is rotated about
the x-axis. Find the volume of the resulting solid.
Solution: The curves y = x and y = x 2 intersect at the points (0, 0)
and (1, 1).
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Volumes by cylindrical shell
The solid of rotation and a cross-section perpendicular to the x-axis are shown in the
following figures
A cross-section in the plane Px has the shape of a washer (an annular ring) with
inner radius x 2 and outer radius x. So, we find the cross-sectional area by subtracting
the area of the inner circle from the area of the outer circle
2
A(x) = πx 2 − π x 2 = π x 2 − x 4 .
Therefore we have
Z 1
V =
A(x)dx =
0
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2π
.
15
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Volumes by cylindrical shell
Example 1.9
The region R enclosed by the curves y = x and y = x 2 is rotated about
the line y = 2. Find the volume of the resulting solid.
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Volumes by cylindrical shell
Example 1.9
The region R enclosed by the curves y = x and y = x 2 is rotated about
the line y = 2. Find the volume of the resulting solid.
Solution: The curves y = x and y = x 2 intersect at the points (0, 0)
and (1, 1). The solid and a cross-section are shown in the following.
Again
the cross-section is a washer,
but this time the inner radius is
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Volumes by cylindrical shell
The cross-sectional area is
A(x) = π 2 − x 2
2
− π (2 − x)2
and so the volume of S is
Z 1
V =
A(x)dx =
0
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8π
.
15
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Volumes by cylindrical shell
Example 1.10
The region R enclosed by the curves y = x and y = x 2 is rotated about
the line x = −1. Find the volume of the resulting solid.
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Volumes by cylindrical shell
Solution
A horizontal cross-section is shown in the following figure
The horizontal cross-section is a washer with inner radius 1 + y and outer radius
√
1 + y . So, the cross-sectional area is
√
A(y ) = π (1 + y )2 − π(1 + y )2 .
The volume is
Z 1
V =
A(y )dy =
0
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π
.
2
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Volumes by cylindrical shell
Exercise 1.1
Find the volume of the solid obtained by rotating the region bounded
by the given curves about the specified line.
(i) y = x + 1, y = 0, x = 0, x = 2; about the x − axis.
(ii) y = x 3 , y = x, x ≥ 0; about the x − axis.
(iii) y 2 = x, x = 2y ; about the y − axis.
(iv) y = x 2 , x = y 2 about y = 1.
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Volumes by cylindrical shell
Some volume problems are very difficult to handle by the methods of
the preceding section.
Example: Let’s consider the problem of finding the volume of the solid
obtained by rotating about the y -axis the region bounded by y = 2x 2 −
x 3 and y = 0.
If we slice perpendicular to the y -axis, we get a washer. But to compute
the inner radius and the outer radius of the washer, we have to solve
the cubic equation y = 2x 2 − x 3 for x in terms of y .
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Volumes by cylindrical shell
Fortunately, there is a method, called the method of cylindrical shells,
that is easier to use in such a case.
The following shows a cylindrical shell with inner radius r1 , outer radius
r2 and height h
Its volume V is calculated by subtracting the volume V1 of the inner
cylinder from the volume V2 of the outer cylinder
r1 + r2
V = V2 − V1 = πr22 h − πr12 h = 2π
h(r2 − r1 ).
2
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Volumes by cylindrical shell
Its volume V is calculated by subtracting the volume V1 of the inner
cylinder from the volume V2 of the outer cylinder
V = V2 − V1 = πr22 h − πr12 h = 2π
r1 + r2
h(r2 − r1 ).
2
If we let
▶ ∆r = r2 − r1 (the thickness of the shell)
r + r2
▶ r= 1
(the average radius of the shell)
2
then this formula for the volume of a cylindrical shell becomes
height
z}|{
V = |{z}
2πr h ∆r → thickness
circumference
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(1.1)
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Volumes by cylindrical shell
Now let S be the solid obtained by rotating about the y -axis the region bounded by
y = f (x) ≥ 0, y = 0, x = a and x = b.
We divide the interval [a, b] into n sub-intervals [xi−1 , xi ] of equal width ∆x and let
x i be the midpoint of the ith sub-intervals.
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Volumes by cylindrical shell
If the rectangle with base [xi−1 , xi ] and height f (x i ) is rotated about the y -axis, then
the result is a cylindrical shell with average radius x i , height f (x i ), and thickness ∆x.
By using (1.1), its volume is
Vi = 2πx i f (x i )∆x.
Therefore an approximation to the volume V of S is given by the sum of the volumes
of these shells
V ≈
n
X
i=1
Vi =
n
X
2πx i f (x i )∆x.
i=1
This approximation appears to become better as n → ∞.
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Volumes by cylindrical shell
From the definition of an integral, we know that
lim
n→∞
n
X
Z b
2πx i f (x i )∆x =
2πxf (x)dx.
a
i=1
Thus, we have the following result
The volume of the solid obtained by rotating about the y -axis the region
under the curve y = f (x) from a to b, is
Z b
V =
2πxf (x)dx,
where 0 ≤ a < b.
a
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Volumes by cylindrical shell
Example 1.11
Find the volume of the solid obtained by rotating about the y -axis the
region bounded by y = 2x 2 − x 3 and y = 0.
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Volumes by cylindrical shell
Example 1.11
Find the volume of the solid obtained by rotating about the y -axis the
region bounded by y = 2x 2 − x 3 and y = 0.
Solution:
We can see that a typical shell has radius x, circumference 2πx and
height f (x) = 2x 2 − x 3 . So, by the shell method, the volume is
Z 2
16
V =
2πx 2x 2 − x 3 dx = π.
5
0
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Volumes by cylindrical shell
Example 1.12
Find the volume of the solid obtained by rotating about the y -axis the
region between y = x and y = x 2 .
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Volumes by cylindrical shell
Example 1.12
Find the volume of the solid obtained by rotating about the y -axis the
region between y = x and y = x 2 .
Solution: The region and a typical shell are shown in the following figure
We see that the shell has radius x, circumference 2πx and height x −x 2 .
So the volume is
Z 1
π
V =
2πx x − x 2 dx = .
6
0
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Volumes by cylindrical shell
Example 1.13
Use cylindrical shells to find the volume of the solid
√obtained by rotating
about the x-axis the region under the curve y = x from 0 to 1.
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Volumes by cylindrical shell
Solution
This problem was solved using disks
√ (see the 2previous example). To use
shells we relabel the curve y = x as x = y in the following figure
For rotation about the y −axis, we see that a typical shell has radius y ,
circumference 2πy and height 1 − y 2 . So, the volume is
Z 1
π
V =
2πy 1 − y 2 dy = .
2
0
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Volumes by cylindrical shell
Example 1.14
Find the volume of the solid obtained by rotating the region bounded
by y = x − x 2 and y = 0 about the line x = 2.
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Volumes by cylindrical shell
Example 1.14
Find the volume of the solid obtained by rotating the region bounded
by y = x − x 2 and y = 0 about the line x = 2.
Solution: The region and a cylindrical shell formed by rotation about
the line x = 2 are shown in the following figures
It has radius 2 − x, circumference 2π(2 − x) and height x − x 2 . The
volume of the given solid is
Z 1
π
2π(2 − x)(x − x 2 )dx = .
V =
2
0
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Volumes by cylindrical shell
Exercise 1.2
Use the method of cylindrical shells to find the volume the solid obtained
by
(a) rotating the region bounded by the given curves about the y −axis
(i) y = x 3 , y = 0, x = 1, x = 2.
(ii) y = 4x − x 2 , y = x.
(b) rotating the region bounded by the given curves about the x−axis
(i) x = 1 + (y − 2)2 , x = 2.
(ii) x + y = 4, x = y 2 − 4y + 4.
(c) rotating the region bounded by the given curves about the specified
axis.
(i) y = x 3 , y = 8, x = 0 about x = 3.
(ii) x = 2y 2 , y ≥ 0, x = 2 about y = 2.
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Arc length
▶ We suppose that a curve C is defined by the equation y = f (x),
where f is continuous and x ∈ [a, b].
▶ We obtain a polygonal approximation to C by dividing the interval
[a, b] into n sub-intervals with endpoints x0 , x1 , ..., xn and equal width
∆x.
▶ If yi = f (xi ) then the point Pi (xi , yi ), lies on C and the polygon with
vertices P0 , P1 , ..., Pn , is an approximation to C .
Figure 1.9
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Arc length
▶ Therefore, we define the length L of the curve C with equation
y = f (x) , x ∈ [a, b], as the limit of the lengths of these inscribed
polygons (if the limit exists)
L = lim
n→∞
n
X
|Pi−1 Pi |.
(1.2)
i=1
▶ If we let ∆yi = yi − yi−1 , then
q
q
2
2
|Pi−1 Pi | = (xi − xi−1 ) + (yi − yi−1 ) = (∆x)2 + (∆yi )2 . (1.3)
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Arc length
▶ By applying the Mean Value Theorem to f on the interval [xi−1 , xi ],
we find that there is a number xi∗ between xi−1 and xi such that
f (xi ) − f (xi−1 ) = f ′ (xi∗ )(xi − xi−1 )
⇔∆yi = f ′ (xi∗ )∆x.
▶ Plugging (1.4) into (1.3), we get
q
2 q
2
|Pi−1 Pi | = (∆x)2 + f ′ (xi∗ )∆x = 1 + f ′ (xi∗ ) ∆x.
(1.4)
(1.5)
▶ According to (1.2) and (1.5), it yields
n q
X
2
L = lim
1 + f ′ (xi∗ ) ∆x.
n→∞
i=1
▶ By the definition of a definite integral, we recognize this expression
as being equal to
Z bq
L=
1 + [f ′ (x)]2 dx.
a
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Arc length
To sum up, we have the following result
Theorem 1.1 (The Arc Length Formula)
If f ′ is continuous on [a, b], then the length of the curve y = f (x), x ∈
[a, b], is
Z bq
1 + [f ′ (x)]2 dx.
L=
(1.6)
a
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Arc length
Example 1.15
Find the length of the arc of the semi-cubical parabola y 2 = x 3 between
the points (1, 1) and (4, 8).
Solution
For the top half of the curve we have
y = x 3/2 = f (x).
So, the arc length formula gives
Z 4q
2
1 + [f ′ (x)] dx
L=
1
Z 4r
9
=
1 + xdx
4
1
√ 1 √
=
80 10 − 13 13 .
27
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Arc length
If a curve has the equation x = g (y ), y ∈ [c, d] and g ′ (y ) is continuous,
then by interchanging the roles of x and y in (1.6), we obtain the
following formula for its length
Z dq
1 + [g ′ (y )]2 dy .
(1.7)
L=
c
Example 1.16
Find the length of the curve x = 1 + 6y 3/2 with y ∈ [0, 1].
Solution
We have
x = g (y ) = 1 + 6y 3/2 .
So, the arc length formula gives
Z 1q
Z 1p
√
2
2
′
L=
1 + [g (y )] dy =
1 + 81y dy =
(82 82 − 1).
243
0
0
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Arc length
Exercise 1.3
Find the exact length of the curve.
3
(i) 36y 2 = x 2 − 4 , 2 ≤ x ≤ 3, y ≥ 0
1
x3
+ ,1≤x ≤2
3
4x
y4
1
(iii) x =
+ 2, 1 ≤ y ≤ 2
8
4y
1√
(iv) x =
y (y − 3), 1 ≤ y ≤ 9
3
(ii) y =
D-T. Nguyen (VNU-HCMUS)
Calculus II
53 / 108
Average Value of a Function
▶ It is easy to calculate the average value of finitely many numbers
y1 , y2 , . . . , yn :
D-T. Nguyen (VNU-HCMUS)
Calculus II
54 / 108
Average Value of a Function
▶ It is easy to calculate the average value of finitely many numbers
y1 , y2 , . . . , yn :
y1 + y2 + . . . + yn
yave =
n
▶ But how do we compute the average temperature during a day if
infinitely many temperature readings are possible?
Figure 1.10: The graph of a temperature function T (t), where t is measured
in hours and T in ◦ C , and a guess at the average temperature, Tave .
D-T. Nguyen (VNU-HCMUS)
Calculus II
54 / 108
Average Value of a Function
In general, let’s try to compute the average value of a function y =
f (x) , a ≤ x ≤ b.
▶ We start by dividing the interval [a, b] into n equal subintervals, each
with length ∆x = (b − a) /n.
▶ Then we choose points x1∗ , x2∗ , . . . , xn∗ in successive subintervals and
calculate the average of the numbers f (x1∗ ) , f (x2∗ ) , . . . , f (xn∗ ):
f (x1∗ ) + f (x2∗ ) + . . . + f (xn∗ )
.
n
D-T. Nguyen (VNU-HCMUS)
Calculus II
55 / 108
Average Value of a Function
In general, let’s try to compute the average value of a function y =
f (x) , a ≤ x ≤ b.
▶ We start by dividing the interval [a, b] into n equal subintervals, each
with length ∆x = (b − a) /n.
▶ Then we choose points x1∗ , x2∗ , . . . , xn∗ in successive subintervals and
calculate the average of the numbers f (x1∗ ) , f (x2∗ ) , . . . , f (xn∗ ):
f (x1∗ ) + f (x2∗ ) + . . . + f (xn∗ )
.
n
For example, if f represents a temperature function and n = 24, this
means that we taketemperature readings every hour and then average
them.
D-T. Nguyen (VNU-HCMUS)
Calculus II
55 / 108
Average Value of a Function
▶ Since ∆x = (b − a) /n, we can write n = (b − a) /∆x and the
average value becomes
f (x1∗ ) + f (x2∗ ) + . . . + f (xn∗ )
b−a
∆x
=
1
[f (x1∗ ) + f (x2∗ ) + . . . + f (xn∗ )] ∆x
b−a
=
1 X
f (xi∗ ) ∆x.
b−a
n
i=1
▶ If we let n increase, we would be computing the average value of a
large number of closely spaced values.
D-T. Nguyen (VNU-HCMUS)
Calculus II
56 / 108
Average Value of a Function
▶ Since ∆x = (b − a) /n, we can write n = (b − a) /∆x and the
average value becomes
f (x1∗ ) + f (x2∗ ) + . . . + f (xn∗ )
b−a
∆x
=
1
[f (x1∗ ) + f (x2∗ ) + . . . + f (xn∗ )] ∆x
b−a
=
1 X
f (xi∗ ) ∆x.
b−a
n
i=1
▶ If we let n increase, we would be computing the average value of a
large number of closely spaced values.
For example, we would be averaging temperature readings taken every
minute or even every second.
D-T. Nguyen (VNU-HCMUS)
Calculus II
56 / 108
Average Value of a Function
The limiting value is
n
Z b
i=1
a
1 X
1
f (xi∗ ) ∆x =
lim
n→∞ b − a
b−a
f (x) dx
by the definition of a definite integral
Therefore we define the average value of f on the interval [a, b] as
1
fave =
b−a
Z b
f (x) dx.
a
For a positive function, we can think of this definition as saying
area
= average height.
width
D-T. Nguyen (VNU-HCMUS)
Calculus II
57 / 108
Average Value of a Function
▶ If T (t) is the temperature at time t, we might wonder if there
is a specific time when the temperature is the same as the average
temperature.
▶ For the temperature function graphed in Figure 1.10, we see that
there are two such times: just before noon and justbefore midnight.
▶ In general, is there a number c at which the value of a function f is
exactly equal to the average value of the function, that is, f (c) = fave ?
D-T. Nguyen (VNU-HCMUS)
Calculus II
58 / 108
Average Value of a Function
The following theorem says that this is true for continuous functions.
Theorem 1.2 (The Mean Value Theorem for Integrals)
If f is continuous on [a, b], then there exists a number c in [a, b] such
that
Z b
1
f (c) = fave =
f (x) dx,
b−a a
that is,
Z b
f (x) dx = f (c) (b − a) .
a
D-T. Nguyen (VNU-HCMUS)
Calculus II
59 / 108
Average Value of a Function
▶ The Mean Value Theorem for
Integrals is a consequence of the
Mean Value Theorem for
derivatives and the Fundamental
Theorem of Calculus.
▶ The geometric interpretation
of the Mean Value Theorem for
Integrals is that, for positive
functions f , there is a number c
such that the rectangle with base
[a, b] and height f (c) has the
same area as the region under the
graph of f from a to b.
D-T. Nguyen (VNU-HCMUS)
Figure 1.11: You can always chop off
the top of a (two-dimensional) mountain at a certain height and use it to
fill in the valleys so that the mountain
becomes completely flat.
Calculus II
60 / 108
Average Value of a Function
Example 1.17
Find the average value of the function f (x) = 1 + x 2 on the interval
[−1, 2].
Since f (x) = 1 + x 2 is continuous on the interval [−1, 2], the Mean
Value Theorem for Integrals says there is a number c in [−1, 2] such
that
Z 2
1 + x 2 dx = f (c) (2 − (−1)) .
−1
In this particular case we can find c explicitly.
We know that fave = 2, so the value of c satisfies
f (c) = fave = 2.
Therefore 1 + c 2 = 2 so c 2 = 1. So in this case there happen to be two
numbers c = ±1 in the interval [−1, 2] that work in the MVT.
D-T. Nguyen (VNU-HCMUS)
Calculus II
61 / 108
Average Value of a Function
Example 1.18
Show that the average velocity of a car over a time interval [t1 , t2 ] is
the same as the average of its velocities during the trip.
Solution
If s (t) is the displacement of the car at time t, then, by definition, the average
velocity of the car over the interval is
s (t2 ) − s (t1 )
∆s
=
.
∆t
t2 − t1
On the other hand, the average value of the velocity function on the interval
is
Z t2
Z t2
1
1
vave =
v (t) dt =
s ′ (t) dt
t 2 − t 1 t1
t2 − t1 t1
1
=
[s (t2 ) − s (t1 )] .
t2 − t1
D-T. Nguyen (VNU-HCMUS)
Calculus II
62 / 108
Average Value of a Function
Exercise 1.4
(a) Find the average value of f on the given interval.
(b) Find c such that fave = f (c).
(c) Sketch the graph of f and a rectangle whose area is the same as
the area under the graph of f .
(i) f (x) = (x − 3)2 , x ∈ [2, 5].
√
(ii) f (x) = 3 x, x ∈ [0, 8].
(iii) f (x) = 2 sin (x) − sin (2x), x ∈ [0, π].
p
(iv) f (x) = x 4 − x 2 , x ∈ [0, 2].
D-T. Nguyen (VNU-HCMUS)
Calculus II
63 / 108
Applications to Physics, Engineering, Economics, and
Biology
▶ Among the many applications of integral calculus, we consider here:
1. Force due to water pressure (physics).
2. Centers of mass (engineering).
3. Consumer surplus (economics).
4. Cardiac output (biology)
▶ As with our previous applications to geometry (areas, volumes, and
lengths) and to work, our strategy is:
1. to break up the physical quantity into a large number of small parts;
2. approximate each small part, add the results (giving a Riemann sum);
3. take the limit;
4. and then evaluate the resulting integral.
D-T. Nguyen (VNU-HCMUS)
Calculus II
64 / 108
Hydrostatic pressure and Force
▶ Deep-sea divers realize that water pressure increases as they dive
deeper. This is because the weight of the water above them increases.
▶ In general, suppose that a thin horizontal plate with area A square
meters is submerged in a fluid of density ρ kilograms per cubic meter
at a depth d meters below the surface of the fluid as in Figure 1.12.
▶ The fluid directly above the plate (think of a column of liquid) has
volume V = Ad, so its mass is m = ρV = ρAd.
▶ The force exerted by the fluid on the plate is therefore
F = mg = ρgAd
where g is the acceleration due to gravity.
Figure 1.12: Surface of fluid
D-T. Nguyen (VNU-HCMUS)
Calculus II
65 / 108
Hydrostatic pressure and Force
The pressure P on the plate is defined to be the force per unit area:
P=
F
= ρgd.
A
The SI unit for measuring pressure is a newton per square meter, which
is called a pascal (abbreviation: 1N/m2 = 1Pa). Since this is a small
unit, the kilopascal (kPa) is often used.
For instance, because the density of water is ρ = 1000kg/m3 , the pressure
at the bottom of a swimming pool 2m deep is
P = ρgd = 1000kg/m3 × 9.8m/s 2 × 2m = 19600Pa = 19.6kPa.
D-T. Nguyen (VNU-HCMUS)
Calculus II
66 / 108
Hydrostatic pressure and Force
An important principle of fluid pressure is the experimentally verified fact
that at any point in a liquid the pressure is the same in all directions.
A diver feels the same pressure on nose and both ears.
Thus the pressure in any direction at a depth d in a fluid with mass
density ρ is given by
P = ρgd = δd.
(1.8)
This helps us determine the hydrostatic force (the force exerted by a
fluid at rest) against a vertical plate or wall or dam. This is not a
straightforward problem because the pressure is not constant but increases as the depth increases.
D-T. Nguyen (VNU-HCMUS)
Calculus II
67 / 108
Hydrostatic pressure and Force
Example 1.19
A dam has the shape of the trapezoid shown in Figure 1.13. The height
is 20m and the width is 50m at the top and 30m at the bottom. Find
the force on the dam due to hydrostatic pressure if the water level is
4m from the top of the dam.
Figure 1.13
D-T. Nguyen (VNU-HCMUS)
Calculus II
68 / 108
Hydrostatic pressure and Force
Solution
Figure 1.14
D-T. Nguyen (VNU-HCMUS)
.
▶ We choose a vertical x-axis
with origin at the surface of the
water and directed downward as
in Figure 1.14.
▶ The depth of the water is 16m,
so we divide the interval [0, 16]
into subintervals of equal length
with endpoints xi and we choose
xi∗ ∈ [xi−1 , xi ].
Calculus II
69 / 108
Hydrostatic pressure and Force
Solution
▶ The i th horizontal strip of the dam is approximated by a rectangle with
height ∆x and width wi , where, from similar triangles in Figure 1.15
10
a
=
16 − xi∗
20
or
a=
16 − xi∗
x∗
=8− i
2
2
and so
.
wi = 2 (15 + a) = 46 − xi∗ .
▶ If Ai is the area of the i th strip,
then
Ai ≈ wi ∆x = (46 − xi∗ ) ∆x.
Figure 1.15
D-T. Nguyen (VNU-HCMUS)
Calculus II
70 / 108
Hydrostatic pressure and Force
Solution
▶ If ∆x is small, then the pressure Pi on the i th strip is almost constant and
we can use (1.8) to write
Pi ≈ 1000gxi∗ .
▶ The hydrostatic force Fi acting on the i th strip is the product of the pressure
and the area:
Fi = Pi Ai ≈ 1000gxi∗ (46 − xi∗ ) ∆x.
▶ Adding these forces and taking the limit as n → +∞, we obtain the total
hydrostatic force on the dam
F = lim
n→+∞
n
X
1000gxi∗ (46 − xi∗ ) ∆x =
Z 16
1000gx (46 − x) dx
0
i=1
Z 16
= 1000(9.8)
0
D-T. Nguyen (VNU-HCMUS)
53
≈ 4.43 × 107 N.
46x − x 2 dx = 9800 × 256 ×
3
Calculus II
71 / 108
Hydrostatic pressure and Force
Example 1.20
Find the hydrostatic force on one end of a cylindrical drum with radius
3ft if the drum is submerged in water 10ft deep.
D-T. Nguyen (VNU-HCMUS)
Calculus II
72 / 108
Hydrostatic pressure and Force
Solution
▶ It is convenient to choose the axes as in
Figure 1.16 so that the origin is placed at
the center of the drum.
▶ Then the circle has a simple equation,
x 2 + y 2 = 9.
Figure 1.16
▶ As in Example 1.19, we divide the circular region into horizontal strips of equal
width.
▶ From q
the equation of the circle, we see that the length of the i th
2
strip is 2 9 − yi∗ and so its area is
q
2
Ai = 2 9 − yi∗ ∆y .
D-T. Nguyen (VNU-HCMUS)
Calculus II
73 / 108
Hydrostatic pressure and Force
Solution
▶ Because the weight density of water is δ = 62.5Ib/ft 3 , the pressure
on this strip is approximately
δdi = 62.5 (7 − yi∗ )
and so the force on the strip is approximately
q
2
∗
δdi Ai = 62.5 (7 − yi ) 2 9 − yi∗ ∆y .
▶ The total force is obtained by adding the forces on all the strips and
taking the limit
q
n
X
2
F = lim
62.5 (7 − yi∗ ) 2 9 − yi∗ ∆y
n→+∞
i=1
Z 3
(7 − y )
= 125
−3
D-T. Nguyen (VNU-HCMUS)
p
7875
9 − y 2 dy = · · · =
πIb.
2
Calculus II
74 / 108
Hydrostatic pressure and Force
Exercise 1.5
A vertical plate is submerged in water and has the indicated shape.
Explain how to approximate the hydrostatic force against one side of
the plate by a Riemann sum. Then express the force as an integral and
evaluate it.
Figure 1.17
D-T. Nguyen (VNU-HCMUS)
Figure 1.18
Calculus II
Figure 1.19
75 / 108
Moments and Centers of Mass
Our main objective here is to find the point P on which a thin plate of
any given shape balances horizontally as in Figure 1.20.
Figure 1.20
This point is called the center of mass (or center of gravity) of the plate.
D-T. Nguyen (VNU-HCMUS)
Calculus II
76 / 108
Moments and Centers of Mass
We first consider the simpler situation illustrated in Figure 1.21.
Figure 1.21: Two masses m1 and m2 are attached to a rod of negligible mass
on opposite sides of a fulcrum and at distances d1 and d2 from the fulcrum.
The rod will balance if
m1 d1 = m2 d2 .
(1.9)
This is an experimental fact discovered by Archimedes and called
the Law of the Lever.
D-T. Nguyen (VNU-HCMUS)
Calculus II
77 / 108
Moments and Centers of Mass
Figure 1.22: Think of a lighter person balancing a heavier one on a seesaw by
sitting farther away from the center.
D-T. Nguyen (VNU-HCMUS)
Calculus II
78 / 108
Moments and Centers of Mass
Now suppose that the rod lies along the x-axis with m1 at x1 and m2
at x2 and the center of mass at x.
Figure 1.23
If we compare Figures 1.21 and 1.23, we see that d1 = x − x1 and
d2 = x2 − x and so Equation (1.9) gives
m1 (x − x1 ) = m2 (x2 − x)
m1 x1 + m2 x2
x=
.
m1 + m2
D-T. Nguyen (VNU-HCMUS)
Calculus II
(1.10)
79 / 108
Moments and Centers of Mass
The numbers m1 x1 and m2 x2 are called the moments of the masses m1
and m2 (with respect to the origin), and Equation (1.10) says that the
center of mass x is obtained by adding the moments of the masses and
dividing by the total mass m = m1 + m2 .
In general, if we have a system of n particles with masses m1 ,
m2 ,· · · , mn located at the points x1 , x2 , · · · , xn on the x-axis, it
can be shown similarly that the center of mass of the system is
located at
Pn
Pn
mi xi
mi xi
i=1
x = Pn
= i=1
(1.11)
m
i=1 mi
where m =
n
X
mi is the total mass of the system, and the sum of the
i=1
individual moments M =
n
X
mi xi is called the moment of the system
i=1
about the origin.
D-T. Nguyen (VNU-HCMUS)
Calculus II
80 / 108
Moments and Centers of Mass
Then Equation (1.11) could be rewritten as
mx = M,
which says that if the total mass were considered as being concentrated
at the center of mass x, then its moment would be the same as the
moment of the system.
D-T. Nguyen (VNU-HCMUS)
Calculus II
81 / 108
Moments and Centers of Mass
Now we consider a system of n particles with masses m1 , m2 , · · · , mn
located at the points (x1 , y1 ) , (x2 , y2 ) , · · · , (xn , yn ) in the xy -plane as
shown in Figure 1.24.
By analogy with the one-dimensional
case, we define the moment of the
system about the y -axis to be
n
X
My =
mi xi ,
(1.12)
Figure 1.24
Then My measures the
tendency of the system to
rotate about the y -axis and
Mx measures the tendency
to rotate about the x-axis.
D-T. Nguyen (VNU-HCMUS)
i=1
and the moment of the system about
the x-axis as
n
X
Mx =
mi yi .
(1.13)
i=1
Calculus II
82 / 108
Moments and Centers of Mass
As in the one-dimensional case, the coordinates (x, y ) of the center
of mass are given in terms of the moments by the formulas
x=
where m =
n
X
My
m
and
y=
Mx
.
m
(1.14)
mi is the total mass.
i=1
Since mx = My and my = Mx , the center of mass (x, y ) is the
point where a single particle of mass m would have the same moments as the system.
D-T. Nguyen (VNU-HCMUS)
Calculus II
83 / 108
Moments and Centers of Mass
Example 1.21
Find the moments and center of mass of the system of objects that have
masses 3, 4, and 8 at the points (−1, 1), (2, −1), and (3, 2), respectively.
Solution
We use Equations (1.12) and (1.13) to
compute the moments:
My = 3(−1) + 4(2) + 8(3) = 29
Mx = 3(1) + 4(−1) + 8(2) = 15
Since m = 3 + 4 + 8 = 15, we use Equation (1.14) to obtain
Figure
1.25:
The center of mass
My
29
Mx
29
x=
=
and y =
= 1 is
,1 .
m
15
m
15
D-T. Nguyen (VNU-HCMUS)
Calculus II
84 / 108
Moments and Centers of Mass
▶ Next we consider a flat plate (called a lamina) with uniform density
ρ that occupies a region R of the plane.
▶ We wish to locate the center of mass of the plate, which is called
the centroid of R.
▶ In doing so we use the following physical principles: The symmetry
principle says that if R is symmetric about a line l, then the centroid
of R lies on l.
If R is reflected about l, then R remains the same so its centroid remains
fixed. But the only fixed points lie on l.
▶ Thus the centroid of a rectangle is its center.
▶ Moments should be defined so that if the entire mass of a region
is concentrated at the center of mass, then its moments remain unchanged.
▶ Also, the moment of the union of two nonoverlapping regions should
be the sum of the moments of the individual regions.
D-T. Nguyen (VNU-HCMUS)
Calculus II
85 / 108
Moments and Centers of Mass
▶ Suppose that the region R is of the type
shown in Figure 1.26.
▶ We divide the interval [a, b] into n subintervals with endpoints x0 , x1 , x2 , · · · , xn
and equal width ∆x.
▶ We choose the sample point xi∗ to be
Figure 1.26: R lies between the
the midpoint x i of the i th subinterval, that
lines x = a and x = b, above the
xi−1 + xi
x-axis, and below the graph of f . is, x i =
.
2
▶ This determines the polygonal approximation to R shown in Figure 1.27.
▶ The centroid of the i th approximating
1
rectangle Ri is its center Ci x i , f (x i ) .
2
▶ Its area is f (x i ) ∆x, so its mass is
ρf (x i ) ∆x.
Figure 1.27
D-T. Nguyen (VNU-HCMUS)
Calculus II
86 / 108
Moments and Centers of Mass
The moment of Ri about the y -axis is the product of its mass and the
distance from Ci to the y -axis, which is x i . Thus
My (Ri ) = [ρf (x i ) ∆x] x i = ρx i f (x i ) ∆x.
Adding these moments, we obtain the moment of the polygonal approximation to R, and then by taking the limit as n → +∞ we obtain the
moment of R itself about the y -axis:
My = lim
n→+∞
D-T. Nguyen (VNU-HCMUS)
n
X
Z b
ρx i f (x i ) ∆x = ρ
xf (x) dx.
a
i=1
Calculus II
87 / 108
Moments and Centers of Mass
In a similar fashion we compute the moment of Ri about the x-axis as
the product of its mass and the distance from Ci to the x-axis (which
is half the height of Ri ):
Mx (Ri ) = [ρf (x i ) ∆x] f (x i ) = ρ ·
1
[f (x i )]2 ∆x.
2
Again we add these moments and take the limit to obtain the moment
of R about the x-axis:
Mx = lim
n→+∞
D-T. Nguyen (VNU-HCMUS)
n
X
i=1
1
ρ · [f (x i )]2 ∆x = ρ
2
Calculus II
Z b
a
1
[f (x)]2 dx.
2
88 / 108
Moments and Centers of Mass
Just as for systems of particles, the center of mass of the plate is defined
so that mx = My and my = Mx . But the mass of the plate is the
product of its density and its area:
Z b
m = ρA = ρ
f (x) dx,
a
and so
Rb
Rb
ρ a xf (x) dx
xf (x) dx
My
x=
= Rb
= Ra b
m
ρ a f (x) dx
a f (x) dx
Rb 1
Rb 1
2
2
ρ a 2 [f (x)] dx
Mx
a 2 [f (x)] xdx
y=
=
=
Rb
Rb
m
ρ f (x) dx
f (x) dx
a
a
Notice the cancellation of the ρ’s.
The location of the center of mass is independent of the density.
D-T. Nguyen (VNU-HCMUS)
Calculus II
89 / 108
Moments and Centers of Mass
In summary, the center of mass of the plate (or the centroid of R)
is located at the point (x, y ), where
1
x=
A
Z b
xf (x) dx
a
D-T. Nguyen (VNU-HCMUS)
and
1
y=
A
Calculus II
Z b
a
1
[f (x)]2 xdx.
2
(1.15)
90 / 108
Moments and Centers of Mass
Example 1.22
Find the center of mass of a semicircular plate of radius r .
Solution
▶ In order to use Equation (1.15) we place
the semicircle
as in Figure 1.28 so that
p
2
f (x) = r − x 2 and a = −r , b = r .
▶ Here there is no need to use the formula
to calculate x because, by the symmetry
principle, the center of mass must lie onFigure 1.28: The center of mass
is located at the point (0, 4r/3π)
the y -axis, so x = 0.
1 2
▶ The area of the semicircle is A = πr , so
2
Z
Z b p
2
1 b1
1
1
4r
y=
[f (x)]2 xdx = 1 2
r 2 − x 2 xdx = · · · =
.
A a 2
3π
a 2
2 πr
D-T. Nguyen (VNU-HCMUS)
Calculus II
91 / 108
Moments and Centers of Mass
If the region R lies between two curves
y = f (x) and y = g (x), where f (x) >
g (x), as illustrated in Figure 1.29, then the
same sort of argument that led to Equation (1.15) can be used to show that the
centroid of R is (x, y ), where
Figure 1.29
Z
1 b
x=
x [f (x) − g (x)] dx
A a
Z
o
1 b1n
y=
[f (x)]2 − [g (x)]2 dx.
A a 2
D-T. Nguyen (VNU-HCMUS)
Calculus II
(1.16)
92 / 108
Moments and Centers of Mass
Example 1.23
Find the centroid of the region bounded by the curves y = cos (x),
y = 0, x = 0, and x = π/2.
Example 1.24
Find the centroid of the region bounded by the line y = x and the
parabola y = x 2 .
D-T. Nguyen (VNU-HCMUS)
Calculus II
93 / 108
Moments and Centers of Mass
Example 1.25
Point-masses mi are located on the x-axis as shown. Find the moment
M of the system about the origin and the center of mass x.
Figure 1.30
Figure 1.31
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Moments and Centers of Mass
Example 1.26
The masses mi are located at the points Pi . Find the moments Mx and
My and the center of mass of the system.
(i) m1 = 4, m2 = 2, m3 = 4; P1 (2, −3), P2 (−3, 1), P3 (3, 5).
(ii) m1 = 5, m2 = 4, m3 = 3, m4 = 6; P1 (−4, 2), P2 (0, 5), P3 (3, 2),
P4 (1, −2).
Example 1.27
Sketch the region bounded by the curves, and visually estimate the
location of the centroid. Find the exact coordinates of the centroid
(i) y = 2x, y = 0, x = 1.
√
(ii) y = x, y = 0, x = 4.
(iii) y = e x , y = 0, x = 0, x = 1.
(iv) y = sin (x), y = 0, 0 ≤ x ≤ π
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Moments and Centers of Mass
We end this section by showing a surprising connection between centroids and volumes of revolution.
Theorem 1.3 (Theorem of Pappus)
Let R be a plane region that lies entirely on one side of a line l in the
plane. If R is rotated about l, then the volume of the resulting solid
is the product of the area A of R and the distance d traveled by the
centroid of R.
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Moments and Centers of Mass
Proof.
We give the proof for the special
case in which the region lies between y = f (x) and y = g (x) as
in Figure 1.32 and the line l is the
y -axis. Using the method of cylindrical shells, we have
Z b
V =
2πx [f (x) − g (x)] dx
a
Z b
= 2π
x [f (x) − g (x)] dx
Figure 1.32
a
= 2π (xA) = (2πx) A = Ad
where d = 2πx is the distance traveled by the centroid during one
rotation about the y -axis.
■
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Moments and Centers of Mass
Example 1.28
A torus is formed by rotating a circle of radius r about a line in the
plane of the circle that is a distance R(R > r ) from the center of the
circle. Find the volume of the torus.
Solution
The circle has area A = πr 2 . By the symmetry principle, its centroid is
its center and so the distance traveled by the centroid during a rotation
is d = 2πR. Therefore, by the Theorem of Pappus, the volume of the
torus is
V = Ad = (2πR) πr 2 = 2π 2 r 2 R.
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Consumer Surplus
▶ Let p (x) is the price that a company has to charge in order to sell x
units of a commodity.
▶ Usually, selling larger quantities requires lowering prices, so the demand
function is a decreasing function.
Figure 1.33: A typical demand curve.
▶ The graph of a typical demand function, called a demand curve.
▶ If X is the amount of the commodity that can currently be sold, then
P = p (X ) is the current selling price.
▶ At a given price, some consumers who buy a good would be willing to pay
more; they benefit by not having to.
▶ The difference between what a consumer is willing to pay and what the
consumer actually pays for a good is called the consumer surplus.
▶ By finding the total consumer surplus among all purchasers of a good,
economists can assess the overall benefit of a market to society.
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Consumer Surplus
Figure 1.34
▶ To determine the total consumer surplus, we look at the demand curve and divide the interval [0, X ] into n subintervals,
each of length ∆x = X/n.
▶ Let xi∗ = xi be the right endpoint of the
i th subinterval, as in Figure 1.34.
▶ According to the demand curve, xi−1
units would be purchased at a price of
p (xi−1 ) dollars per unit.
▶ To increase sales to xi units,the price would have to be lowered to
p (xi ) dollars.
▶ In this case, an additional ∆x units would be sold (but no more).
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Consumer Surplus
▶ In general, the consumers who would have paid p (xi ) dollars placed
a high value on the product; they would have paid what it was worth
to them.
▶ So in paying only P dollars they have saved an amount of
(saving per unit) (number of units) = [p (xi ) − P] ∆x.
▶ Considering similar groups of willing
consumers for each of the subintervals and
addingthe savings, we get the total savings
n
X
[p (xi ) − P] ∆x.
i=1
Figure 1.35
D-T. Nguyen (VNU-HCMUS)
This sum corresponds to the area enclosed
by the rectangles in Figure 1.35.
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Consumer Surplus
If we let n → +∞, this Riemann sum approaches the integral
Z X
[p (x) − P] dx
(1.17)
0
which economists call the consumer surplus for the commodity.
The consumer surplus represents
the amount of money saved by
consumers in purchasing the commodity at price P, corresponding
to an amount demanded of X .
Figure 1.36 shows the interpretation of the consumer surplus as
the area under the demand curve
and above the line p = P.
Figure 1.36
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Consumer Surplus
Example 1.29
The demand for a product, in dollars, is
p = 1200 − 0.2x − 0.0001x 2 .
Find the consumer surplus when the sales level is 500.
Solution
Since the number of products sold is X = 500, the corresponding price is
P = 1200 − (0.2)(500) − (0.0001)(500)2 = 1075.
Therefore, from (1.17), the consumer surplus is
Z X
Z 500
[p (x) − P] dx =
0
1200 − 0.2x − 0.0001x 2 − 1075 dx
0
= $33, 333.33.
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Cardiac output
Figure 1.37: The human cardiovascular system.
D-T. Nguyen (VNU-HCMUS)
▶ Blood returns from the body
through the veins, enters the
right atrium of the heart, and is
pumped to the lungs through the
pulmonary arteries for oxygenation.
▶ It then flows back into the
left atrium through the pulmonary
veins and then out to the rest of
the body through the aorta.
▶ The cardiac output of the
heart is the volume of blood
pumped by the heart per unit
time, that is, the rate of flow into
the aorta.
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Cardiac output
▶ The dye dilution method is used to measure the cardiac output.
▶ Dye is injected into the right atrium and flows through the heart into
the aorta.
▶ A probe inserted into the aorta measures the concentration of the
dye leaving the heart at equally spaced times over a time interval [0, T ]
until the dye has cleared.
▶ Let c (t) be the concentration of the dye at time t.
▶ If we divide [0, T ] into subintervals of equal length ∆t, then the
amount of dye that flows past the measuring point during the subinterval
from t = ti−1 to t = ti is approximately
(concentration) (volume) = c (ti ) (F ∆t)
where F is the rate of flow that we are trying to determine.
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Cardiac output
Thus the total amount of dye is approximately
n
X
c (ti ) F ∆t = F
i=1
n
X
c (ti ) ∆t,
i=1
and, letting n → +∞, we find that the amount of dye is
Z T
A=F
c (t) dt.
0
Thus the cardiac output is given by
F = RT
0
A
(1.18)
c (t) dt
where the amount of dye A is known and the integral can be approximated from the concentration readings.
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Cardiac output
Example 1.30
A 5-mg bolus of dye is injected into a right atrium. The concentration
of the dye (in milligrams per liter) is measured in the aorta at one second
intervals as shown in the table. Estimate the cardiac output.
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Cardiac output
Solution
Here A = 5, ∆t = 1, and T = 10. We use Simpson’s Rule to approximate the integral of the concentration
Z 10
1
c (t) dt ≈ [0 + 4(0.4) + 2(2.8) + 4(6.5) + 2(9.8) + 4(8.9)
3
0
+2(6.1) + 4(4.0) + 2(2.3) + 4(1.1) + 0]
= 41.87
Thus (1.18) gives the cardiac output to be
F = R 10
0
D-T. Nguyen (VNU-HCMUS)
A
c (t) dt
≈
5
= 0.12L/s = 7.2L/min.
41.87
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