Announcements
Topics:
- sections 7.1 (differential equations), 7.2
(antiderivatives), and 7.3 (definite integrals)
* Read these sections and study solved examples in your
textbook!
Work On:
- Practice problems from the textbook and assignments
from the coursepack as assigned on the course web
page (under the link “SCHEDULE + HOMEWORK”)
Differential Equations
A differential equation is an equation that
involves an unknown function and one or
more of its derivatives.
Examples:
y"2xy  x 2
y' 2  y


y' x 2  e x
Differential Equations
A solution of a differential equation is a function
that, along with its derivatives, satisfies the DE.
Example:
x 3
Show that y  2  e is a solution of the
differential equation y'3x 2 y  6x 2 and initial
condition y(0)  3.



Solutions for General DEs

Algebraic Solutions


Geometric Solutions


an explicit formula or algorithm for the solution (often,
impossible to find)
a sketch of the solution obtained from analyzing the DE
Numeric Solutions

an approximation of the solution using technology and
and some estimation method, such as Euler’s method
Graphical Solutions of Pure-Time DEs
1. Graph the derivative.
2. Create a chart relating information about the
derivative to information about the solution.
3. Sketch the solution using the initial condition
to ‘anchor’ the graph.
Graphical Solutions of Pure-Time DEs
Example:
Sketch the graph of the solution to s'(t)  3 t
given the initial condition s(0) 1.



Euler’s Method
What information does an initial value problem
tell us about the solution?
Example:
dy
DE:
xy
dx
IC:
y(0)  1
slope of the solution curve y(x)
an exact value of the solution
Euler’s Method
Euler’s Idea:
First, using the initial
condition as a base point,
approximate the solution
curve y(x) by its tangent
line.
First Euler approximation
Euler’s Method
Next, travel a short
distance along this line,
determine the slope at the
new location (using the
DE), and then proceed in
that ‘corrected’ direction.
Euler’s approximation with step size x  0.5

Euler’s Method
Repeat, correcting your
direction midcourse using
the DE at regular intervals
to obtain an approximate
solution of the IVP.
By increasing the number
of midcourse corrections,
we can improve our
estimation of the solution.
Euler approximation with step size x  0.25

Euler’s Method
Summary:
An approximate solution to the IVP
dy
 G(t, y), y(t 0 )  y 0
dt
is generated by choosing a step size t
and computing values according to the

algorithm
tn 1  t n  t

y n 1  y n  G(tn , y n )t
Euler’s Method
Algorithm:
tn 1  t n  t
y n 1  y n  G(tn , y n )t
Algorithm In Words:
next
 time = current time + step size
next approximation = current approximation + rate of
change at current values x step size
Example
Consider the IVP
y'  t  3, y(0)  2
Approximate the value of the solution at t=1 by
applying
Euler’s method and using a step size of 0.25.

Example
Calculations:
Table of Approximate Values for the
Solution y(t) of the IVP
tn
yn
t0 = 0
y0 = 2
Example
Graph of Approximate
Solution:
Plot points and connect with
straight line segments.
6
5

4

tn
yn
t0 = 0
y0 = 2
t1 = 0.25
y1 = 2.75
t2 = 0.5
y2 = 3.5625
t3 = 0.75
y3 = 4.4375

t4 = 1
y4 = 5.375

3

2


1
0

0.25


0.5
0.75

1
Determining Properties of a Solution
Example #36, p. 417
A population P(t) of
caribou is modeled by the
autonomous DE
  P(t) 
P'(t)  2P(t)1
, P(t)  0.
 2500 
Analyze this equation to
describe the behaviour of
the population of caribou.
Solving Pure-Time DEs
The general form of a pure-time differential
equation is
dF
 f (x)
dx
where F(x) is the unknown state variable and f (x)
is the measured rate of change.
Examples:

 (a) dF  4x 3  1
dx
(b)
dy  x
1
 5e 
dx
1 x 2
Solving Pure-Time DEs
Solve each by “guess and check”.
Ask yourself:
“What function has this as its derivative?”
Examples:
dF
(a)
 4x 3  1
dx
dy
1
x
(b)
 5e 
dx
1 x 2
Antiderivatives/Indefinite Integrals
An antiderivative (or indefinite integral) of a
function f (x) is a function F(x) with derivative
equal to f (x) .


F'(x)  f (x)  
F(x) 
 f (x)dx
An antiderivative F(x) is a solution to the puretime differential equation

dF
 f (x).

dx
Initial Value Problems
A differential equation has a
whole family of solutions.
Example:
y' 2x  4
y  x2  4x  C
Initial Value Problems
An initial value problem
provides an initial condition
so you can find a particular
solution.
Example:
y' 2x  4, y(0)  3
y  x2  4x  3
Initial Value Problems
An initial value problem
provides an initial condition
so you can find a particular
solution.
Example:
y' 2x  4, y(0)  3
y  x2  4x  3
Antiderivatives/Indefinite Integrals
Theorem 6.1:
If F(x) is an antiderivative of f(x), then the most
general antiderivative of f(x) is F(x)+C; i.e.,
 f (x)dx  F(x)  C
where C is a real number.
If an 
initial value of the solution is given, then we
can solve for C to find a specific or particular
antiderivative of f(x).
Rules for Antiderivatives
The Power Rule for Integrals

n 1
x
x n dx 
C
n 1
for n  1
Example:

Integrate each.
(a)
 x dx
7
(b)

1
dt
4
t
(c)

xdx
Rules for Antiderivatives
The Constant Product Rule for Integrals
Suppose
Then
 f (x)dx  F(x)  C.
 af (x)dx  aF(x)  C'. for any constant a.

The Sum Rule for Integrals

Suppose
Then
 f (x)dx  F(x)  C
and

 g(x)dx  G(x)  C'.
 [ f (x)  g(x)]dx   f (x)dx   g(x)dx  F(x)  G(x)  C''.

Some More Examples
Example 1:
Integrate.
(a)
3
 (5x  x )dx
4
(b)
Example 2:
Solve the differential equation
x
1
f '(x)  2 
2 x
with initial condition f (0)  0.

2
4x
(sec
x

e
)dx

Application:
A Differential Equation for AIDS
During the early years of the AIDS epidemic, the
number of new AIDS cases per year was  523.8t 2,
where t is measured in years since the beginning
of 1981.

rate at which new AIDS cases were reported:
dA
2
 523.8t
dt
Application:
A Differential Equation for AIDS
Surveys indicated that about 340 people had
been infected at the beginning of 1981.
Initial condition: A(0)  340
Solution:
the number of AIDS cases t
A(t) represents

years after 1981.
A(t) 174.6t 3  340
Summary Of Some Basic
Integration Formulas

n 1
x
x n dx 
C
n 1
x
 cos xdx  sin x  C


 sin xdx  cos x  C
 sec
2
xdx  tan x  C
for n  1


1
dx 

1
dx  ln x  C
x
x
x
e
dx

e
C

1
 1 x 2 dx  arctan x  C
Area
How do we calculate the
area of some irregular
shape?
For example, how do we
calculate the area under
the graph of f on [a,b]?
Area  ?

Area
Approach:
We approximate the area using rectangles.
number of
rectangles:
n4
width of each
rectangle:
ba
x 
n

x0 
x1
x2
x3
 x4
Area
Left-hand estimate:
Let the height of each rectangle be given by the value
of the function at the left endpoint of the interval.
x0 
x1
x2
x3
 x4


Area
Left-hand estimate:
Area  f (x0 )x  f (x1)x  f (x2 )x  f (x3 )x
 ( f (x0 )  f (x1)  f (x2 )  f (x3 ))x
3
  f (x i )x
i 0
Riemann Sum
Area
Right-hand estimate:
Let the height of each rectangle be given by the value
of the function at the right endpoint of the interval.
x0 
x1
x2
x3
 x4


Area
Right-hand estimate:
Area  f (x1)x  f (x2 )x  f (x3 )x  f (x4 )x
 ( f (x1)  f (x2 )  f (x3 )  f (x4 ))x
4
  f (x i )x
i1
Riemann Sum
Area
Midpoint estimate:
Let the height of each rectangle be given by the value
of the function at the midpoint of the interval.
x1
x2
x3
x4
Area
Midpoint estimate:
Area  f (x )x  f (x )x  f (x )x  f (x )x
*
1
*
2
*
3
*
4
 ( f (x )  f (x )  f (x )  f (x ))x
*
1
*
2
*
3
*
4
4
  f (x )x
*
i


i1
Riemann Sum
Area
How can we improve our estimation?
Increase the number of rectangles!!!
16
Area   f (t )t
*
i
i1
How do we make it exact?

Let the number of rectangles go to infinity!!!
Area
How can we improve our estimation?
Increase the number of rectangles!!!
16
Area   f (x )x
*
i
i1
How do we make it exact?

Let the number of rectangles go to infinity!!!
Area
How can we improve our estimation?
Increase the number of rectangles!!!
How do we make it exact?
Let the number of rectangles go to infinity!!!
Area
How can we improve our estimation?
Increase the number of rectangles!!!
16
Area   f (x )x
*
i
i1
How do we make it exact?

Area
How can we improve our estimation?
Increase the number of rectangles!!!
16
Area   f (x )x
*
i
i1
How do we make it exact?

Let the number of rectangles go to infinity!!!
Riemann Sums and the
Definite Integral
Definition:
The definite integral of a function f on the
interval from a to b is defined as a limit of the
Riemann sum
b

n
*

f (x)dx  lim  f (x i )x
n 
a
*
i
i1
where x is some sample point in the interval
[x i1, x i ] and x  b  a .

n
The Definite Integral
Interpretation:
If f  0 , then the definite
integral is the area under
the curve y  f (x) from
a to b.
Area 
b
 f (x)dx
a


Example
Estimate the following definite integrals using
left-endpoints, midpoints, and right-endpoints
and the indicated number of intervals.
5
(a)  (0.5t  2)dt,
1
n5
(b)  e dt,
0
0

t 2
n4
5
Estimating  (0.5t  2)dt Using Left-Endpoints
0


50
t 
1
5
5
Estimating  (0.5t  2)dt Using Left-Endpoints
0

50
t 
1
5

L5  f (0) 1 f (1) 1 f (2) 1 f (3) 1 f (4) 1
 2  2.5  3  3.5  4
 15
5
Estimating  (0.5t  2)dt Using Right-Endpoints
0


50
t 
1
5
5
Estimating  (0.5t  2)dt Using Right-Endpoints
0

50
t 
1
5

R5  f (1) 1 f (2) 1 f (3) 1 f (4) 1 f (5) 1
 2.5  3  3.5  4  4.5
 17.5
5
Estimating  (0.5t  2)dt Using Midpoints
0
50
t 
1

5

5
Estimating  (0.5t  2)dt Using Midpoints
0
50
t 
1

5

M 5  f (0.5) 1 f (1.5) 1 f (2.5) 1 f (3.5) 1 f (4.5) 1
 2.25  2.75  3.25  3.75  4.25
16.25
1
Estimating  e
t 2
dt
0
t 
1 0
 0.25
4 
L4  ( f (0)  f (0.25)  f (0.5)  f (0.75))  0.25
 (e0  e0.25  e0.5  e0.75 )  0.25
 0.822
2
2
2
2
Using Left-Endpoints
1
Estimating  e
t 2
dt
0
t 
1 0
 0.25
4 
L4  ( f (0)  f (0.25)  f (0.5)  f (0.75))  0.25
 (e0  e0.25  e0.5  e0.75 )  0.25
 0.822
2
2
2
2
Using Left-Endpoints
1
Estimating  e
t 2
dt
0
t 
1 0
 0.25
4 
R4  ( f (0.25)  f (0.5)  f (0.75)  f (1))  0.25
 (e0.25  e0.5  e0.75  e1 )  0.25
 0.664
2
2
2
2
Using Right-Endpoints
1
Estimating  e
t 2
dt
0
t 
1 0
 0.25
4 
R4  ( f (0.25)  f (0.5)  f (0.75)  f (1))  0.25
 (e0.25  e0.5  e0.75  e1 )  0.25
 0.664
2
2
2
2
Using Right-Endpoints
1
Estimating  e
t 2
dt
Using Midpoints
0
t 
1 0
 0.25
4 
M 4  ( f (0.125)  f (0.375)  f (0.625)  f (0.875))  0.25
 (e0.125  e0.375  e0.625  e0.875 )  0.25
 0.749
2
2
2
2
1
Estimating  e
t 2
dt
Using Midpoints
0
t 
1 0
 0.25
4 
M 4  ( f (0.125)  f (0.375)  f (0.625)  f (0.875))  0.25
 (e0.125  e0.375  e0.625  e0.875 )  0.25
 0.749
2
2
2
2
The Definite Integral
Interpretation:
If f is both positive and
negative, then the definite
integral represents the
NET or SIGNED area, i.e.
the area above the x-axis
and below the graph of f
minus the area below the
x-axis and above f
4

1
f (x)dx  net area
Definite Integrals and Area
Example:
Evaluate the following integrals by interpreting
each in terms of area.
3
1
(a)

1  x 2 dx
(b)
0
0
(c)


 sin x dx

 (x 1) dx

Properties of Integrals
Assume that f(x) and g(x) are continuous
functions and a, b, and c are real numbers such
that a<b.
a
(1)
 f (x) dx  0
a
(2)
(3)
b
a
a
b
 f (x) dx    f (x) dx
b
b
a
a
 c f (x) dx  c  f (x) dx
Properties of Integrals
Assume that f(x) and g(x) are continuous
functions and a, b, and c are real numbers such
that a<b.
(4)
b
b
b
a
a
a
  f (x) g(x)dx   f (x) dx   g(x) dx
b
(5)
 c dx  c(b  a)
a
Summation Property of the
Definite Integral
(6) Suppose f(x) is continuous on the interval
from a to b and that a  c  b.

Then
b
c
 f (x) dx
 f (x) dx
a
c
b 
c
a
a

b
 f (x) dx   f (x) dx   f (x) dx .
c
Properties of the Definite Integral
(7) Suppose f(x) is continuous on the interval
from a to b and that m  f (x)  M.

Then m(b  a) 
b
 f (x) dx  M(b  a).
a