Antiderivatives and Indefinite Integration Definition of an Antiderivative A function F is an antiderivative of f on an interval I if Fx f x for all x in I Theorem Representation of ID: 920481
Download Presentation The PPT/PDF document "Chapter 6 Integration Section 6.1" is the property of its rightful owner. Permission is granted to download and print the materials on this web site for personal, non-commercial use only, and to display it on your personal computer provided you do not modify the materials and that you retain all copyright notices contained in the materials. By downloading content from our website, you accept the terms of this agreement.
Slide1
Chapter 6
Integration
Slide2Section 6.1
Antiderivatives
and Indefinite Integration
Slide3Definition of an
Antiderivative
A function F is an
antiderivative
of
f
on an interval I if F’(x) =
f
(x) for all x in I.
Slide4Theorem – Representation of
Antiderivtives
If
F
is an
antiderivative
of
f
on an interval
I
, then
G
is an
antiderivative
of
f
on the interval
I
if and only if
G
is on the form:
G(x) = F(x) + C
, for all
x
in
I
where
C
is a constant
Slide5Terminology
C
- is called the
constant of integration
G(x) = x
2
+ c is the
general solution of the
differential equation G’(x) = 2x
Integration is the “inverse” of Differentiation
Slide6Differential Equation
Is an equation that involves
x, y
and derivatives of
y.
EXAMPLES
y’ = 3x and y’ = x
2
+ 1
Slide7Notation for
Antiderivatives
When solving a
differential
equation of the form
dy
/
dx
= f(x)
you can write
dy
=
f(x)
dx
and is called
antidifferentiation
and is denoted by an integral sign
∫
Slide8Integral Notation
y =
∫
f(x)
dx
= F(x) + C
f(x) – integrand
dx
– variable of integration
C – constant of integration
Slide9Basic Integration Rules
y =
∫
F’(x)
dx
= F(x) + C
f(x) – integrand
dx
– variable of integration
C – constant of integration
Slide10Basic Integration Rules
Differentiation formula
Integration formula
d/
dx
[c
] = 0
d/
dx
[
kx
] = k
d/
dx
[
kf
(x)] =
kf
’(x)
d/
dx
[f(x) ± g(x)] = f’(x) ± g’(x)d/dx [xn] = nxn-1
∫ 0
dx
= c
∫
kdx
=
kx
+ c
∫
kf
(x)
dx
= k ∫ f(x)
dx
∫ f(x) ± g(x)]
dx
= ∫ f(x)
dx
± ∫ g(x)
dx
∫
x
n
dx
= (x
n+1
)/(n+1) + c, n≠ -
1
(Power Rule)
Slide11Examples
∫ 3x
dx
∫ 1/x
3
dx
∫ (x + 2)
dx
∫ (x + 1)/√x
dx
Slide12Finding a Particular Solution
EXAMPLE
F’(x) = 1/x
2
, x > 0 and find the particular solution that satisfies the initial condition F(1) = 0
Slide13Finding a Particular Solution
Find the general solution by integrating, -
1/x
+ c. x > 0
Use initial condition F(1) = 0 and solve for c, F(1) = -1/1
+ c
, so c = 1
Write the particular solution F(x) = - 1/x + 1, x > 0
Slide14Solving a Vertical Motion Problem
A ball is thrown upward with an initial velocity of 64 ft/sec from an initial height of 80 ft.
Find the position function giving the height
s
as a function of the time
t
When does the ball hit the ground?
Slide15Solution
Let t = 0 represent the initial time; s(0) = 80 and s’(0) = 64
Use -32 ft/sec as the acceleration due to gravity, then s”(t) = - 32
∫ s”(t)
dt
= ∫ -32
dt
= -32 t + c
s’(0) =64 = -32(0) = c, so c = 64
s(t) = ∫ s’(t) = ∫ (-32t + 64)
dt
= -16t
2
+ 64t + C
s(0) = 80 = -16(0
2
) + 64(0) + C, hence C = 80
s(t) = -16t
2
+ 64t + 80 = 0, solve and t = 5
Slide16Section 6.2
Area
Slide17Sigma Notation
The sum of
n
terms
a
1
, a
2
, a
3
…,a
n
is written as
n
∑
a
i
= a
1
+ a
2
+ a
3
+ …+ a
n
i
= 1
Where
i
the
index of summation
,
a
1
is the
i
th
term
of the sum, and the
upper and lower bounds of summation
are n and 1, respectively.
Slide18EXAMPLES
6
∑
i
= 1 + 2 + 3 + 4 + 5 + 6
i
= 1
5
∑ (
i
+ 1)= 1 + 2 + 3 + 4 + 5 + 6
i
= 0
7
∑
j
2
= 9 + 16 + 25 + 36 + 49
j = 3
Slide19SUMMATION FORMULAS
n
∑
c =
cn
i
= 1
n
∑
i
= n(n + 1)/2
i
= 1
n
∑
i
2
= n(n + 1)(n + 2)(2n + 1)/6
i
= 1
n
∑
i
3
= n
2
(n + 1)
2
/4
i
= 1
Slide20PROPERTIES OF SUMMATION
n
n
∑
ka
i
= k
∑
a
i
i
= 1
i
= 1
n
n
n
∑
(
a
i
±
b
i
) =
∑
a
i
±
∑
b
i
i
= 1
i
= 1
i
= 1
Slide21EXAMPLE
Find the
sum for n = 10 and n = 100
n
∑ (
i
+ 1)/n
2
i
= 1
Slide22AREA OF A PLANE REGION
Find the area of the region lying between the graph of f(x) = - x
2
+ 5 and the x-axis between x = 0 and x = 2 using five rectangles to find an approximation of the area. You should use both inscribed rectangles and circumscribed rectangles. In doing so you will be able to find a lower and upper sum.
Slide23Limits of the Lower and Upper Sums
Let f be continuous and nonnegative on the interval [
a,b
]. The limits as n→∞
of
both the lower and upper sums exist and are equal to each other. That is,
n
lim
s(
n
) =
lim
∑
f
(m
i
)
x and
n→∞
n→∞
i
= 1
Slide24Limits of the Lower and Upper Sums
n
lim
s(
n
) =
lim
∑
f
(M
i
)
x
n→∞
n→∞
i
= 1
n
lim
S(
n
) =
lim
∑
f
(Mi)
x
n→∞ n→∞
i
= 1
Slide25Definition of the Area of a Region in the Plane
Let
f
be continuous and nonnegative on the interval [
a,b
]. The area of the region bounded by the graph of
f
, the x-axis, and the vertical lines x= a and x = b is
n
Area =
lim
∑ f
(
c
i
)
x, x
i -1
c
i
x
i
n→∞
i
= 1
Where
x = (b-a)/n
Slide26EXAMPLE
Fine the area of the region bounded by the graph of f(x) =x
3
, the x-axis, and the vertical lines x=0 and x =1
.
Partition the interval [0,1] into n subintervals each of width 1/n =
x
Simplify using the formula below and A = 1/4
n
Area =
lim
∑ f
(
c
i
)
x, x
i -1
c
i
x
i
n→∞
i
= 1
Where
x = (b-a)/n
Slide27Section 6.3
Riemann Sums and Definite Integrals
Slide28Definition of Riemann Sum
Let
f
be defined on the closed interval [
a, b
] and let
be a partition of [
a,b
] given by
a = x
o
< x
1
< x
2
< …<x
n-1
<
x
n
=b
Where xi is the width of the ith subinterval. If c
i
is
any
point in the
i
th
subinterval, then the sum
n
f
(
c
i
) x
i
, x
i-1
c
i
x
i
is called a
Riemann
i
= 1
sum
of
f
for the partition
Slide29Definition of the Norm
The width of the largest subinterval of a partition
is the
norm
of the partition
and is denoted by
. If every subinterval is of equal width, the partition is regular and the norm is denoted by
= x = (b – a)/n
Slide30Definite Integrals
If
f
is defined on the closed interval [
a, b
] and the
limit
n
lim
∑ f
(
c
i
)
x
→ 0
i
= 1
exists, then
f
is
integrable
on [
a,b
] and the limit is
b
∫
f(x)
dx
a
The number
a
is the lower limit of integration, and the number
b
is the upper limit of integration
Slide31Continuity Implies Integrability
If a function f is continuous on the closed interval [
a,b
], then f is
integrable
on ]
a,b
]
Slide32Evaluating a Definite Integral
Evaluate the definite integral
1
∫
2xdx
-2
Slide33The Definite Integral as the Area of a Region
If
f
is
continuous and nonnegative closed
interval [
a, b
]
the
the
area of the region bounded by the graph of
f
, the x-axis, and the vertical lines x =a and x = b is given by
b
∫
f(x)
dx
a
Slide34Examples
Evaluate the Definite Integral
0
∫
(
x + 2)
dx
3
Sketch the region and use formula for trapezoid
Slide35Additive Interval Property
If
f
is
integrable
on the three closed intervals determined by
a
,
b
and
c
, then ,
b
c
b
∫
f(x)
dx
= ∫
f(x)
dx
=
∫
f(x)
dx
a
a
c
Slide36Properties of Definite Integrals
If
f
and
g
are
integrable
on [
a,b
] and
k
is a constant, then the functions of
kf
and
f ± g
are
integrable
on [
a,b
], and
b b 1. ∫kf dx
= k
∫
f(x)
dx
a
a
Slide37Properties of Definite Integrals
If
f
and
g
are
integrable
on [
a,b
] and
k
is a constant, then the functions of
kf
and
f ± g
are
integrable
on [
a,b
], and
b b b1. ∫[f(x) ± g(x)]
dx
=
∫
f(x)
dx
±
∫
f(x)
dx
a
a
a
Slide38Section 6.4
THE FUNDAMENTAL THEOREM OF CALCULUS
Slide39The Fundamental Theorem of Calculus
If a function
f
is continuous on the closed interval [
a,b
] and F is an
antiderivative
of
f
on the interval [
a,b
], then
b
∫
f(x)
dx
= F(b) – F(a)
a
Slide40Using the Fundamental Theorem of Calculus
Find the
antiderivative
of
f
if possible
Evaluate the definite integral
Example: ∫ x
3
dx on the interval [1,3]
Slide41Using the Fundamental Theorem of Calculus to Find Area
Find the area of the region bounded by the graph of y = 2x
2
– 3x +2, the x-axis, and the vertical lines x=0 and x= 2.
Graph
Find the
antiderivative
Evaluate on your interval
Slide42Mean Value Theorem for Integrals
If
f
is continuous on the closed interval [
a,b
], then there exists a number
c
in the closed interval [
a,b
] such that
∫
f(x)
dx
= f(c)(b-a)
Slide43Average Value of a Function on an Interval
If
f
is
integrable
on the closed interval [
a,b
], then the
average value
of
f
on the interval is
b
1/(b-a)
∫
f(x)
dx
a
Slide44The Second Fundamental Theorem of Calculus
If
f
is continuous on an open interval
I
containing
a
, then, for every
x
in the interval
x
d/
dx
[
∫
f(t)
dt
]
= f(x)
a
Slide45Section 6.5
INTEGRATION BY SUBSTITUTION
Slide46Antidifferentiation of a Composite Function
Let
g
be a function whose range is an interval
I
, and let
f
be a function that is continuous on
I
. If
g
is differentiable on its domain and
F
is an
antiderivative
of
f
on
I
, then
∫ f(g(x))g’(x)dx = F(g(x)) + CIf u = g(x), then du = g’(x)dx and ∫ f(u)du = F(u) + C
Slide47Change in Variable
You completely rewrite the integral in terms of
u
and
du
.
This is useful technique for complicated
intergrands
.
∫
f(g(x))g’(x)
dx
=
∫
f(u) du = F(u)+ C
Slide48Example
Find
∫
(2x – 1)
.5
dx
Let u = 2x - 1, then du/
dx
= 2dx/
dx
Solve for
dx
and substitute back to obtain the
antiderivative
.
Check your answer.
Slide49Power Rule for Integration
If
g
is a differentiable function of
x
, then,
∫
((g(x))
n
g
’(x)
dx
=
∫
(g(x))
n+1
/(n+1) + C
Slide50Change of Variables for Definite Integrals
If the function
u = g(x)
has a continuous derivative on the closed interval [
a,b
] and f is continuous on the range of
g
, then,
b g(b)
∫
(g(x)g’(x)
dx
=
∫
f(u)du
a g(a)
Slide51Integration of Even and Odd Functions
Let
f
be
integrable
on the closed interval [ -
a,a
].
If
f
is an
even
function, then
a
a
∫
f(x) dx=2 ∫ f(x) dx -a 0
Slide52Integration of Even and Odd Functions
Let
f
be
integrable
on the closed interval [ -
a,a
].
If
f
is an
odd
function, then
a
∫
f(x)
dx= ∫ f(x) dx = 0 -a
Slide53Section 6.6
NUMERICAL INTEGRATION
Slide54Trapezoidal Rule
Let f be continuous on [
a,b
]. The trapezoidal Rule for approximating
∫
f(x)
dx
(b-a)/2n [f(x
0
) = 2(f(x
1
) +…..+2f(x
n-1
) + f(
x
n
)]
Slide55Simpson’s Rule
If p(x) = Ax2 +
Bx
+ c, then
b
∫
p(x)
dx
=
a
(b-a)/6 [p(a) + 4p[(
a+b
)]/2) + p(b)]
Slide56END