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Today’s Lecture
•Recalls
•Improper integrals
•Introduction to vectors
•Dot product of vectors
•Cross product of vectors
When evaluating integrals of the form • If the power of sine is odd save one sine factor. • Use sin2x = 1 - cos2x to express the remaining factors in terms of
cosine:
• Then, substitute . • If the powers of both sine and cosine are even, use the half-angle
identities: • Sometimes, it is helpful to use the identity
• This also works for integrals of the form
and .
2 1 2
2
sin cos (sin ) cos sin
(1 cos ) cos sin
k n k n
k n
x x dx x x x dx
x x x dx
12sin cos sin 2x x x
Recalls
Example: Evaluate.
Solution: We use the identity . Then, substitute
so that .
6 4 6 2 2
6 2 2
6 2 6 8
7 9
7 91 17 9
tan sec tan sec sec
tan (1 tan )sec
(1 ) ( )
7 9
tan tan
x x dx x x x dx
x x x dx
u u du u u du
u uC
x x C
Recalls
We use the following identities to evaluate special of trigonometric integrals.
Integral Identity
12
sin cos
sin( ) sin( )
A B
A B A B
12
sin sin
cos( ) cos( )
A B
A B A B
12
cos cos
cos( ) cos( )
A B
A B A B
Recalls
Sometimes, it is helpful to use following trigonometric substitution.
In each case, the restriction on is imposed to ensure that the function that defines the substitution is one-to-one.
Recalls
How to find partial fractions:• Linear factors
• Power of a linear factor
• Quadratic factor, we break it down to partial fractions as follows:
2 20
( 2)( 5) 2 2
x A B
x x x x
3 2 3
2 20
( 2) 2 ( 2) ( 2)
x A B C
x x x x
2 2
2 20
( 2)( 1) 2 1
x A Bx C
x x x x
Recalls
Algorithm for integrating rational functions:
Integration of a rational function , where and are polynomials can be performed as follows.
• If perform polynomial division and write , where and are polynomials with
• Integrate the polynomial .
• Factorize the polynomial .
• Perform partial fraction decomposition of .
• Integrate the partial fraction decomposition.
Recalls
In defining a definite integral we dealt with a function defined on a finite interval and we assumed that does not have an infinite discontinuity.
Definition (informal): An integral is called improper integralif either• The interval is infinite, or • Function has an infinite discontinuity in .
One of the most important applications of this idea is in probability distributions.
Improper integrals
Improper integrals
Example: Find the area of the region bounded under the curve , above the axis and by the line to the right.
Solution: The area of region S that lies to the left of the line is:
Notice that no matter how large is chosen.
211
1 1 1( ) 1
tt
A t dxx x t
Improper integrals
Cont: We also observe that
• The area of the shaded region approaches as .• So, we say that the area of the infinite region S is equal to 1 and
we write:
1lim ( ) lim 1 1t t
A tt
2 21 1
1 1lim 1
t
tdx dx
x x
Improper integralsImproper integral of type I:
If exists for every number , then
provided this limit exists (as a finite number).
If exists for every number , then
provided this limit exists (as a finite number).
Improper integralsImproper integral of type I:
An improper integral is called convergent if the corresponding limit exists. Otherwise it is divergent.
If both and are convergent, then we define:
where can be any real number.
Improper integralsExample: Determine whether the integral is convergent or divergent.
Solution: By definition
The limit does not exist as a finite number. So, the integral is divergent.
1 1 1
1 1lim lim ln
lim(ln ln1)
lim ln
tt
t t
t
t
dx dx xx x
t
t
Improper integralsExample: Evaluate
Solution: By definition
We integrate by parts with and
We know that et→ 0 as t → -∞ therefore by l’Hôspital rule:
0 0limx x
ttxe dx xe dx
0 00
1
x x x
tt t
t t
xe dx xe e dx
te e
Improper integralsConti..
Therefore, 0lim ( 1 )
0 1 0
1
x t t
txe dx te e
lim lim
1lim
lim ( )
0
ttt t
tt
t
t
tte
e
e
e
Improper integralsExample: Evaluate
Solution: By definition
We must evaluate both integrals separately. i.e.
0
2 2 20
1 1 1
1 1 1dx dx dx
x x x
20
20
1
0
1 1
1
1
1
lim1
lim tan
lim(tan tan 0)
lim tan
2
t
t
t
t
t
t
dxx
dx
x
x
t
t
0
2
0
2
01
1 1
1
1
lim1
lim tan
lim (tan 0 tan )
02
2
tt
t t
t
dxx
dx
x
x
t
Improper integralsConti..
Since both these integrals are convergent, the given integral is convergent and
As , the given improper integral can be interpreted as the area of the infinite region that lies under the curve and above the x–axis.
2
1
1 2 2dx
x
Improper integralsLetbe the unbounded region under the graph of and above the axis between and .
The area of the part of between and is:
If it happens that approaches a definite number as , then we say that the area of the region is and we write:
provided this limit exists (as a finite number).
Improper integralsImproper integral of type II:
If is continuous on and is discontinuous at , then
if this limit exists (as a finite number).
If is continuous on and is discontinuous at , then
provided this limit exists (as a finite number).
Improper integralsImproper integral of type II:
If is discontinuous at and if both and , are convergent then
Improper integralsExample: Evaluate
Solution: First, we note that the given integral is improper because has the vertical asymptote .
By definition:
Thus, the given improper integral is convergent.
5 5
2 2
5
2
2
lim2 2
lim 2 2
lim 2( 3 2)
2 3
tt
tt
t
dx dx
x x
x
t
Improper integralsExample: Determine whether converges or diverges
Solution: Note that the given integral is improper because
By definition:
Thus, the given improper integral is divergent. This is because
and as
/ 2
0 0( / 2)
( / 2) 0
( / 2)
sec lim sec
lim ln sec tan
lim ln(sec tan ) ln1
t
x
t
x
x
x dx x dx
x x
t t
Improper integralsRemark: Sometimes, it is impossible to find the exact value of an improper integral and yet it is important to know whether it is convergent or divergent.
Theorem: Suppose andare continuous functions with
for .• If convergent, then is also convergent.• If is divergent, then is also divergent.
Improper integralsExample: Show that is convergent.
Solution: We write .• We observe that the first integral on the right-hand side is just an
ordinary definite integral.• In second integral, we use the fact that, for , we have .• So, and, therefore, .
• It follows that is convergent.
1 1
1
1
lim
lim( )
tx x
t
t
t
e dx e dx
e e
e
Introduction to vectorsDistance formula in three dimensions:
The distance │P1P2│between the points P1(x1,y1,z1) and P2(x2,y2,z2) is
212
212
21221 )()()( zzyyxxPP
Introduction to vectorsDefinition: The term vector is used by scientists to indicate a quantity that has both magnitude and direction, such as displacement or velocity or force.
The displacement vector v, shown below, has initial point A (the tail) and terminal point B (the tip) and we indicate this by writing .
Notice that the vector has the same length and the same direction as , even though it is in a different position. We say that and are equivalent (or equal) and we write .
We have .
Introduction to vectorsDefinition: If and are vectors positioned so the initial point of is at the terminal point of , then the sum is the vector from the initial point of to the terminal point of .
By the difference u - v of two vectors we mean
Introduction to vectorsDefinition: If is a scalar and is a vector, then the scalar multiple is the vector whose length is times the length of and whose direction is the same as if and is opposite to if . If or , then .
Introduction to vectorsDefinition: The length of the three-dimensional vector is given by .
Properties of Vectors: Ifandare vectors in and and are scalars, then
Introduction to vectorsDefinition: A vector is called unit vector if it has unit length i.e. . The unit vectors along axes are respectively denoted by and
.
These vectors , andare
called the standard basis vectors.
Any vector can be written as
Definition: If and , then the dot product of and is the number given by
Properties of dot production: Ifandare vectors in and and scalar, then
2aaa
abba
cabacba )(
)()()( dbabadbda 00 a
Dot product of vectors
Dot product of vectorsTheorems: • If is the angle between the vectors and then
• Two vectors aand bare orthogonal if and only if • If S is the foot of the perpendicular from R to the line containing
PQ, then the vector with representation PSis called the vector projection of b onto a and is denoted by prjoab
• and
cosbaba
Determinant:
Definition: If and , then the dot product of and is the number given by
In determinant form
Cross product of vectors
21
213
31
312
32
32
321
321
321
cc
bba
cc
bba
cc
bba
ccc
bbb
aaa
321
321
bbb
aaa
kji
ba
Cross product of vectorsTheorems: • If is the angle between the vectors and then
• Two vectors aand bare parallel if and only if
• and
sinbaba
Cross product of vectorsProperties of Cross product: Ifand are vectors in and and is scalar, then
abba )()()( dbabadbda
cabacba )(
cbcacba )(
cbacba )()(
cbabcacba )()()(