(MTH 250)

Lecture 21

Calculus

Previous Lecture’s Summary

•Trigonometric integrals

•Trigonometric substitutions

•Partial fractions

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 vectorsCoordinate axes and planes.:

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: Given the points and the vector a with representation =.

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 )()()(

Lecture Summary

•Recalls

•Improper integrals

•Introduction to vectors

•Dot product of vectors

•Cross product of vectors