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More on Integration

Calculus Module C19

LAST REVISED May, 2008

Module C19 − More on Integration

1

Introduction to More on Integration Statement of Prerequisite Skills Complete all previous TLM modules before beginning this module.

Required Supporting Materials Access to the World Wide Web. Internet Explorer 5.5 or greater. Macromedia Flash Player.

Rationale Why is it important for you to learn this material?

Learning Outcome When you complete this module you will be able to…

Learning Objectives 1. Evaluate the constant of integration. 2. Find the equation of a curve given a point on the curve. 3. Evaluate the definite integral. 4. Solve a simple differential equation. 5. Integrate using tables.

Connection Activity

Module C19 − More on Integration

OBJECTIVE ONE When you complete this objective you will be able to… Evaluate the constant of integration.

Exploration Activity The Constant of Integration C Up to this point, we have been integrating terms and polynomials to obtain an indefinite integral containing a constant. The constant C can be found after integration if the values for the independent and dependent variables at one point are known. This process is illustrated below.

EXAMPLE 1 Find the constant of integration C if y = 10 when x = 3 for and express

y as a particular function of x. ( )2 5y x d= +∫ ,x

SOLUTION: ( )2 5y x= +∫ dx

2 5y x dx dx= +∫ ∫

22 52 1x xy C= + +

2 5y x x C= + +

Substituting and 3x = 10y = into the indefinite integral and solving for C we get:

( )210 3 5 3 C= + + 10 9 15

14CC= − −= −

Therefore from the indefinite integral: 2 5y x x C= + + The definite integral becomes: 2 5 14y x x= + −

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Module C19 − More on Integration

EXAMPLE 2

3

dFind the definite integral of if ( )23 6 1y x x= − −∫ x 2x = − when 5y = .

SOLUTION:

( )23 6 1y x x= − −∫ dx

1 x

23 6y x dx xdx d= − −∫ ∫ ∫

3 23 6 13 2 1x x xy C= − − +

3 23y x x x C= − − + Substituting and into the indefinite integral and solving for C we get:

2x = − 5y =

( ) ( ) ( )3 25 2 3 2 25 8 12 25 18

23

CC

CC

= − − − − − +

= − − + += − +=

Therefore from the indefinite integral 3 23y x x x C= − − + The definite integral becomes: 3 23 2y x x x= − − + 3

Module C19 − More on Integration

Experiential Activity One PART A Determine the particular integral in each of the following if y = 9 when x = 3: 1. ( )26 6 1y x x= − +∫ dx

2. ( )22 5 1y x x= + +∫ dx

3. ( )( )2 1 1y x x d= + −∫ x

dx

4. ( )( )25 2 1y x x= + −∫5. ( )224 3y x= −∫ dx

NOTE: In questions 4 and 5 above, the constants 5 and 4 respectively can be taken

outside the integral sign ∫ before integration takes place.

In general: ( ) ( )4 6 4 6K x dx k x dx+ = +∫ ∫ where K is a real number constant.

PART B Determine C in each of the following:

1. 3 2

20 5 3 9y x dx x yx x

⎛ ⎞= + − ; = , =⎜ ⎟⎝ ⎠∫

2. ( )33 9 3 3y x x x dx x y= − + ; = ,∫ 9=

8

0

7

3. 3 1 2s dt t s= ; = . , =∫4. ( )1 2 1 2 4 1r s s ds s r/ − /= − ; = , =∫5. ( )3 22 3v t t dt t v− −= − ; = , =∫

Experiential Activity One Answers PART A

1. 3 22 5 34 53 2

y x x x= + + − .

2. 3 22 3 2y x x x= − + − 1

3. 4 3 2

3 754 3 2x x xy x= − + − − .

4. 4 2 35 5 10 10 129 75

4 2 3x x xy x= − + − − .

5. 5

34 8 36 775xy x x= − + − .4

PART B

1. 7118

C =

2. 3 75C = − . 3. 4 4C = .

4. 263

C =

5. 11518

C =

4

Module C19 − More on Integration

OBJECTIVE TWO When you complete this objective you will be able to… Find the equation of a curve given a point on the curve.

Exploration Activity The Constant of Integration C Up to this point, we have been integrating terms and polynomials to obtain a general indefinite integral containing a constant. The constant C can be found after integration if the values for the independent and dependent variables are known. This process is illustrated below.

EXAMPLE 1 Find the particular equation of the curve passing through the point (1 2),− and having a slope given by . 23 4m x x= − − 6 SOLUTION:

Step 1: 23 4dym x xdx= = − − 6

x

therefore 2(3 4 6)dy x x dx= − − and ( )23 4 6y x x d= − −∫ Step 2 Integrating we get: equation 1 3 22 6y x x x C= − − + Step 3 Substituting the boundary conditions of: 1x = and into equation 1 2y = − 2 1 2 6 C− = − − + therefore 5C = step 4 Substituting into equation 1 we get the equation of the curve: 5C = 3 22 6y x x x= − − + 5

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Module C19 − More on Integration

6

Experiential Activity Two Find the equation of the particular curve whose slope is shown by the given function of x, if the curve passes through the indicated point: 1. ; point 2m x= − (0 7), 2. ; point (24 4m x= − 11),

3. ; point (323m x= − 2 21),

4. ; point 23 8m x x= + + 2 ( 1 6)− ,

5. ; point 32 6m x−= − + x (1 12), 6. ; point (3 14m x x− /= − 2 4 259),

Experiential Activity Two Answers 1. 2 6 7y x x= − +

2. 22 4 1y x x= − + 1

5

3. 3 2y x x= −

4. 3 24 2y x x x= + + +

5. 2 23 8y x x−= + +

6. 4 1 22 7y x x /= − +

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Module C19 − More on Integration

OBJECTIVE THREE When you complete this objective you will be able to… Evaluate the definite integral.

Exploration Activity THE DEFINITE INTEGRAL

The definite integral of a function ( )f x is:

( ) ( ) ( )b

af x dx F b F a= −∫

where ( )F x′ = ( )f x .

We call this a definite integral because the final result of integrating and evaluating is a number. The numbers a and b are called the lower limit and the upper limit respectively. We can see the value of a definite integral is found by evaluating the integral of the integrand at the upper limit and subtracting the value of this same integral at the lower limit. The next example illustrates this procedure: Note: We do not have to add + C to the integral to calculate the definite integral of this type.

EXAMPLE 1

Evaluate the following integral: 1

0(3 4)x dx+∫

SOLUTION:

Integrating we have:

1

0(3 4)x dx+∫

= 12

0

3 42x x+

Here the vertical bar is used to the right of the integrated expression with

the limits placed on it to tell us to evaluate:

23 4 1

2x x at x+ =

and subtract from it the value of:

23 4 0

2x x at x+ =

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Module C19 − More on Integration

Therefore we have:

( ) ( ) ( ) ( )

( )

1 2 22

0

3 1 3 03 4 4 12 2 2

3 4 02

112

x x⎛ ⎞ ⎛⎜ ⎟ ⎜+ = + − +⎜ ⎟ ⎜⎝ ⎠ ⎝⎛ ⎞= + −⎜ ⎟⎝ ⎠

=

4 0⎞⎟⎟⎠

EXAMPLE 2

Evaluate the following integral: 2 4

0x dx∫

SOLUTION:

( ) ( )

252 4

005 5

5

2 05 5

325

xx dx =

= −

=

∫

Note: 1. When evaluating definite integrals there is NO CONSTANT of integration.

Why? It is because the constant would simply cancel so there is no reason to add it.

2. In words 2 4

0x dx∫ is read " the definite integral of 4x from x = 0 to x = 2"

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Module C19 − More on Integration

EXAMPLE 3

Evaluate the following integral. ( )4 2

2x x dx

−+∫

SOLUTION:

( ) ( ) ( ) ( )

43 24 2

22

3 2 3 2

( )3 2

4 4 2 23 2 3 2

64 16 8 43 2 3 2

64 88 23 3

72 63

24 630

x xx x dx−

−

+ = +

⎛ ⎞ ⎛ − −⎜ ⎟ ⎜= + − +⎜ ⎟ ⎜⎝ ⎠ ⎝

−⎛ ⎞ ⎛ ⎞= + − +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= + + −

= +

= +=

∫

⎞⎟⎟⎠

NOTE:

In this example 43 2

23 2x x

−

+ means the value of 3 2

3 2x x

+ at x = 4 minus the value of

3 2

3 2x x

+ at x = −2.

ALWAYS substitute the upper value first then the lower value.

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Module C19 − More on Integration

Experiential Activity Three Evaluate the following equations:

1. 1 4

1x dx

−∫ 9. 3

06 2xdx−∫

2. 1

02xdx∫ 10. ( )2 22

02 9 2x x dx−∫

3. 4 5 2

1x dx/∫ 11. ( )9

41x x d−∫ x

4. 1 1 3

1(1 )x dx/

−−∫ 12. ( )2 5 3

13 2x x dx−∫

5. 2 2

03x dx∫ 13. ( )2 32

12 4x x dx−∫

6. ( )2 3 22 3

12x x d

/+∫ x 14. ( )0 3 4

11 2x x dx

−−∫

7. 4

0 2 9xdx dxx +

∫ 15. 37

2

1x dx∫

8. 1 2 3

0(3 1)x x d−∫ x 16.

2

14 2x dx+∫

Experiential Activity Three Answers 1. 2

5

2. 1

3. 2547

4. 1.8899

5. 8

6. 40.0852

7. 2

8. 0.625

9. 2 6

10. 3643

11. 79.4

12. 24

13. 814

14. 0

15. 1.1843

16. 2.8210

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Module C19 − More on Integration

OBJECTIVE FOUR When you complete this objective you will be able to… Solve a simple differential equation.

Exploration Activity Differential Equations A differential equation is an equation that contains derivatives (or derivatives disguised as differentials) Examples of simple differential equations are:

5 and 7dydy x dx x

dx= = 4+

When we solve a differential equation and use extra information to solve for C it is called a particular solution. (This extra information is called a boundary condition)

EXAMPLE 1 Find the general solution of the following differential equations and then find the particular solution from the boundary conditions:

2 7 4 when 2dy x y xdx

= − ; = − =

SOLUTION: Step 1 (2 7)dy x dx= −

( )2 7y x= −∫ dx General Solution 2 7y x x C= − + Step 2 Substituting x = 2 and y = − 4 into the General Solution we get: 4 4 14 C− = − + 6C = Step 3 Particular Solution 2 7y x x= − + 6

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Module C19 − More on Integration

EXAMPLE 2 Find the particular solution of the following differential equation and describe the curve

passing through the point 4 213

⎛ ⎞,⎜ ⎟⎜ ⎟

⎝ ⎠.

49

dy xdx y

−=

SOLUTION:

This is called a separable equation. We convert it to two differentials by an operation that resembles standard algebra.

49

dy xdx y

−=

9 4ydy xdx= −

Integrating by sides we get:

9 4y dy x dx= −

2 29 42 2y x C−

= +

Multiplying through by 2 we get:

2 29 4 2y x= − + C

12

2 24 9x y C+ =

Now we substitute the coordinates of the point on the curve and get:

( )22 4 234(1) 9C = +

C = 36.

The particular solution is 4 92 2 3x y 6+ = , an ellipse.

2C and C are the same because we are simply looking for the constant to be added.

Module C19 − More on Integration

Experiential Activity Four Find the particular solution of the following differential equations for the curve passing through the given point:

1. 2 6 7 when 0dy x y xdx

= − ; = =

2. 4 4 11 when 2dy x y xdx

= − ; = =

3. 23 8 2 6 whendy x x y xdx

= + + ; = = −1

4. 3 1 24 259 when 4dy x x y xdx

− /= − ; = =

5. ( )5 13dy x pointdx y

= ; ,

6. ( )23 2 1dx y point

dy x= ; ,

Experiential Activity Four Answers

1. 2 6 7y x x= − + 2. 22 4 1y x x= − + 1

5 3. 3 24 2y x x x= + + + 4. 4 1 22 7y x x /= − + 5. 2 2 12x y− = 6. 2 32x y=

13

Module C19 − More on Integration

OBJECTIVE FIVE When you complete this objective you will be able to… Integrate using tables.

Exploration Activity INTEGRATION WITH TABLES The process of integration can become very difficult with certain functions. This text basically covers three types of integration problems: Individual terms and polynomials integrated by the power rule for terms:

1

1n naax dx x C

n+= +

+∫

Functions (or powers of polynomials) for which a constant may or may not need being compensated for; integrated by the power rule for functions:

1

1

nn uu du C

n

+

= ++∫

Functions for which a variable may NOT be compensated for; or trigonometric, exponential, or logarithimic functions for which TABLES OF INTEGRALS are required for integration. To simplify the solution of those problems in category 3 above, we refer to Appendix II which contains a short TABLE OF INTEGRALS. These formulae cover a wide variety of functions, and it is a matter of searching through the table until a formula is found which is in the same form as the integral we are trying to solve. There are well in excess of 725 integration formulae available. These tables can also be used for the problems done so far in this unit. With sufficient practice they can become a very straight forward and efficient method of integrating any problem you will encounter in this course. The following examples illustrate the use of these integration tables.

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Module C19 − More on Integration

EXAMPLE 1

Integrate 2 42x x dx+∫ using tables.

SOLUTION:

This question CANNOT be done using any of the previous methods discussed. Simplifying the original expression we get:

1 2 42

x x d= +∫ x

We now look for a formula in the Table of Integrals that will match the format of the simplified expression. Formula #16 will work. Notice that the following format is the same as the given question:

( ) ( )3 22

2 3 215

x ax b dx ax b ax b Ca

/+ = + − +∫

Now we substitute the values for a and b into the formula. In our case a = 2 and b = 4,

( )( ) ( )

( ) ( )

( )( )

3 22

3 2

3 2

1 1 22 4 2 4 3 2 2 42 2 15 2

1 2 4 6 8601 3 4 2 430

x x dx x x

x x C

x x C

/

/

/

+ = ⋅ + ⋅ ⋅ − ⋅ +

+ − +

− + +

∫ C

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Module C19 − More on Integration

EXAMPLE 2 Integrate cos dθ θ⋅∫ using tables.

SOLUTION:

From the Table of Integrals we find formula # 9:

cos sinu du u C= +∫

In our case u θ= Therefore: cos sind Cθ θ θ= +∫

NOTE: In using the tables, we must NOT forget to include the constant of integration, C.

EXAMPLE 3

Evaluate the definite integral 2 3

0

te dt−∫ SOLUTION:

From the tables we find formula #13.

1ax axe dx e Ca

= +∫

In our case a = −3 and x = t substituting into the formula we get:

( )( ) ( )( )

( )

( )

22 3 3

00

3 2 3 0

6

6

13

1 13 3

1 1 13 3

1 1 13 3

t te dt e

e e

e

e

− −

−

−

−

=−

= −− −

= +−

= +−

∫

QUESTION: Why is C not included in the solution to this example? NOTE: e is the base of the natural logarithm, (e = 2.7182818…)

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Module C19 − More on Integration

Experiential Activity Five Complete each of the following using tables:

1. 2 3 4x x d+∫ x 8. 32 xx e dx−⋅∫

2. 5x x dx−∫ 9. 2 3dxx +∫

3. 0

sin dπ

θ θ⋅∫ 10. ( )43 2x dx−∫

4. 2te dt∫ 11. ( )21

dxx +∫

5. 2 3 5x x dx+∫ 12. ( )32 1

dxx +∫

6. ( )53 4x x d+∫ x 13. 4 3 1x dx−∫

7. 2cos 3xdx∫ 14. 3 1xdxx +∫

Experiential Activity Five Answers

1. ( ) ( )3 24 3 4 9 8135

x x C/+ − + 8. ( )32 3 19

xe x− C−+ +

2. ( ) ( )3 22 5 3 1015

x x C/− + + 9. 2 3x C+ +

3. 2 10. ( )51 3 210

x C−− +

4. 212

te C+ 11. ( )

11

Cx−

++

5. ( ) (3 2 22 3 5 27 36 40567

)x x x/+ − + C+ 12. ( )2

14 2 1

Cx−

++

6. ( ) ( )7 61 23 4 3 463 27

x x C+ − + + 13. ( )3 28 3 19

x C/− +

7. sin 62 12x x C+ +

14.( )( )1 22 3 2 3 1

27x x C/− + +

17

Module C19 − More on Integration 18

Practical Application Activity Complete the More on Integration assignment in TLM.

Summary The constant of integration was evaluated. We found the equation of a curve knowing a point on the curve. The definite integral was evaluated. Simple differential equations were introduced. The use of tables of integrals was included in order to provide a method of solving more involved integrals. It should be emphasized that integration tables provide a means of solving any integral. The student should develop a proficiency at using them.

Module C19 − More on Integration

Appendix TABLE of INTEGRALS

A constant of integration should be supplied with each formula by the student. This table is designed to be used over the real number system only. When the use of a formula involves the square root of a negative number, the formula does NOT apply. Some of the basic formula’s include:

1. ( ) ( )d f x f x C⎡ ⎤ = +⎣ ⎦∫

2. adu a du=∫ ∫3. ( )du dv dw du dv dw+ + + = + + +∫ ∫ ∫ ∫

4. 1

1

nn uu du

n

+

=+∫

5. u dv uv v dv= −∫ ∫6. 1 lnu

u=∫

7. u ue du e=∫8. sin cosudu u= −∫9. cos sinudu u=∫10. ( )tan ln secudu u=∫

11. 2 sin 2sin2 4x axaxdx

a= −∫

12. 2 sin 2cos2 4x axaxdx

a= +∫

13. 1ax axe dx ea

=∫

14. ( )

2

1axax e ax

xe dxa

−=∫

15. ( )3 223

ax b dx ax ba

/+ = +∫

16. ( ) (3 22

2 3 215

)x ax b dx ax b ax ba

/+ = + −∫

17. ( ) ( )3 22 23

2 15 12 8105

2 2x ax b dx ax b a x abx ba

/+ = + − +∫

18. ( ) ( ) ( )( )111

n nax b dx ax ba n

++ = ++∫

19. ( ) ( ) ( )( )

22

2 22

n

n ax bax b dx

a n

±

± / ++ =

±∫

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Module C19 − More on Integration

20. ( ) ( ) ( )( )

( ) ( )( )2 12 2

12 1

n n bx ax b dx ax b ax ba n a n

n+ ++ = + − ++ +∫

21. ( ) ( ) ( )( )

( ) ( )( )2 12 2

12 1

n n ax a bx dx a bx a bxb n b n

n+ ++ = + − ++ +∫

22. ( ) ( )( ) ( )( ) ( )( )3 22 2

3

1 23 2

n nn ax b ax b ax b

x ax b dx b ba n n n

+ + 1

1

n+⎡ ⎤+ + +⎢ ⎥+ = − +

+ + +⎢ ⎥⎣ ⎦∫

23. ( ) ( )2

1dxa ax bax b

= −++∫

24. ( ) ( )3 2

12

dxax b a ax b

= −+ +∫

25. ( ) ( ) ( )3 2 22

12

x dx ba ax bax b a ax b

= −++ +∫

26. 2dx ax baax b+

=+∫

27. ( )

2

2 23

ax bx dx ax baax b−

= ++∫

28. ( )2 2 22

3

2 3 4 815

a x abx bx dx ax baax b

− += +

+∫

29. ( )3 3 2 2 2 33

4

2 5 6 8 1635

a x a bx ab x bx dx ax baax b

− + −= +

+∫

30. 2 2

22 2 2

dx a xa xx a x

− −=

−∫

31. 2 2

22 2 2

dx a xa xx a x

− +=

+∫

20