Chapter 5 Integration 2012

HELP Matriculation Centre – Foundation in Science/Arts – FDCAL001 Calculus – Version 1.0/2012

CHAPTER 5: INTEGRATION

5.1 Concept of Anti-derivatives

A function F is an anti-derivative of f on an interval I if F ' ( x )= f (x) for all x in I.

Thus, an anti-derivative of a function f is a function whose derivative is f. For example, F ( x )=x2

is an anti-derivative of f ( x )=2 x because:

F ' ( x )= ddx

( x2 )=2 x=f (x)

And F ( x )=x3+2 x+1 is an anti-derivative of f ( x )=3 x2+2 because:

F ' ( x )= ddx

( x3+2 x+1 )=3 x2+2=f (x )

Example 5.1: Let F ( x )=13

x3

−2 x2+x−1. Show that F is an anti-derivative of F ( x )=x2−4 x+1.

Example 5.2: Let F ( x )=x , G ( x )=x+2 and H ( x )=x+C, where C is a constant. Show that F, G

and H are all anti-derivatives of the function defined by f ( x )=1.

Example 5.2 shows that once an anti-derivative G of a function f is known, then another anti-derivative of f may be found by adding an arbitrary constant to the function G.

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THEOREM I

Let G be an anti-derivative of a function f. Then every anti-derivative F of f must be of the form F(x) = G(x) + C, where C is a constant.

The Indefinite Integral

The process of finding all anti-derivatives of a function is called anti-differentiation or

integration. We use the symbol ∫❑, called the integral sign, to indicate that the operation of

integration is to be performed on some function f. Thus,

∫ f (x )dx=F ( x )+C

[Read as: “the indefinite integral of f(x) with respect to x equals to F(x) plus C”] tell us that the

indefinite integral of f is the family of functions given by F(x) + C, where F ' ( x )=f (x).

The function f to be integrated is called the integrand and the constant C is called a constant of integration. The expression dx following the integrand f(x) reminds us that the operation is

performed with respect to x. If the independent variable is t, we write ∫ f (t)dt instead. Bear in

mind integration and differentiation are reverse operations.

5.2 Rules of Integration

Rule 1: The Indefinite Integral of a Constant

∫ k dx=kx+C

Example 5.3: Find each of the following indefinite integrals:

a) ∫2999 dx

b) ∫ π25 dx

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Rule 2: The Power Rule

∫ xn dx= 1n+1

xn+1+C (n ≠−1)


a) ∫ x3 dx

b) ∫ 1

x3/2dx

Rule 3: The Indefinite Integral of a Constant Multiple of a Function

∫ cf (x ) dc=c∫ f (x)dx (c isa constant )


a) ∫5 x3 dx

b) ∫−3 t−2 dt

Rule 4: The Sum Rule

∫ [ f ( x )± g ( x )] dx=∫ f (x )dx±∫ g(x )dx

Example 5.6: Find the indefinite integral of ∫(3 x5+4√ x−2 3√x+ 1

x2)dx .

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Rule 5: The Indefinite Integral of the Exponential Function

∫ ex dx=ex+C

Example 5.7: Find the indefinite integral ∫(2 ex−x3)dx.

Rule 6: The Indefinite Integral of the Function f ( x )=x−1.

∫ x−1 dx=∫ 1x

dx=ln¿ x∨¿+C ¿

Example 5.8: Find the indefinite integral ∫(2 x+3x+

4

x2 )dx.

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Rule 7: The Indefinite Integral of the Trigonometric Functions

∫(sin x¿)dx=−cos x+C ¿

∫(cos x¿)dx=sin x+C ¿

∫(sec 2 x¿)dx=tan x+C ¿

∫(csc2 x¿)dx=−cot x+C ¿

∫(sec x¿∙ tan x)dx=sec x+C ¿

∫(csc x¿∙cot x )dx=−csc x+C ¿

Rule 8: The Indefinite Integral of the Trigonometric Functions (Additional)

∫( tan x¿)dx=−ln∨cos x∨¿+C ¿¿

∫(sec x¿)dx=ln∨sec x+ tan x∨¿+C ¿¿

∫(csc x¿)dx=ln∨csc x−cot x∨¿+C ¿¿

∫(cot x¿)dx=ln∨sin x∨¿+C ¿¿

Example 5.9: Find the indefinite integral ∫ [ (csc2 x )+sin x ] dx.

Example 5.10: Find the indefinite integral ∫¿¿.

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a) ∫¿¿

b) ∫¿¿

Exercise 4.1: 1→ 27, 35 → 40, 57, 59, 63 5.3 Evaluating Definite Integrals

5.3.1 Differential Equation

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Given f ' ( x )=2 x−1 and we wish to find f (x). From what we now know, we can find f by

integrating the equation. Thus,

f ( x )=∫ f ' (x)dx=∫(2 x−1)dx=x2−x+C

where C is an arbitrary constant. Thus, infinitely many functions have the derivative f ', each differing from the other by a constant.

The equation f ' ( x )=2 x−1 is called a differential equation. Equation f ( x )=x2−x+C is called

the general solution of the differential equation f ' ( x )=2 x−1.

5.3.2 Initial Value Problems

Although there are infinitely many solutions to the differential equation f ' ( x )=2 x−1, we can

obtain a particular solution by specifying the value the function must assume at a certain value of x.

For example, suppose we stipulate that the function f under consideration must satisfy the

condition f (1 )=3, then using the condition of the general solution f ( x )=x2−x+C , we find that:

f (1 )=(12 )−1+C=3

C=3

Thus, the particular solution isf ( x )=x2−x+3. The condition f (1 )=3 is an example of an initial

condition.

Example 5.12: Find the function f it is known that f ' ( x )=3 x2−4 x+8 and f (1 )=9.

Example 5.13: Find the function f it is known that f '( x)=sin x−cos x and f ( π )=3.

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Example 5.14: In a test run of a maglev along a straight elevated monorail track, data obtained from reading its speedometer indicate that the velocity of the maglev at a time, t can be described by the velocity function

v (t )=8 t (0≤ t ≤30)

Find the position function of the maglev. Assume that initially the maglev is located at the origin of a coordinate line.

5.3.3 Fundamental Theorem of Calculus

Let f be continuous on [a, b]. Then,

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∫a

b

f (x )dx=F (b )−F(a)

where F is any anti-derivative of f; that is, F '(x )=f (x)

When applying the fundamental theorem of calculus, it is convenient to use the notations;

F( x)|ab=F (b )−F(a)

Example 5.15:Evaluate ∫1

3

(3 x2+ex)dx.

Example 5.16: Evaluate ∫1

2

( 1x− 1

x2 )dx .

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Properties of the Definite Integral

Let f and g be integrable function; then

1) ∫a

a

f ( x )dx=0

2) ∫a

b

f ( x )dx=−∫b

a

f ( x ) dx

3) ∫a

b

cf (x ) dx=c∫a

b

f ( x ) dx (c isa constant )

4) ∫a

b

[ f ( x )± g (x) ]dx=∫a

b

f ( x )dx ±∫a

b

g ( x ) dx

5) ∫a

b

f ( x )dx=∫a

c

f ( x )dx+∫c

b

f (x ) dx (a<c<b)

Property 5 states that if c is a number lying between a and b so that the interval [a, b] is divided into intervals [a, c] and [c, b], then the integral of f over interval [a, b] may be expressed as the sum of the integral of f over the interval [a, c] and the integral of f over the interval [c, b].

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Exercise 4.4: 27, 28, 29, 30; Exercise 4.5: 13 → 22; Exercise 6.1: 71 → 83Exercise 6.3: 83 → 99; Exercise 6.4: 43 → 495.4 Integration by Substitution

The method of integration called the method of substitution is closely relates to the chain rule for differentiating functions. When used in conjunction with the rules of integration developed earlier, the method of substitution becomes a powerful tool for integrating a large class of functions.

Integration by Substitution

1) Let u=g (x), where g(x ) is part of the integrand, usually the “inside function” of the

composite function f ( g (x)).2) Find d u=g ' (x)dx.3) Use substitution u=g (x) and d u=g ' (x)dx to convert the entire integral into one involving

only u.4) Evaluate the resulting integral.5) Replace u by g(x) to obtain the final solution as a function of x.

Note: Sometimes we need to consider choices of g for the substitution u=g (x)in order to carry out step 3 and/or step 4.

Example 5.17: Find ∫2 x (x2+3)4 dx.

Example 5.18: Find ∫ e−3 x dx.

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Example 5.19: Find ∫¿¿¿¿.

Example 5.20: Find ∫sin 3 x dx.

Example 5.21: Find ∫1

21

(3−5 x)2 dx .

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2

x e2x2

dx.


π

¿¿¿.

Example 5.24: A study prepared by the marketing department of Universal Instruments forecasts that after its new line of Galaxy Home Computer is introduced into the market, sales will grow at the rate of

2000−1500 e−0.05t (0 ≤ t ≤ 60)

units per month. Find an expression that gives the total number of computers that will sell t months after they become available on the market. How many computer will Universal sell in the first year they are on the market?

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Exercise 4.2: 1 → 49, 51, 53, 55, 57Exercise 4.5: 33 → 45

5.5 Integration by Parts

The method of integration called the integration by parts is closely relates to the product rule of differentiation. This formula is useful since it enables us to express on indefinite integral in terms of another that may be easier to evaluate.

Integration by Parts Formula

∫u dv=uv−∫ vdu

Example 5.25: Evaluate ∫ x ex dx.

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The success of the method of integration by parts depends on the proper choice of u and dv. In general, we can use the following guidelines.

Guidelines for choosing u and dv.

Choose u and dv so that

1) du is simpler than u.2) dv is easy to integrate.

Example 5.26: Evaluate ∫ x ln x dx .

Example 5.27: Evaluate ∫ x ex

(x+1)2 dx .

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ln 2

x ex dx


π2

x cos2 xdx

Example 5.30: The estimated rate at which oil will be produced from a certain oil well t years after production has begun is given by

R ( t )=100 t e−0.1 t

thousand barrels per year. Find an expression that describes the total production of oil at the end of year t.

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Exercise 7.1:1→25, 31, 32, 33, 34, 35, 36, 37, 39, 45, 51, 53, 55

5.5 Integration by Partial Fractions

In algebra, we learned how to combine two or more rational expressions (fractions) into a single expression by putting them together over a common denominator. For example,

2x−3

−1

x+1=

2 (x+1 )−1(x−3)( x−3 )(x+1)

=x+5

( x−3 )(x+1)=

x+5

x2−2x−3

The expression on the left hand of the example above is called the partial fraction decomposition

of x+5

x2−2 x−3, and each terms is called a partial fraction.

The method of partial fractions used this basic algebra technique to integrate any rational function. The form the partial fraction decomposition of the rational function R(x )/Q(x) takes depends on the form of Q(x ) and can be illustrated through examining four cases (Fortunately, we are going to discuss only two cases in Calculus).

Case 1: Distinct Linear Factors

IfR(x )Q(x)

=R (x)

(a1 x+b1 ) (a2 x+b3 ) …(an x+bn)

where all the factors ak x+bk , k=1 ,2 , …, n , are distinct, then there exist constants A1 , A2 ,…, An such that

R(x )Q(x)

=A1

a1 x+b1

+A2

a2 x+b2

+…+An

an x+bn

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Example 5.31: Find ∫ x+5

x2−2 x−3dx.

Example 5.32: Find ∫ 4 x2−4 x+6x3−x2−6 x

dx.

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Case 2: Repeated Linear Factors

If Q(x ) contains a factor (ax+b)r with r>1, then the partial fraction decomposition of R(x )/Q(x) contains a sum of r partial fractions of the form

R(x )Q(x)

=A1

(ax+b)+

A2

(ax+b)2+…+

Ar

(ax+b)r

where each Ak is a real number.

Example 5.33:∫ 2 x2+3 x+7x3+ x2−x−1

dx

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Example 7.4: 7→ 27

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Chapter 5 Integration 2012

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