Chapter 6 INTEGRATION

•An overview of the area problem

•The indefinite integral

•Integration by substitution

•The definition of area as a limit; sigma notation

•The definite integral

•The fundamental theorem of calculus

•Evaluating definite integrals by substitution

6.1 An overview of the area problem

•The rectangle method for finding areas

•The antiderivative method for finding areas

The rectangle method for finding areas

•Divide the interval [a, b] into n equal subintervals, and over each subinterval construct a rectangle that extends from the x-axis to any point on the curve y=f(x) that is above the subinterval.

nAA

Alim nn

A A

We will call this the rectangle method for computing .A

•For each n, the total area of the rectangles can be viewed as an approximation to the exact area under the curve over the interval [a, b]. If denotes the exact area under the curve and denotes the approximation to using n rectangles, then

Figure 6.1.4 (p. 351)

The antiderivative method for finding areas

It’s shown that if is a nonnegative continuous function on the interval [a, b], and If denotes the area under the graph of over the interval [a, x], where x is anyPoint in the interval [a, b], then

'( ) ( )A x f x

ff( )A x

Example: For each of the function f, find the area A(x) between the graph of f andInterval [a, x]=[-2, x], and find the derivative A’(x) of this area function.

(a) f(x)=3 (b) f(x)=x+2 (3) f(x)=2x+5

Solution (a): From figure, we see

A(x)=3(x-(-2))=3(x+2)=3x+3

is the area of a rectangle of height 3 and base x+2. For this area function

A’(x) = 3 = f(x)

Solution (b): From figure, we see

is the area of an isosceles right triangle with base and height equal to x+1. For this area function,

A’(x) = x+2 = f(x).

21( ) ( 2)( 2) 2 2

2 2

xA x x x x

Example: For each of the function f, find the area A(x) between the graph of f andInterval [a, x]=[-2, x], and find the derivative A’(x) of this area function.

(a) f(x)=3 (b) f(x)=x+2 (3) f(x)=2x+5

Example: For each of the function f, find the area A(x) between the graph of f andInterval [a, x]=[-2, x], and find the derivative A’(x) of this area function.

(a) f(x)=3 (b) f(x)=x+2 (3) f(x)=2x+5

Solution (c): From figure, we see

is the area of a trapezoid with parallel sides of length 1 and 2x+5 and with altitude x-(-2)=x+2. For this area function,

A’(x) = 2x+5 = f(x).

21( ) ((2 5) 1)( ( 2)) 5 6

2A x x x x x

•The antiderivative method is usually the more efficient way to compute areas;

•The rectangle method is used to formally define the notion of area, thereby allow us to prove mathematical results about areas.

Two Methods Compared

6.2 The indefinite integral

For example: are all antiderivatives of 4 4 4 4

, 3, 7, 54 4 4 4

x x x x 3( )f x x

In fact:THEOREM: If F(x) is any antiderivatve of f(x) on an interval I, then for any constantC the function F(x)+C is also an antiderivative on that interval. Moreover, each Antiderivative of f(x) on the interval I can be expressed in the form F(x)+C by choosingThe constant C appropriately.

The process of finding antiderivatives is called antidifferentiation or integration.

Thus, if , then it can be recast using integral notations .

where C is understood to represent an arbitrary constant.

[ ( )] ( )dF x f x

dx

( ) ( )f x dx F x C

[ ( ) ] ( )d

f x dx f xdx

Note:

( )f x dx is called an indefinite integral. Is called an integral sign, the function

f(x) is called the integrand, and the constant C is called the constant of integration.

dx in serves to identify the independent variable. If it is different from x,

then the notation must be adjusted appropriately. Thus,

[ ] [ ]d

and dxdx

[ ( )] ( ) ( ) ( )dF t f t and f t dt F t C

dt

are equivalent statements.

For example:

3 2 2 3[ ] 3 3dx x is equivalent to x dx x C

dx

2 2[tan ] sec sec tand

t t is equivalent to tdt t Cdt

2 2

1

1

dx can be written as dx

dxdx can be written as

x x

Note:

Integration Formulas

Example:

54

5

xx dx C

23

3 2

1 1

2 2

xdx x dx C C

x x

111 32

2 211

2

2

3

xxdx x dx C x C

Properties of the indefinite integral

The statements in Theorem 6.2.3 can be summarized by the following formulas:

Example: Evaluate (a) 5sin xdx

(b) 4( 3 )x x dx

Solution (a):

5sin 5 sin 5cosxdx xdx x C

Solution (b):

5 24 4 4 3

( 3 6) 3 6 3 6 65 2

x xx x dx x dx xdx dx x dx xdx dx x C

Example: (a) 2

sin

cos

xdx

dx

(b) 3 4 5t t

dtt

Solution (a):2

sin sin 1tan sec sec

cos cos cos

x xdx dx x xdx x C

dx x x

Solution (b):3 3

2

3

4 5 4 5

14 5

4 5ln | |3

t t t tdt dt dt dt

t t t t

t dt dt dtt

tt t C