6.5 The Definite Integral

In our definition of net signed area, we assumed that for each positive number n, theInterval [a, b] was subdivided into n subintervals of equal length to create bases forthe approximating rectangles. For some functions, it may be more convienct to userectangles with different width.

Then the net signed area A between the graph of y=f(x) and the interval [a, b] is

*

max 01

lim ( )k

n

k kx

k

A f x x

intkwhere x is the length of k th sub erval

Here is called as the Riemann sum, and the definite integral is Sometimes called the Riemann integral.

*

1

( )n

k kk

f x x

Example: Sketch the region whose area is represented by the definite integral, and Evaluate the integral using an appropriate formula from gemoetry.

6 1 1 2

2 1 0) 3 , ) ( 2) , ) 1a dx b x dx c x dx

Solution (a): The graph of the integrand is the horizontal line y=3. so the region is aRectangle of height 3 extending over the interval from 2 to 5. Thus,

6

2) 3a dx = the area of the rectangle = 4(3)=12

Example: Sketch the region whose area is represented by the definite integral, and Evaluate the integral using an appropriate formula from gemoetry.

Solution (b): The graph of the integrand is the line y=x+2. so the region is aTrapezoid whose base extends from x= -1 to x=1. Thus,

1

1) ( 2)b x dx

Area of trapezoid =1/2(1+3)2=4

6 1 1 2

2 1 0) 3 , ) ( 2) , ) 1a dx b x dx c x dx

Example: Sketch the region whose area is represented by the definite integral, and Evaluate the integral using an appropriate formula from gemoetry.

Solution (c): The graph of is the upper semicircle of radius 1, centeredAt the origin, so the region is the right quarter-circle extending from x=0 to x=1. Thus

21y x

1 2

0) 1c x dx Area of quarter-circle= 21

(1 )4 4

6 1 1 2

2 1 0) 3 , ) ( 2) , ) 1a dx b x dx c x dx

Example: Evaluate 4 2

0 0) ( 2) ) ( 2)a x dx b x dx

Solution: from the figure of y=x-2, we can see that triangular region above and belowThe x-axis is both 2. Over the interval [0, 4], the net signed area is 4-4=0, and over The interval [0, 2], the net signed area is -2. Thus,

4

0

2

0

) ( 2) 0

) ( 2) 2

a x dx

b x dx

Properties of the Definite Integral

Example: 2 3

2) 0a x 0 12 2

1 0) 1 1

4b x dx x dx

Example: Evaluate 1 2

0(4 2 1 )x dx

1 1 1 1 12 2 2

0 0 0 0 0(4 2 1 ) 4 2 1 4 2 1x dx dx x dx dx x dx Solution:

The first integral can be interpreted as the area of a rectangle of height 4 and base 1,So its value is 5, and from previous example, the value of the second integral isThus,

/ 4

1 2

0(4 2 1 ) 4 2( ) 4

4 2x dx

The Fundamental Theorem of Calculus

It is standard to denote the difference F(b) - F(a) as

( )] ( ) ( ) ( ) ( ) ( )bb

a aF x F b F a or F x F b F a

Then (2) can be expressed as ( ) ( )]b b

aaf x dx F x

We will sometimes writeWhen it is important to emphasize that a and b are value for the variable x.

( )] ( ) ( )bx aF x F b F a

The Relationship between Definite and Indefinite Integrals

For purposes of evaluating a definite integral we can omit the constant of integrationin

( ) [ ( ) ]b b

aaf x dx F x C

And express as

Which relates the definite and indefinite integrals.

( ) ( )]b b

aaf x dx F x ( ) ( ) ]

b baa

f x dx f x dx

Example: 9 1/2 9 3/2 9

1 11

2 2 52] ] (27 1)

3 3 3xdx x dx x

Example: (a )9 2

4x xdx

9 5/2 7/2 944

2 2 4118] (2187 128)

7 7 7x dx x

0

cos

4

xdx

0

sin 1 1] [sin sin 0] [0 0] 0

4 4 4

x

1

2edxx

12ln | |] 2[ln | | ln |1|] 2[1 0] 2ex e

Solution:

(b)

Solution:

(c)

Solution:

The FTC can be applied to definite integrals in which the lower limit of integration isGreater than or equal to the upper limit of integration.

Example: 2 3

2x dx42 4 42

1] [2 2 ] 0

4 4

x

1

2xdx21 2 22

1 3] [1 2 ]

2 2 2

x

Solution:

Solution:

Example:

To integrate a continuous function that is defined piecewise on an interval [a, b], split This interval into subintervals at the breakpoints of the function, and integrate Separately over each subinterval in accordance with Theorem 6.5.5.

Example: Evaluate if 8

2( )f x dx

3 , 4( )

2 3, 4

x xf x

x x

Solution:8 4 8

2 2 4

4 83

2 4

44 2 82 4

( ) ( ) ( )

(2 3)

] [ 3 ]4(64 4) (88 28)

120

f x dx f x dx f x dx

x dx x dx

xx x

If f is a continuous function on the interval[a, b], then we define the total area betweenThe curve y=f(x) and the interval [a, b] to be

Total area = | ( ) |b

af x dx

To compute total area using the above Formula,

• begin by dividing the interval of Integration into subintervals on which f(x)Does not change sign.

• On the subintervals for which 0<=f(x), replace |f(x)| by f(x), and on the subintervals for which f(x)<=0 replace |f(x)| by –f(x).

• Adding the resulting integrals then yields the total area.

Example: Find the total area between the curve and the x-axis over the Interval [0, 2].

21y x

Solution: from the graph of , the area is given by 21y x

2 1 22 2

0 0 1

3 31 20 1

|1 | (1 ) (1 )

[ ] [ ]3 3

2 4( ) 2

3 3

A x dx x dx x dx

x xx x