section 4.2 the definite integral...the definite integral of f from a to b is, provided that it...

1

Section 4.2 The Definite Integral

In the previous section we saw a limit of the form

using sample points xi* .

It arises when we compute- the area under the curve, or- the distance traveled by an object

limn→∞

(Δ x f (x1∗ )+Δ x f (x2

∗ )+…+Δ x f (xn∗ ))=lim

n→∞∑i=1

n

Δ x f (xi∗ )

2

The Definite Integral

In the previous section we saw a limit of the form

using sample points xi* .

The same limit occurs in a wide variety of situations even when f is not necessary positive function.

limn→∞

(Δ x f (x1∗ )+Δ x f (x2

∗ )+…+Δ x f (xn∗ ))=lim

n→∞∑i=1

n

Δ x f (xi∗ )

a b+

-

3


[Def] If the function f is defined on a x b, and the interval [a,b] is divided into n sub-intervals of the equal

width , : let x0 = a, x

1, x

2, …, x

n= b be the

endpoints of these sub-intervals, and x1*, x

2*, …, x

n* be

any sample points in the sub-intervals, xi*[x

i-1, x

i] , then

the definite integral of f from a to b is

, provided that it exists and

gives the same value for all possible choices of sample points.If it does exist, we say that f is integrable on [a,b].

Δ x=b−an

Δ x

∫a

b

f (x )dx=limn→∞

∑i=1

n

Δ x f (xi∗ )

4


The precise meaning of the integral is as follows:

for any number > 0, N Z+ such that

for every integer n > N and every choice of xi*[x

i-1, x

i] .

Symbol was introduced by Leibniz, and is called integral sign.

|∫a

b

f (x)dx−∑i=1

n

Δ x f (xi∗ )| <

∫

5


The process of calculation of an integral is called integration.

|∫a

b

f (x)dx−∑i=1

n

Δ x f (x i∗ )| <

∫a

b

f (x )dx

upper limit

integrandlower limit

dx indicates that x is independent variable

The definite integral is a number

∑i=1

n

Δ x f (x i∗ ) is called Riemann sum, after the German

mathematitian

6


If f(x) 0 on [a, b] then

the Riemann sum is the area

is the under the curve from

sum of all the areas a to b of rectangles

a b a b

∑i=1

n

Δ x f (xi∗ )

∫a

b

f (x )dx

7


If f(x) takes both positive and negative values on [a, b], then

∫a

b

f (x )dx=A1−A2

a b+

- a b+

- + +

the Riemann sum is the sum of areas of rectangles that lie above the b-axis and negatives of the areas of rectangles that lie below.

A definite integral can be interpreted as a net area

A1 - area above,

A2 – area below (negative)

8


[Theorem] If f(x) is continuous on [a,b], or if f has only a finite number of jump discontinuities, then f is integrable

on [a,b] , that is the definite integral exist.∫a

b

f (x )dx

9



on [a,b] , that is the definite integral exist.

Note that points xi*can be chosen any way; to simplify

calculation of the integral very often we take them to be the endpoints, i.e. x

i* = x

i , so

∫a

b

f (x )dx

10



on [a,b] , that is the definite integral exist.

Note that points xi*can be chosen any way; to simplify

calculation of the integral very often we take them to be the endpoints, i.e. x

i* = x

i , so

[Theorem 4] If f(x) is integrable on [a,b], then

, where ,

∫a

b

f (x )dx

∫a

b

f (x )dx=limn→∞

∑i=1

n

Δ x f (xi) Δ x=b−an

xi=a+iΔ x

11


Example: express the limit as a definite integral

on [2,6].limn→∞

∑i=1

n 1−xi2

4+xi2 Δ x

∫a

b

f (x )dx=limn→∞

∑i=1

n

Δ x f (xi)

12


Example: If , , find

the Riemann sum with n = 5 correct to 6 decimal places, taking sample points to be midpoints.What does the sum represent? Illustrate.

f (x)=√x−2f (x)=√x−2 1≤x≤6

13


Example: , , find the Riemann sum with n = 5 correct to 6 decimal places, sample points are midpoints.

f (x)=√x−2 1≤x≤6

∑i=1

5

Δ x f (xi)

14

Some Useful Summations

1+2+3+…+n=∑i=1

n

i=n(n+1)2

arithmetic series / progression

12+22+32+…+n2=∑i=1

n

i2=n(n+1)(2n+1)

6

13+23+33+…+n3=∑i=1

n

i3=(n(n+1)2 )2

c+c+c+…+c=∑i=1

n

c=nc

n c’s

(1)

(2)

(3)

(4)

15

Some Useful Summations

∑i=1

n

cai=c∑i=1

n

ai

∑i=1

n

(ai+bi)=∑i=1

n

ai+∑i=1

n

bi

(5)

(6)

(7)

, i . e . ca1+ca2+…+can=c (a1+a2+…+an)

, i . e .(a1+b1)+(a2+b2)+…

+(an+bn)=(a1+a2+…+an)+(b1+b2+…+bn)

∑i=1

n

(ai−bi)=∑i=1

n

ai−∑i=1

n

bi

16


Example: Use Theorem 4 to evaluate the integral

∫0

2

(2 x−x3)dx

17


Example: Use Theorem 4 to evaluate the integral

∫0

2

(2 x−x3)dx

18


The Midpoint Rule

Where , - midpoint of [xi-1

, xi].

Midpoint works good for approximation.

∫a

b

f (x )dx≈∑i=1

n

Δ x f ( x̄ i)=Δ x (f ( x̄1)+f ( x̄2)+…+ f ( x̄n))

Δ x=b−an

x̄i=xi−1+xi2

19


Example: Use The Midpoint Rule to approximate the

integral , n = 4.∫0

π2

cos(4 x)dx

20

Properties of definite integral

∫b

a

f (x)dx=−∫a

b

f (x)dx because x changes from

to

(1)

(2)

(3)

(4)

b−an

a−bn

∫a

a

f (x)dx=0 because x = 0

∫a

b

c dx=c (b−a) because c is a constant,

and b−an

⋅c=c (b−a) , if n=1

∫a

b

( f (x)±g(x))dx=∫a

b

f (x)dx ± ∫a

b

g(x)dx

21


Example: assume we have y = 5, then the area under y from 3 to 10 is 5(10-3) = 35.

∫3

10

5dx=5(10−3)=5⋅7=35

3 10

5

22


Example: Use the properties of the integral to

evaluate ∫2

5

(4−2 x)dx

23


Example: Use the properties of the integral to

evaluate ∫2

5

(4−2 x)dx

O 1 2 3 4 5

-6

24


Combining integrals

Comparison properties of integrals

∫a

c

f (x )dx=∫a

b

f (x)dx+∫b

c

f (x )dx , a<b<c

if f (x)≥0 for a≤x≤b , then ∫a

b

f (x)dx≥0

if f (x)≥g (x) for a≤x≤b , then ∫a

b

f (x)dx≥∫a

b

g (x)dx

then m(b−a)≤∫a

b

f (x)dx≤M (b−a)if m≤f (x)≤M for a≤x≤b ,

25


Example: Use the properties of the integral to estimate

∫0

2

(x3−3 x+3)dx

section 4.2 the definite integral...the definite integral of f from a to b is, provided that it...

Documents