Chapter 7 integration

Question 1

Done

Answer 1

Done

Question 2

Definition:
Characteristic function

Step function

Answer 2

Let a,b in R with a less than b. Let A be a subset of [a,b]. Then we define the characteristic function of A, χ_A:[a,b] to R.
As,
χ_A (t)

=
{1 t in A
{ 0 t not in A

A function s:[a,b] to R is called a step function if we have constants alpha_i in R
And
Bounded intervals I_i subset of [a,b] such that

s(t) = Σ i=1 to n
Of
α_i χ_I_i (t), For all t∈(a,b)

Step function only finitely many steps

Question 3

Definition: 7.1.2 integral of a step function

Answer 3

By adding areas of a step function..

Let s:[a,b] to R be a step function.

s(t) = sum from I=1 to n

Of α_i χ_I_i (t).

Then we define the integral of s to be

Integral from a to b of { s(t).dt} = sum from I=1 to n

Of α_i•length (I_i)

We can also write the same step function in different ways

Question 4

Example 7.1.3

Define step functions r,s : [0,5] to R

By
r(t) =χ_1,3 + 2χ[2,4]

s(t) = χ_1,2 + 3χ[2,3] (t) + 2χ_(3,4] (t)

r(t)= s(t) for all t

Calculate Integrals of each function from 0 to 5

Answer 4

Sol

Integral of r(t).dt = 12 + 22 =6

Integral of s(t) .dt = 11 + 31+ 2*1 =6

Question 5

Proposition 7.1.4: for Integrals of step functions

Answer 5

Let r,s: [a,b] to R be step functions and
α,β∈R. Then αr+ βs : [a,b] to R
Is a step function and

Integral from a to b
Of αr(t)+ βs(t) .dt
= α• integral from a to b of r(t).dt + β •integral from a to b of s(t).st

Question 6

Proposition 7.1.5:

Step functions r(t)≤ s(t) for all t in [a,b].

Answer 6

Let r,s: [a,b] to R be step functions with

r(t)≤ s(t) for all t in [a,b].

Then

Integral from a to b of r(t) .dt ≤
Integral from a to be of s(t) .dt

Question 7

Corollary 7.1.6:

Modulus of a step function

Answer 7

Let s:[a,b] to R be a step function. Then | s(t)| : [a,b] to R is also a step function and

Proof: easily a step function.

By showing that -|s(t)| ≤ s(t) ≤ |s(t)|

and using the Integrals of step functions in inequalities theorem

integral from a to b of s(t) .dt | ≤
Integral from a to b of MOD{s(t)} .dt

Question 8

Proposition 7.1.7: for a step function and a restriction ie partion of intervals for its integral

Answer 8

Let s: [a,b] to R be a step function.

Let a ≤ c ≤ b. Then the restriction of a to each of [a,c] and [c,b] is again a step function and

Integral from a to b of s(t) .dt
= sum of the Integrals of s(t) over both [a,c] and [c,b]

Question 9

A bounded function and approximating from below by a step function

Answer 9

Let f: [a,b] to R be a bounded function. So here exist constant m and M st m ≤ f(t) ≤ M for all t in [a,b]

Then if s(y) is a step function and s(t) ≤ f(t) then s(t) ≤ M for all t so

 Integral from a to b 
Of s(t) .dt ≤  M(b-a)

Question 10

A bounded function and aproximatinf from above by a step function

Answer 10

Let f: [a,b] to R be a bounded function. So here exist constant m and M st m ≤ f(t) ≤ M for all t in [a,b]

Then if s(y) is a step function and s(t) ≥ f(t) then s(t) ≥ m for all t so

 Integral from a to b 
Of s(t) .dt ≥  m(b-a)

Question 11

Definition 7.2.1: approximating a nouned function by its UPPER and LOWER Integrals

Answer 11

Let f: [a,b] to R be a bounded function.

The LOWER INTEGRAL of f:

L ∫ from a to b of f(t).dt =
Sup{ ∫ from a to b of s(t).dt | s is a step function, s(t) ≤ f(t) }

The UPPER INTEGRAL of f:

U∫ from a to b of f(t).dt =
Inf{ ∫ from a to b of s(t).dt | s is a step function, s(t) ≥ f(t) }

Note the steps are non empty and bounded so integrals are real. By definition of approximating above and below by step functions each set is bounded with m(b-a) and M(b-a) thus have sup and inf.

Question 12

Property of lower and upper Integrals of a step function

Answer 12

The integral from a to b of a step function 
Equals 
The lower Integral
Of the step function
Equals
The upper integral 
Of the step function 

Step functions are Riemann integrable

Question 13

Definition 7.2.2: Riemann integrable

Answer 13

We call a bounded function f:[a,b] to R
Riemann integrable if

Lower integral from a to b of f(t)

Equals

Upper Integral from a to b of f(t)
And the integral equals these upper and lower ones.

Not every bounded function is RI

Question 14

Example 7.2.3 define f:[0,1] to R by

f(t) =
{ 0 t In Q

{ 1 if t not in Q

Show that f is not Riemann integrable

Answer 14

Note that every interval however small contains both rationasl and irrationals .

So if s:[0,1] to R is a step function if s(t) ≥ f(t) for all t

Then s(t) ≥ 1

And if s(t) ≤ f(t) for all t then s(t) ≤ 0.

Hence lower integral of f(t) is 0

And upper integral of f(t) is 1 and since these aren’t equal not RI

Question 15

Proposition 7.2.4 bounded functions that are Riemann integrable ***

Answer 15

Let the bounded functions f,g:[a,b] to R be Riemann integrable

• Let α,β∈R. Then αf+ βg : [a,b] to R
Is RIEMANN INTEGRABLE and

Integral from a to b
Of αf(t)+ βg(t) .dt
= α• integral from a to b of f(t).dt + β •integral from a to b of g(t).dt

• suppose that f(t)≤ g(t) for all t in [a,b] then

Integral from a to b of f(t).dt ≤ integral from a to b of g(t) .dt

•a ≤ c ≤ b. Then the restriction to each of [a,c] and [c,b]

Integral from a to b of f(t) .dt
= sum of the Integrals of f(t) over both [a,c] and [c,b]

• the function |f(x)| : [a,b] to R is RIEMANN INTEGRABLE
And
| integral from a to b of f(t) .dt| ≤

Integral from a to b of |f(t)| .dt

Question 16

Corollary 7.2.5: bounded function that is Riemann integral and it’s Integrals Lower and upper bounds

Answer 16

Let the bounded function f: [a,b] to R be Riemann integrable.

Let m= inf{ f(x) | x in [a,b]}
M = sup{ f(x) | x in [a,b]}

Then

m(b-a) ≤ integral from a to b of f(t) ≤ M(b-a)
Proof:

By taking Integrals of inequality of m ≤ f(t) ≤ M

Question 17

Theorem 7.3.1: the fundamental theorem of calculus

Answer 17

Let f: [a,b] to R be continuous

Define F:[a,b] to R by

F(x) =
Integral from a to x of f(t).dt

(Only upper bound x variable!)
Then F is differentiable and
F’(x) = f(x) for all x in [a,b]

Proof: by manipulating inequalities of the integral and taking limits

Question 18

Corollary 7.3.2: continuous function and differentiable function F’(x) = f(x)

Answer 18

Let f: [a,b] to R be a continuous function. Suppose we have a differentiable function
F: [a,b] to R such that
F’(x) =f(x) for all x in [a,b]

then

Integral from a to b
Of f(t).dt
= F(b) - F(a)

Proof: by the fundamental theorem of calc