Unit 2 | unit 2 riemann integration

TRF

Unit - 2

Riemann integration

2.1 Riemann integration & inequalities of upper and lower sums

The definite integral is the key tool in calculus for defining and calculating quantities like volumes, areas, length of curved paths, probabilities and weights of many objects.

Historically the idea of integration was initially given by Archimedes in 287-212 BC.

This can be extended to the area of wide class of planar region such as the region R = { (x, y): } under a curve y = f(x), where f: [a, b] is a bounded function. Such that .

A formal setup for these is the theory of Riemann integration.

Suppose [a, b] be a closed interval.

The partition of [a, b] means a finite set P of points where

The partition P consists of n+1 points.

Example: P = {a, b} is a partition of [a, b]

P = {0, 1/3, 2/3, 1} is the partition of [0, 1].

Note- the consecutive points of a partition are not necessary of equidistant.

The i’th sub interval is denoted by

Let f be a bounded real function on [a, b].

Evidentally f is bounded on each sub-interval corresponding to each partition P,

be the supremum and infimumof f in .

These are called the upper and lower sums of f corresponding to the partition P.

If M and m are the bounds of f in [a, b], we have,

Put I = 1,2,….n and adding all inequalities, we obtain

Now each partition gives rise to a pair of sums, the upper and lower sums, by considering all partitions of [a, b], we get a set U of upper sums and a set L of lower sums.

The above inequality shows that both these sets are bounded and so each set has the supremum and infimum.

The infimum of the set of upper sums is called the upper integral and supremum of the set of lower sums is called the lower integral of f over [a, b]. Thus

The above integrals may or may not be equal.

Note- Lower sums increases and upper sums decreases as the partition becomes finer and finer.

Let f: [a, b] be a bounded function

If P is the partition of [a, b] and p* is a refinement of P, which means *

Then

And

2. If are any two partitions of [a, b], then

3. Let P be any partition of [a, b] then

Definition-

Let f: [a, b] be a bounded function and P = {a = be a partition of [a, b].

The lower Riemann integral of f on [a, b] is defined as Sup {L(P, f) | P is a partition of [a, b]} and is defined by

i.e.

The upper Riemann integral of f on [a, b] is defined as Sup {U(P, f) | P is a partition of [a, b]} and is defined by

Definition of Riemann integrable

A bounded function f: [a, b] is said to be “Riemann integrable” over [a, b] if

For each for every partition of P of [a, b] such that

And it is denoted by

Note-

If f: [a, b] is a bounded function, then

2. A bounded function f is Riemann integrable on [a, b]

3. If bounded function f is such that

Then f is not Riemann integrable on [a, b].

[x] = the greatest integer not greater than x.

Since x belongs to [0, 3], f(x) = [x] is bounded on [0, 3], further f(x) = [x] is discontinuous at x = 1, 2

So that f(x) = [x] is bounded and has finite number of points of discontinuity.

Example: A constant function is Riemann integrable on [a, b].

Sol:

Let f(x) = k for every x, where be a constant.

Clearly f is bounded on [a, b] and inf f = sup f = k

Let be a partition on [a, b].

Let and be the inf and sup of f on , where

Since f(x) = k for every x

So that

And

Hence

Similarly

So that f is Riemann integrable on [a, b].

Note-

A function need not be Reimann integrable on [a, b], even though it is bounded on [a, b].
If f is Riemann integrable on [a, b] and m and M are the infimum and supremum of f on [a, b], then

3. If f: [a, b] is a bounded function, then for each

(i) U(P, f) < and

(ii) L(P, f) >

For each P- set of all partitions on [a, b] with ||P|| < .

Note- This result is known as Darboux’s theorem.

4. Let If f: [a, b] be a bounded function. As the norm of a partition, ||P||, becomes small, the number of partition points becomes large in such a way that n and ||P|| . Hence

Similarly

Example: Show that the function f defined by

Is not integrable on any interval.

Sol.

Suppose we consider a partition P of an interval [a, b].

= sup{}

= 0

Thus

Ex: Show that (3x + 1) is integrable on [1, 2] and

Key takeaways:

The partition of [a, b] means a finite set P of points where

2. Lower sums increases and upper sums decreases as the partition becomes finer and finer.

3. The lower Riemann integral of f on [a, b] is defined as Sup {L(P, f) | P is a partition of [a, b]} and is defined by

4. The upper Riemann integral of f on [a, b] is defined as Sup {U(P, f) | P is a partition of [a, b]} and is defined by

5. A bounded function f: [a, b] is said to be “Riemann integrable” over [a, b] if

2.2 Riemann conditions of integrability

We know that a bounded function is said to be integrable when the upper and lower integrals are equal.

Here we will give the necessary and sufficient conditions for the integrability of a function in two forms.

Form-1:

The necessary and sufficient condition for the integrability of a bounded function f is that to every , there corresponds such that for every partition P of [a, b] with norm

The condition is necessary-

The bounded function f is integrable,

Let be any positive number.

By the Darboux’stheoem there exists such that for every partition P with norm

From (1) and (3), we get on adding,

For every partition P with norm

The condition is sufficient –

Let be any positive number. For any partition P with norm , where

Also for any partition P. We know that-

Since is an arbitrary positive number, so that we see that a non-negative number is less than every positive number, so that it must be equal to zero.

So that f is integrable.

Form-2:

A bounded function f is integrable on [a, b] if and only if for every there exists a partition P such that

The condition is necessary

Let the function f is integrable, so that

Let be any positive number.

Since the upper and lower integral are the infimum and the supremum respectively of the upper and lower sums, therefore partitions and such that

Let P be the common refinement of .

Thus a partition P such that,

The condition is sufficient

Let be any positive number, P be a partition for which

For any partition

The non-negative number, being less than every positive numbers must be zero

So that f is integrable.

Example: Prove that f(x) = is integrable on [0, a] and

Sol:

f(x) = is bounded on [0, a].

Let us consider the partition P = {0, a/n, 2a/n,….,ra/n,…,na/n}

Length of each sub-interval =

Since f(x) = is increasing function in [0, a],

So that

And

f(x) = is integrable on [0, a] and

Key takeaways:

The necessary and sufficient condition for the integrability of a bounded function f is that to every , there corresponds such that for every partition P of [a, b] with norm

2. A bounded function f is integrable on [a, b] if and only if for every there exists a partition P such that

2.3 Riemann sum and definition of Riemann integral through Riemann sums, equivalence of two definitions

Integral as a limit of sums (Riemann sums)

Corresponding to a partition P of [a, b], suppose we chose points such that , let consider the sum

The above sum is called a Riemann sum of f over [a, b] relative to P.

Definition of integrability (Second)

A function f is said to be integrable on[a, b] if lim S(P, f) exists as , and then

Theorem- if and are two bounded and integrable functions on [a, b] then f = is also integrable on [a, b] and

Proof:

Here f is bounded on [a, b].

Suppose be any partition of [a, b] and ; be the bounds of and f respectively in

and are the rough upper and lower bounds whereas and are the supremum and infimum of of f in .

So that

Multiply by and adding all these inequalities for i = 1, 2, 3,…,n, we obtain

Suppose be a positive number.

Since are integrable therefore we can choose such that for any partitions P with norm we have

Thus for any partition P with norm , we have, from equation (2) and (3)

Thus the function f is integrable.

Since are integrable and is any positive number, therefore by using Darboux’s theorem, for every such that for all partitions P whose norm , we have

Also

Since is arbitrary,

Proceeding with in place of respectively, we get

From equation (5) and (6)

Theorem- if and are two bounded and integrable functions on [a, b] then f = is also integrable on [a, b] and

Proof:

Let f = + so that f is bounded on [a, b].

Suppose be any partition of [a, b] and ; be the bounds of and f respectively in

The bounds of

So that are

Multiply by and adding all these inequalities for i = 1, 2, 3,…,n, we obtain

Suppose be a positive number.

Since are integrable therefore we can choose such that for any partitions P with norm we have

Thus for any partition P with norm , we have, from equation (2) and (3)

Thus the function f is integrable.

Since are integrable and is any positive number, therefore by using Darboux’s theorem, for every such that for all partitions P whose norm , we have

Also

Since is arbitrary,

Proceeding with in place of respectively, we get

Equivalence of two definitions

We have two definitions of integrability, now we will show the equivalency of these two definitions.

Suppose f is the bounded function and integrable, so that

Suppose be any positive number.

By using Darboux’s theorem, there exists such that for every partition P with norm ,

And

If is any point of , we have

From (1), (2) and (3), we obtain that for any , such that for every partition with norm ,

Thus the function is integrable according to the second definition also.

Applications:

If the functions where are bounded and integrable on [a, b], then

Proof:

Let be any positive number and

Since are integrable, therefore for every such that for every partition with norm and for every choice of points in ,

Thus

Similarly we can prove for .

2. If a function f is bounded and integrable on each of the intervals [a, c], [c, b], [a, b] where c is a point of [a, b], then

Proof:

Let be given

As f is integrable on each of the intervals [a, c], [c, b], [a, b], there exists such that for every partition containing the point c with norm and for every choice of points in .

But

There we deduce

Which gives

Key takeaways:

If and are two bounded and integrable functions on [a, b] then f = is also integrable on [a, b] and

2. If and are two bounded and integrable functions on [a, b] then f = is also integrable on [a, b] and

3. A function f is said to be integrable on[a, b] if lim S(P, f) exists as , and then

2.4 Riemann integrability of monotone and continuous functions

If f: is monotonic on [a, b], then f is integrable on [a, b].
If the set of points of discontinuity of a bounded function is finite, then f is Riemann integrable on [a, b].
If the set of points of discontinuity of a bounded function f: has a finite number of limit points, then f is integrable on [a, b].

Theorem- Every monotonic function is integrable.

Proof:

We will prove the theorem for the case where I: [a,b] R is a monotonically increasing function. The function is bounded. f(a) and f(b) being g.l.b. And l.u.b. Let Î > 0 be given number, Let n be a positive integer such that

Divide the interval [a,b] into n equal sub-intervals, by the partition P = of [a, b]. Then

This proves that f is integrable.

Theorem: Every continuous function is integrable.

Proof:

Here we will prove that a function f which is continuous on [a, b] is also integrable on [a, b].

Suppose be given

Let we choose a positive number , such that

Since f is continuous on [a, b], therefore it is bounded and is uniformly continuous on [a, b].

, such that

Now choose a partition P with norm

Then by equation (1), we get

Hence

Thus f is integrable.

Note- Continuity is sufficient but not necessary condition for integrability.

Theorem: A bounded function f, having a finite number of points of discontinuity on [a, b] is integrable on [a, b].

Proof:

Suppose M and m be the bounds of f.

Suppose be given

Let there be p points of discontinuity of f on [a, b].

Here we consider a partition P of [a, b] such that all the points of the sum of whose length .

The oscillation of f in each of these sub-intervals being , their total contribution to {U(P, f) - L(P, f)} is less than

The function f is continuous in the remaining portion of [a, b], which means in the (p+1) sub-intervals of [a, b] excluding the sub-intervals taken above.

The contribution to the {U(P, f) - L(P, f)} from each of these (p+1) sub-intervals can be made , so that the total contribution to {U(P, f) - L(P, f)} yby these (p+1) sub-intervals is less than

Thus, for the partition P of [a, b].

U(P, f) - L(P, f)}

Hence the function f is integrable.

Key takeaways:

Every monotonic function is integrable.
Every continuous function is integrable.
A bounded function f, having a finite number of points of discontinuity on [a, b] is integrable on [a, b].

2.5 Properties of the Riemann integral

Property-1: if f and g are integrable on [a, b], then so is f + g, and

Proof:

Any Riemann sum of f+g over a partition P = of [a, b]can

Be written as

Here are the Riemann sums for f and g.

Above definition(property), implies that if > 0 there are positive numbers and such that

So the conclusion follows from the above definition.

Property-2: If f is integrable on [a, b] and c is a constant, then cf is integrable on [a, b] and

Property-3: If f and g are integrable on [a, b] and

Then

Proof:

Since , every lower sum of g – f over any partition of [a, b] is non-negative, therefore,

Hence

Property-4: if f is integrable on [a, b], then so is |f|, and

Key takeaways:

If f and g are integrable on [a, b], then so is f + g, and

2. If f is integrable on [a, b] and c is a constant, then cf is integrable on [a, b] and

3. If f and g are integrable on [a, b] and

Then

2.6 Definition and integrability of piecewise continuous and monotone functions

Definition of piecewise continuous functions-

A piecewise continuous function is a function that is continuous except at a finite number of points in its domain.

We usually write piecewise continuous functions by defining them case by case on different intervals. For example,

Is a piecewise continuous function

Note- Let f(x) be a real valued function. Suppose that f(x) has both a left limit S and a right limit R at x = a. Suppose further that f(x) is defined at x = a, and that R and S are both equal to f(a). We write this in limit notation as

And f(x) is continuous at x = a.

If we approach x = a from either the left or the right, f(x) approaches f(a).

In this section, we will show that functions f : [a, b] → R which are either piecewise monotonic or piecewise continuous are integrable.

f : [a, b] → R is increasing if f(x) ≤ f(y) whenever x < y for any x, y ∈ [a, b]. Similarly, f : [a, b] → R is decreasing if f(x) ≥ f(y) whenever x < y for any x, y ∈ [a, b]. The function f : [a, b] → R is called monotonic if it is either increasing or decreasing.

Theorem: If f : [a, b] → R is monotonic, then f ∈ R(a, b).

Here we are going to prove the theorem for the case that f is increasing, as the argument for the case that f is decreasing is almost identical.

Since f is increasing, by definition, f(a) ≤ f(x) ≤ f(b) for all x ∈ [a, b]. It follows that f is bounded on [a, b] by the value f(b). In order to prove that f ∈ R(a, b), we will show that the Cauchy criterion is satisfied. To this end, for > 0, we choose

And choose any partition Pδ. Then

Theorem: If f : [a, b] → R is continuous, then f ∈ R(a, b).

Here we will use the Cauchy criterion. Since [a, b] is closed, f is uniformly continuous on [a, b]; therefore, for any > 0, we can choose δ > 0 such that

Whenever x, y ∈ [a, b] and |x − y| < δ

Let Pδ denote any partition of [a, b]. Since f is continuous, we can replace the inf and sup with min and max, respectively, in the definitions of mi and Mi , so that for each i = 1, ..., N(δ),

For each i = 1, ..., N(δ)

So that

Definition of monotone functions-

A function f is monotone increasing on (a, b) if f(x) f(y) whenever x < y. A function f is monotone decreasing on (a, b) if f(x) f(y) whenever x < y.

A function f is called monotone on (a, b) if it is either always monotone increasing or monotone decreasing.

Note-

If f is a monotonic function defined on an interval I, then f is differentiable almost everywhere on I, i.e. the set of numbers x in I such that f is not differentiable in x has Lebesgue measure zero.
If f is a monotonic function defined on an interval [a, b], then f is Riemann integrable.
A function is unimodal if it is monotonically increasing up to some point (the mode) and then monotonically decreasing.

Integrability of monotone function

Let the partition of the interval-

If we divide the interval into n equal parts then the width of each rectangle will be

We calculate an upper bound of integral as

We consider the decreasing function

With the same partition this is a lower bound of the integral

We consider the increasing function

If this two sequences converges towards one and same limit, we can call this limit the definite integral, and we write

When we consider an increasing function which is monotonic, we have anincreasing sequence { and decreasing sequence {.

Then the question is only whether we can verify

We can calculate

Then the limit

We conclude that if f is monotonic in the interval [a, b], then the definite integral exists.

Key takeaways:

A function f is called monotone on (a, b) if it is either always monotone increasing or monotone decreasing.
If f is a monotonic function defined on an interval [a, b], then f is Riemann integrable.

2.7 Intermediate value theorem for Integrals

Intermediate Value Theorem

Let f be a continuous function on an interval containing a and b.

If K is any number between f(a) and f(b) then there is a number c, a c S b such that f(c) = K

Proof:

Either f(a) = f(b) or f(a) < f(b) or f(b) < f(a). If f(a) = f(b) then K = f(a) = f(b) and so c can be taken to be either a or b. We will assume that f(a) < f(b). (The other case can be dealt with similarly.) We can, therefore, assume that f(a) < K < f(b).

Let S denote the collection of all real numbers x in [a, b] such that f(x) < K. Clearly S contains a, so S and b is an upper bound for S. Hence, by completeness property of R, S has least upper bound and let us denote this least upper bound by c. Then a c b. We want to show that f(c) = K. Let S denote the collection of all real numbers x in [a, b] such that f(x) < K. Clearly S contains a, so S and b is an upper bound for S. Hence, by completeness property of R, S has least upper bound and let us denote this least upper bound by c. Then a c b. We want to show that f(c) = K.

Since f is continuous on [a, b], f is continuous at c. Therefore, given > 0, there exists a 6 > 0 such that whenever x is in [a, b] and |x – c| < 6, |f(x) – f(c) ( < G,

i.e., f(c) – < f(x) < f(c) + .

If c b, we can clearly assume that c + 6 < b. Now c is the least upper bound of S. So c – is not ‘an upper bound’ of S. Hence, there exists a y in S such that c – 6 < y c. Clearly |y – c| < and so by (4) above, we have

f(c) – < f(y) < f(c) + .

Since y is in S, therefore f(y) < K. Thus, we get

f(c) – S < K

If now c = b then K – < K < f(b) = f(c), i.e., K < f(c) + E. If c b, then c < b; then there exists an x such that c < x < c + 6, 6, x [a, b] and for this x, f(x) < f(c) + by (4) above. Since x > c, K f(x), for otherwise x would be in S which will imply that c is not an upper bound of S. Thus, again we have K f(x) < f(c) + E.

In any case,

K < f(c) + ...(6)

Combining (5) and (6), we get for every > 0

f (c) – < K < f(c) +

Which proves that K = f(c), since is arbitrary while K, f(c) are fixed. In fact, when f(a) < K < f(b)

And f(c) = K, then a < c < b.

Note-

If f is a continuous function on the closed interval [a, b] and If f(a) and f(b) have opposite signs (i.e., f(a) f(b) < 0), then there is a point x0 in ]a, b[ at which f vanishes. (i.e., f() = 0).
Let f be a continuous function defined on a bounded closed interval [a, b] with values in [a, b]. Then there exists a point c in [a, b] such that f(c) = c. (i.e., there exists a fixed point c for the function f on [a, b]).

Example: The equation has a real root lying between 1 and 2.

Sol:

The function is a continuous function on the closed interval [1, 2], f(1) = –8 and f(2) = 9. Hence, by Corollary 1, there exists an such that ) = 0.

Which means is the real root of the equation lying in the interval ]1, 2[.

Key takeaways:

Let f be a continuous function on an interval containing a and b. If K is any number between f(a) and f(b) then there is a number c, a c S b such that f(c) = K
If f is a continuous function on the closed interval [a, b] and If f(a) and f(b) have opposite signs (i.e., f(a) f(b) < 0), then there is a point x0 in ]a, b[ at which f vanishes. (i.e., f() = 0).