Unit 2 | unit 2 riemann integration

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TRF

Unit - 2

Riemann integration

Q1) Define partition

A1)

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.

Q2) What are supremum and infimum?

A2)

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.

Q3) Define lower Riemann integral and upper Riemann integral.

A3)

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

Q4) What do you understand by Riemann integrability?

A4)

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

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].

Q5) A constant function is Riemann integrable on [a, b].

A5)

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].

Q6) Show that the function f defined by

Is not integrable on any interval.

A6)

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

= sup{}

= 0

Thus

Q7) Give the necessary condition for Riemann integrability.

A7)

The bounded function f is integrable,

Let be any positive number.

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

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

For every partition P with norm

Q8) What is the sufficient condition for Riemann integrability.

A8)

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.

Q9) Prove that f(x) = is integrable on [0, a] and

A9)

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

Q10) if and are two bounded and integrable functions on [a, b] then f = is also integrable on [a, b] and

A10)

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)

Q11) if and are two bounded and integrable functions on [a, b] then f = is also integrable on [a, b] and

A11)

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

Q12) What do you understand by equivalency of two definitions of integration?

A12)

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.

Q13) If the functions where are bounded and integrable on [a, b], then

A13)

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 .

Q14) Prove that every monotonic function is integrable.

A14)

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.

Q15) Prove that every continuous function is integrable.

A15)

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.

Q16) if f and g are integrable on [a, b], then so is f + g, and

A16)

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.

Q17) If f and g are integrable on [a, b] and

Then

A17)

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

Hence

Q18) What do you understand by monotone function.

A18)

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.

Q19) State and prove intermediate value theorem.

A19)

Statement-

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.