1 Riemann Integration

Definition 1 (Partition of an interval). Let a, b R with a < b. By a partition of aninterval [a, b] R, we mean a finite set P = {x0, x1, , xn} such that

a = x0 < x1 < < xn1 < xn = b.Examples 2. 1. The simplest partition P = {a, b} is called trivial partition.

2. let n N. For 0 i n, define xi = a + i(b a)n

. Then the partition Pn ={x0, x1, , xn} divides the interval [a, b] into n equal subintervals.

Definition 3. If a point ti is selected from each subinterval [xi, xi+1] then we say thepartition is tagged. It will be denoted by P . Tags can be selected in any way.

P = (x0 < x1 < < xn) and P ={(

[xi, xi+1], ti)

: i = 0, . . . , n 1}.Definition 4 (Refinement of a partition). Given a partition P of [a, b], we say that apartition P is a refinement of P if every point of P is also a point of P.Definition 5. Given partitions P1 and P2 of [a, b], the partition P consisting entirelyof the points of P1 and the points of P2 is called the common refinement of P1 and P2.Definition 6. Let f : [a, b] R be a bounded function. We know that inf and supare attained in [a, b]. Define

m(f) = inf{f(x) | x [a, b]} and M(f) = sup{f(x) | x [a, b]}.Let P = {x0, x1, , xn} be a partition of [a, b]. Then for i = 0, 1, , n 1, definemi(f) = inf{f(x) | x [xi1, xi]} and Mi(f) = sup{f(x) | x [xi1, xi]}.

Result 1.

m(f) mi(f) Mi(f) M(f), for all i = 0, 1, , n 1.Definition 7 (Lower Sum and Upper Sum). We define the lower sum as follows:

L(P , f) =ni=1

mi(f)(xi xi1)

Similarly we define the upper sum as

U(P , f) =ni=1

Mi(f)(xi xi1)

Remark 1. Note that mi(f) is nothing but the height of the inscribed rectangle and Mi(f)is the height of the circumscribed rectangle. The term (xi xi1) gives the width of therectangle.

Thus L(P , f) gives the sum of all such inscribed rectangles and U(P , f) gives the sumof all such circumscribed rectangles.

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Result 2. Let f : [a, b] R be a bounded function. Then for any partition P of [a, b] wehave,

m(f)(b a) L(P , f) U(P , f) M(f)(b a).Outline of the proof. Let P = {x0, x1, , xn} be a partition of [a, b]. Then note thatni=1

(xi xi1) = (b a). Then the inequality follows from

m(f) mi(f) Mi(f) M(f).

Definition 8. Define

L(f) = sup{L(P , f) | P is a partition of [a, b]}U(f) = inf{U(P , f) | P is a partition of [a, b]}

Result 3. Let f : [a, b] R be a bounded function. If P is a partition of [a, b] and P isa refinement of P then

L(P , f) L(P, f)U(P , f) U(P, f)

and consequently,U(P, f) L(P, f) U(P , f) L(P , f).

Proof. Let P = {x0, x1, , xn} be a partition of [a, b]. First let us assume that arefinement P of P contains just one additional point x . Then xi1 x xi for somei {1, , n}. Using the abbreviations ` and r for left and right respectively, define

M` = sup{f(x) : x [xi1, x]} and Mr = sup{f(x) : x [x, xi]}.Clearly M` Mi(f) and Mr Mi(f). We also have xi xi1 = (xi x) + (x xi1)and therefore,

U(P , f) U(P, f) = Mi(f)(xi xi1)M` (x xi1)Mr (xi x)= (Mi(f)M` )(x xi1) + (Mi(f)Mr )(xi x) 0 + 0 = 0

If a refinement P of P contains several additional points, we repeat the above argumentseveral times and again obtain U(P, f) U(P , f).

The proof of L(P , f) L(P, f) is similar. Subtracting, we obtainU(P, f) L(P, f) U(P , f) L(P , f).

Result 4. If P1 and P2 are partitions of [a, b], then L(P1, f) U(P2, f).Proof. Let P denote the common refinement of partitions P1 and P2. Then in view of(i) above,

L(P1, f) L(P, f) U(P, f) U(P2, f).

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Result 5.L(f) U(f)

Proof. Let us fix a partition P0 of [a, b]. By (ii) above, we have L(P0, f) U(P , f) for anypartition P of [a, b]. Hence L(P0, f) is a lower bound of the set {U(P , f) | P is a partition of [a, b]}.Since U(f) is the greatest lower bound of this set, we have L(P0, f) U(f).

Now, since P0 is an arbitrary partition of [a, b], we see that U(f) is an upper bound ofthe set {L(P0, f) | P0 is a partition of [a, b]}. Again, since L(f) is the least upper boundof this set, we have

L(f) U(f).

Definition 9 (Integrable Functions). A bounded function f : [a, b] R is said to beIntegrable (on [a, b]) if

L(f) = U(f).

The value L(f) is called Riemann Integral of f on [a, b].Notation: b

a

f(x)dx = L(f) = U(f) or

ba

f

Remark 2. If f is a non-negative function then the area of the region enclosed by thecurve y = f(x) and the lines x = a and x = b is defined to be the Riemann Integral of f .Let Rf = {(x, y) R2 | a x b, 0 y f(x)} then

Area(Rf ) = ba

f(x)dx.

Remark 3. For a non-positive function, Area(Rf ) is defined to be the negative of areacorresponding to f . i.e.

Area(Rf ) = Area(R(f)).

In general, for any function f , let f+ and f be positive and negative parts. Then

Area(Rf ) = Area(R(f+))Area(R(f)).

Exercise:If f : [a, b] R is integrable, then are f+ and f integrable? Why?Result 6. Let f : [a, b] R be integrable function. Suppose there exist , R suchthat f(x) for all x [a, b]. Then

(b a) ba

f (b a).

In particular, if |f(x) for all x [a, b], then ba f (b a).

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Proof. Since f(x) , we get m(f) and M(f) . Consider P = {a, b}, thetrivial partition. Then we have

(b a) m(f)(b a) = L(P , f) and U(P , f) = M(f)(b a) (b a).

(b a) = L(f) = ba

f = U(f) (b a).

If |f | , then the result follows by setting = .Example 10. Let f : [a, b] R defined by f(x) = for all x [a, b].

For any partition P = {x0, x1, , xn} of [a, b], we have mi(f) = = Mi(f) for alli = 1, , n and so

L(P , f) = U(P , f) =ni=1

(xi xi1) = (b a).

Therefore L(f) = (b a) = U(f) and hence f is integrable and ba

f = b a.

Example 11. Let f : [a, b] R defined by f(x) = x for all x [a, b].Let P = {x0, x1, , xn} be a partition of [a, b]. Therefore mi(f) = xi1 and Mi(f) =

xi. Therefore

U(P , f) =ni=1

xi (xi xi1) and L(P , f) =ni=1

xi1 (xi xi1)

U(P , f) L(P , f) =ni=1

(xi xi1)2

and U(P , f) + L(P , f) =ni=1

x2i x2i1 = b2 a2

This shows that U(P , f) = b2 a2

2+

1

2

ni=1

(xi xi1)2

and L(P , f) = b2 a2

2 1

2

(xi xi1)2

Note that the above arguments hold for all partitions P of [a, b]It can be shown that (How? See the exercise below)

U(f) b2 a2

2and L(f) b

2 a22

.

But L(f) U(f). HenceL(f) = U(f) =

b2 a22

.

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Thus, f is integrable and ba

f =b2 a2

2.

Exercise:Show that given > 0, there exists a partition P = {x0, x1, , xn} of [a, b] such thatn

i=1(xi xi1)2 < .Brief outline:Let P = Pn = {x0, x1, , xn} denote the partition of [a, b] into n equal parts, where

n N. (i.e. xi = a+ i(ba)n for 1 i n.)Choose n such that (ba)

2

n< . Then

ni=1

(xi xi1)2 =ni=1

(b a)2n2

=(b a)2

n< .

This shows that given > 0, there exists N N such that, U(Pn, f) < b2a22 + forall n > N .

Now U(f) = inf{U(P , f) | P is a partition of [a, b]}. Note that for any partition P ,there exists n N such that Pn is a refinement of P . Therefore U(f) = inf{U(Pn, f) | n N}. Hence U(f) b2a2

2.

Similarly, L(f) b2a22

.

Result 7 (Riemann Condition). Let f : [a, b] R be a bounded function. Then f isintegrable if and only if for every > 0 there exists a partition P of [a, b] such that

U(P, f) L(P, f) < .Proof. Let > 0. Suppose there exists a partition P of [a, b] such that U(P, f) L(P, f) < . Then

0 U(f) L(f) U(P, f) L(P, f) < .But > 0 is arbitrary. Therefore U(f) = L(f) and hence f is integrable.

Conversely, suppose f is integrable. Thereforef = U(f) = L(f). Choose > 0.

Now since U(f) and L(f) are supremum and infimum respectively, there exists partitionsQ and Q such that

U(Q, f) < U(f) + and L(Q, f) > L(f) .Let P be a common refinement of Q and Q. Then

L(f) < L(Q, f) L(P, f) U(P, f) U(Q, f) < U(f) + .But f is integrable, therefore U(f) = L(f). Thus,

U(P, f) L(P, f) < 2.

Let f, g : [a, b] R be bounded integrable functions and , R. Then

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1. f + g is integrable on [a, b] and ba

(f + g) =

ba

f +

ba

g.

2. If f(x) g(x) for all x [a, b] then ba

f ba

g.

3. Let c (a, b) then, ba

f =

ca

f +

bc

f.

Moreover f is integrable over [a, b] if and only if f is integrable over both [a, c] and[c, b] for any point c (a, b).

Proof. Easy but longgg exercise :-)

Example 12. Below are few classes / examples of Riemann Integrable functions:

1. Step function on [a, b]

2. Any continuous function f : [a, b] R3. Any function f : [a, b] R which is monotonic on [a, b]4. If f is integrable on [a, b] then for any subinterval [c, d] [a, b], f is integrable on

[c, d]

Definition 13. Let f : [a, b] R be integrable and , [a, b] with < . Then wedefine

f =

f and

f = 0.

Definition 14 (Riemann Sum). Let f : [a, b] R be a bounded function and P ={x0, x1, , xn} be a partition of [a, b]. Let {t1, , tn} be tags of P (i.e. ti [xi1, xi]).Then the Riemann Sum of f w.r.t. P is defined as

S(P , f) =ni=1

= f(ti)(xi xi1).

Definition 15 (Norm of a partition). Let P = {x0, x1, , xn} be a partition of [a, b].Then norm of P , denoted by P is length of largest subinterval of the partition.

P = max{xi xi1 | 1 i n}.

Result 8. Let f : [a, b] R be a bounded function. Then f is integrable if and only iflimP0

S(P , f) exists.Moreover, if f is integrable, then

limP0

S(P , f) = ba

f.

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Proof. () Note that for any partition P ,

L(P , f) S(P , f) U(P , f).

But f is integrable, so L(f) = U(f). Therefore by Sandwich theorem, limP0

S(P , f)

exists and is equal to

ba

f .

() Not-so-?Easy exercise :-)

Result 9. Let f : [a, b] R be integrable. Suppose there exists a function F : [a, b] Rsuch that F (x) = f(x) for all x [a, b]. Then b

a

f(x)dx = F (b) F (a).

Proof. Let P = {x0, x1, , xn} be a partition of [a, b]. Then

F (b) F (a) =1in

(F (xi) F (Xi1))

= F (ci)(xi xi1) Applying MVT to F[xi1,xi]

= f(ci)(xi xi1) = S(P , f).

Therefore F (b) F (a) = limP0

S(P , f) = ba

f(x)dx.

Result 10. Let f : [a, b] R be integrable. Define F : [a, b] R by F (x) = xaf(t)dt.

Then F is continuous.Moreover there exists R such that |F (x) F (y)| |x y| for all x, y [a, b].

(This is called Lipschitz condition.)

Remark 4. if f is bounded, i.e. |f(x)| M for some M R. Then one can take = M .

Result 11 (Second Fundamental Theorem of Integral Calculus). If f is continuous atc [a, b], then F is differentiable at c and F (c) = f(c).

In general, if f is continuous on [a, b] then F is differentiable on [a, b]. In such case,F is called an antiderivative of f on [a, b].)

Result 12 (Change of Variables). Let f : [a, b] R be continuous on [a, b]. Let g :[c, d] [a, b] be such that g(c) = a and g(d) = b. Suppose g is integrable on [c, d]. Then b

a

f(x)dx =

dc

f(g(t)) g(t)dt.

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Proof. Let F (x) =

xa

f(t)dt. Define : [c, d] R by (t) = F (g(t)). Then isdifferentiable on [c, d] with (t) = F (g(t)) g(t) = f(g(t))g(t). Also is integrable on[c, d]. Therefore applying First FTIC, b

a

f(x)dx = F (b) F (a)= F (g(d)) F (g(c))= (d) (c)

=

dc

(t)dt = dc

f(g(t)) g(t)dt.

Wish you all the best

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Riemann Integration