the lengths of the respective subintervals, given in order. Obviously,these lengths of subintervals need not be equal.
(c) The simplest partition of [a,b] is {a,b}. If we go on adding more andmore pointsto this partition thenthe lengths of the subintervals will go on decreasing.
(ii) Norm of a Partition: LetP¼{x0,x1,x2,. . .,xnja¼x0<x1<x2. . .<xn¼b}be a partition of[a,b].The greatest of the lengths x1–x0,x2–x1,. . .,x3–x2,. . .,xn–xn–1, of the subintervals formed by the partitions, is calledthe norm of the partitionand is denoted byjjPjj(read as the norm of the partitionP). Now, we describe thenotion of a Riemann sum.
5.5.1 Riemann Sums
Consider a functionfdefined on aclosed interval[a,b]. It may have bothpositiveandnegative values on the interval andit does not even need to be continuous. Its graph might look something like the one in Figure 5.14.
Consider a partitionPof the interval [a,b] intonsubintervals (not necessarily of equal length) by means of pointsa¼x0<x1<x2<. . .<xn–1<xn¼b, and letDxi¼xi–xi–1(i«N).
On each subinterval [xi–1,xi],choose a perfectly arbitrary point xi(which could even be an end point of the subinterval). We call it asample pointfor theith subinterval. An example of this construction is shown in Figure 5.15, forn¼6.
Now, form the sumRp¼P6
i¼1
fðxiÞDxi. y
x y = f (x)
b a
FIGURE 5.14
Δx1 Partition
points Sample points
A partition of [a,b] with sample points Xi a = x0 x1
x1
x2 x3 x4 x5 x6= b
Δx2 Δx3 Δx4 Δx5 Δx6
x2 x3 x4 x5 x6
FIGURE 5.15
We callRpaRiemann sumforf, corresponding to the partitionP. Its geometric interpretation is shown in Figure 5.16.
Note that,the contributions from the rectangles below the x-axis are the negatives of their areas.Thus, aRiemann sumis interpreted asan algebraic sum of areas.(8)
X6
i¼1
fðxiÞDxi¼A1þ ðA2Þ þ ðA3Þ þ ðA4Þ þA5þA6
Riemann sums corresponding to various choices ofx1;x2;. . .;xncan be different from one another. However, all Riemann sums must lie between the lower sum and the upper sum. An important feature of a Riemann sumPn
i¼1fðxiÞDxiis that it approximates the definite integral Ðb
afðxÞdx.
Therefore, we writeÐb
afðxÞdxPn
i¼1fðxiÞDxi.(9)
Suppose now thatP,Dxiandxihave the meanings discussed above. Also, letjjPjjbe the norm(i.e., the length of the longest subinterval of the partitionP). Then, we give the following definition.
5.5.2 Definition: The Definite Integral
Letfbe a function that is defined on theclosed interval[a,b]. If limjjPjj !0
Pn
i¼1fðxiÞDxi exists, then we saythat f is integrable on [a,b]. Moreover,Ðb
afðxÞdx, [called thedefinite integral(or Riemann integral) offfromatob], is then given by(10)
y
∑ f(xi)Δxi = A1 + (–A2) + (–A3) + (–A4) + A5 + A6 A Riemann sum interpreted as an algebraic sum of areas
i = 1 6
A1
A2
A3 A4
A5
A6 y = f(x)
x4 x5 x6= b x a=x0
x1
x5 x6
x1
x2 x2 x3 x3 x4
FIGURE 5.16
(8)Riemann sums are named after the nineteenth century mathematician Georg Bernhard Riemann (1826–1866), who clarified the concept of the integral while employing such sums. The first formal definition of the integral is attributed to him.
(9)For detailed discussion, seeCalculus with Analytic Geometry(Alternate Edition) by Robert Ellis and Denny Gulick, Chapter 5.
(10)Calculus with Analytic Geometry(Fifth Edition) by Edwin J. Purcell and Dale Varberg (pp. 234–235), Prentice-Hall, Inc, New Jersey.
RIEMANN SUMS AND THE ANALYTICAL DEFINITION OF THE DEFINITE INTEGRAL 153
ðb
a
fðxÞdx¼ lim
jjPjj !0
Xn
i¼1fðxiÞDxi The heart of the definition is the above line.
5.5.3 Concept of the Definite Integral
The concept captured in the above equation grows out of our area of discussion. However, we have considerably modified the notion presented here. For example,
(i) We now allowfto benegative on part or all of[a,b],
(ii) We use partitions withsubintervalsthat may be ofunequal lengths, and (iii) We allowxito beany point on the ith subinterval.
Since we have made these changes, it is importantto state precisely how the definite integral relates to area.
Note (1):In general,Ðb
afðxÞdxgives thesigned area of the region trapped between the curve y¼f(x) and the x-axis,on the interval[a,b], meaning that a plus sign is attached to areas of parts above thex-axis and a minus sign is attached to areas of parts below thex-axis. In symbols, Ðb
afðxÞdx¼AupAdown, whereAupandAdownare the areas corresponding to the þ and regions as shown in Figure 5.17.
Note (2):The meaning of limit in the definition of the definite integral is more general than in earlier usage, and should be explained.
The equality limjjPjj !0Pn
i¼1fðxiÞDxi¼Lmeans that, corresponding to each«>0, there is ad>0, such that Pn
i¼1fðxiÞDxiL
<«, forall Riemann sumsPn
i¼1fðxiÞDxiforfon [a,b], for which the normjjPjjof the associated partition is less thand. In this case, we say that the indicated limit exists and has the valueL.
Note (3):In the symbolÐb
afðxÞdx, most authors use the terminology “a”, as thelower limitof integration and “b”, as theupper limitof integration, which is fine provided we realize that this usage of the wordlimithas nothing to do with its more technical meaning.
5.5.4 Further Modification in the Notion of Definite Integral:
Removal of One More Restriction In our definition ofÐb
afðxÞdx, sincef(x) is defined on the interval [a,b], weimplicitly assumed that a<b. If we omit the conditiona<b, and assume thata>b,we can still retain our
+
y y
a b
x x
0 0 π 2π
y = f (x) y = sin x
+ –
+
–
FIGURE 5.17
arithmetical definition of definite integral;the only change is thatwhen we traverse the interval from a to b, the differencesDx, are negative. Thus, we get the following relation
ðb
a
fðxÞdx¼ ða
b
fðxÞdx ðIÞ
which holds for all values ofaandb(a6¼b). From the above relation, it follows that ða
a
fðxÞdx¼0 ðIIÞ
This must be treated as a definition. Our definition (of the definite integral) immediately gives the basic relation
ðb
a
fðxÞdxþ ðc
b
fðxÞdx¼ ðc
a
fðxÞdx ðIIIÞ
fora<b<c(see Figure 5.18).
Remark: By means of the preceding relations, we at once find that the equation (III) is also true for any position of the pointa,b, andcrelative to one another. Also, we obtain a simple but important fundamental rule by considering the functionc f(x),where c is a constant. From the definition of the definite integral, we immediately obtain,
ðb
a
cfðxÞdx¼c ðb
a
fðxÞdx ðIVÞ
Further, we assert the following addition rule:
If fðxÞ ¼ðxÞ þcðxÞthen, ðb
a
fðxÞdx¼ ðb
a
ðxÞdxþ ðb
a
cðxÞdx ðVÞ
This can be easily proved from the definition of definite integral using Riemann sums.
a b c
y
x 0
R1
y = f(x)
R2
FIGURE 5.18
RIEMANN SUMS AND THE ANALYTICAL DEFINITION OF THE DEFINITE INTEGRAL 155
5.5.5 An Important Remark About the Variable of Integration We have written the definite integral in the formÐb
afðxÞdx. For evaluating the integral, it does not matter whether we use the letterxor any other letter to denote the independent variable. The particular symbol we use for the variable of integration is therefore not important; instead of Ðb
afðxÞdx, we could as well writeÐb
afðtÞdtorÐb
afðuÞduor any other expression. The importance of this remark will be realized shortly in applications, when we prove thesecond fundamental theorem of Calculus, in the next chapter.
5.5.6 What Functions Are Integrable?
Not every function is integrable.
For example, the unbounded function fðxÞ ¼ 1=x2 ifx6¼0 1 ifx¼0
in Figure 5.19, is not integrable on[2, 2].
y¼fðxÞ ¼ 1=x2 ifx6¼0 1 ifx¼0 (
This is because the contribution of any Riemann sum for the subinterval containingx¼0 can be made arbitrarily large by choosing the corresponding sample pointxi, sufficiently close to zero. In fact, this reasoning shows that “any function that is integrable on[a,b], must be bounded in the interval[a,b]”. In other words, there must exist a constantMsuch that jf(x)j Mfor allxin [a,b].
Remark: Even some bounded functions can fail to be integrable. For example, the function fðxÞ ¼ 1 ifxis rational
0 ifxis irrational (
isnot integrableon [0,1].(11)
0
x y =
y
1 x2
FIGURE 5.19
(11)This is so because betweenanytwo real numbers, there are infinite number of rationals and also infinite number of irrationals. Hence, no matter how small the norm of partition (i.e.,jjPjj), the Riemann sumPn
i¼1fðxiÞDxicannot have a unique value. Of course, it can have the valueeither0or1.
By all odds, the theorem given in Section 5.5.7 isthe most important theoremabout integrability of a function.
5.5.7 Integrability Theorem
Iffisboundedon [a,b] and if it iscontinuousthere,except at a finite number of points, then fis integrable on [a,b]. We do not prove it here.
In particular,if f is continuous on the whole interval[a,b],it is integrable on[a,b].(12) As a consequence of this theorem, the following functions are integrable on every closed interval [a,b]
(1) Polynomial functions.
(2) Sine and cosine functions.
(3) Rational functions (provided the interval [a,b] contains no points where denominator is 0).
From the definition of the definite integral,Ðb
afðxÞdx,as the limit of a sum,we have ðb
b
fðxÞdx¼ lim
jjPjj !0
Xn
i¼1
fðxiÞDxi ðAÞð13Þ We know that, iffis integrable on [a,b], then the limit on the right-hand side of equation (A) must exist. Forthe functions that we shall consider here, this limit will always exist. We now suggest the following so that the method for evaluating a definite integral, as the limit of a sum, is simplified to some extent.
(i) We shall consider aconvenient partitionof [a,b], that is, choosex1;x2;x3;. . .;xnin the subintervalssuitablyas explained below, so that the limit of the sum on the right-hand side of equation (A) can be evaluated easily. Themost convenient partitionis that which divides [a,b]
into subintervals ofequal length, sayh.Such a partition is called as a regular partition.
Ifa¼x0,x1,x2,x3,. . .,xn¼bare the points of division such thata¼x0<x1<x2<x3
<. . .<xn–1<xn¼b, thenx1–x0¼(x1–a)¼h,x2–x1¼h,x3–x2¼h,. . .,xn–xn–1¼h.
Hence, the points of subdivision are,a,aþh,aþ2h,aþ3h,. . .,aþnh.
There aren subintervalsin [a,b]each of length h.
)The sum of the lengths of these subintervals must beb–a.
) nh¼ba ) h¼ba n ¼ jjPjj
Now, asn! 1,h!0 (i.e., the length of each subinterval tends to zero, so thatjjPjj !0).
(ii) We choose the pointsx1;x2;x3;. . .;xnasthe right-hand end point of each subinterval, in computing a sum.(14)
Thus,x1¼aþh¼x1;x2¼aþ2h¼x2;x3¼aþ3h¼x3;. . .;xn¼aþnh¼xn.
(12)Unfortunately, the proof of this theorem is not simple. We, therefore, accept the theorem without proof. For the proof of this theorem, advanced texts on Calculus may be referred to.
(13)The meanings ofP,jjPjj,Dxi, andxihave already been explained earlier, in the text.
(14)By choosing each subintervalDxiof equal length, and the pointsx1;x2;x3;. . .;xnas the right-hand end point of each subinterval, helps in computing the sumPn
i¼1fðxiÞDxi, easily.
RIEMANN SUMS AND THE ANALYTICAL DEFINITION OF THE DEFINITE INTEGRAL 157
Then, we have ðb
a
fðxÞdx¼ lim
jjPjj !0
Xn
r¼1
fðxrÞDxr
¼ lim
n! 1
Xn
r¼1
fðxrÞ h
¼ lim
n! 1
Xn
r¼1
fðaþrhÞ h; whereh¼ba n
¼ lim
n! 1
Xn
r¼1
hfðaþrhÞ; whereh¼ba n ðb
a
fðxÞdx¼ lim
n! 1
ba n
Xn
i¼1
f aþr ba n
ðBÞð15Þ
By using the formula (B), we can compute the definite integralÐb
afðxÞdx,as the limit of a sum.
For solving problems on “integral as the limit of a sum”, we shall require the following results, studied in earlier classes.
(1) Pn
r¼1
abr
¼Pn
r¼1
arPn
r¼1
br (2) Pn
r¼1k ar¼kPn
r¼1ar, wherekis a constant, independent ofr.
(3) Pn
r¼1
k¼nk, wherekis a constant.
(4) Pn
r¼1r¼nðnþ21Þ (5) Pn
r¼1r2¼nðnþ1Þð2nþ1Þ 6
(6) Pn
r¼1
r3¼n2ðnþ1Þ4 2
(7) IfSndenotes the sum of firstnterms of a G.P. whose first term is “a” and common ratio is r, then
Sn¼a 1rn 1r
;ifr<1
and Sn¼a rn1 r1
;ifr>1:
At this stage, we state the second fundamental theorem of Calculus, which links definite integralÐa
afðxÞdx to the antiderivative of f(x).
(15)If we choose the left-hand endpoint of each subinterval, then we will have the equation Ðb afðxÞdx¼
n! 1lim ba
n
Xn
i¼1 aþ ðr1Þ ba n
, which is comparatively not so convenient. Of course, the result remains the same in both the cases.
It states that Ðb
afðxÞdx¼ðbÞ ðaÞ, where Ð
fðxÞdx¼ðxÞ (which means is the antiderivative off).
The proof is not given here. It is introduced in Chapter 6a, and its applications are discussed in Chapter 7a of Part II.
Remarks:
(i) It is convenient to introduce a special symbol forðbÞ ðaÞ. We write ðbÞ ðaÞ ¼½ðxÞba:
Thus, we haveÐb
afðxÞdx¼½ðxÞba¼ðbÞ ðaÞ.
(ii) The concepts of the slope of the tangent line (derivative) and the area of the curved region (definite integral) were known long back. Historically, the basic concepts of the definite integral were used by the ancient Greeks, principally Archimedes’ (287–212 BC), more than 2000 years ago. That was many years before differential Calculus was discovered in the seventeenth century by Newton and Leibniz. The fact being that the concepts ofderivatives and definite integralwere known prior to the period of Newton and Leibniz, and that a number of mathematicians had contributed toward the development of the subject, the question is:Why then do Newton and Leibniz figure so prominently in the history of Calculus?
They do so because they understood and exploited the intimate relationship that exists between antiderivatives and definite integrals. It is this relationship that enables us to compute easily the exact values of many definite integrals without ever using Riemann sums. This connection is so important that it is called the second fundamental theorem of Calculus.(16)
Now, we proceed to evaluate some definite integrals by two methods: first as the limit of a sum and second by applying thesecond fundamental theorem of Calculus,which provides a very simple method to calculate definite integrals.
Illustrative Examples Example (1): ExpressÐ3
2xdxasthe limit of a sumand hence evaluate.
Solution:Divide the interval [2,3] inton equal parts. The length of each subinterval so obtained isðð32Þ=nÞ ¼ ð1=nÞ, and the partition formed by the points is given byP¼f2;ð2þ ð1=nÞÞ;
ð2þ ð2=nÞÞ;. . .;ð2þ ððn1Þ=nÞÞ;3g.
Method (I):Let the sample pointsxr¼xr¼ ð2þ ðr=nÞÞ(r¼1, 2, 3,. . .,n) (i.e., we choose the right-hand endpointof each subinterval as thesample point xrfor computing the sum).
(16)In fact, based on the definition of area functionA(x) (to be introduced in the next chapter), there are two basic fundamental theorems to be discussed later. To understand the (second) fundamental theorem we have to go through the (first) fundamental theorem of Calculus.
RIEMANN SUMS AND THE ANALYTICAL DEFINITION OF THE DEFINITE INTEGRAL 159
Then, we have, ðb
a
fðxÞdx¼ lim
n! 1
Xn
r¼1
fðxrÞ h;wherexr¼2þr
nandh¼ba n ¼1
n: )Recall the Formula (B). We have
ðb a
fðxÞdx ¼ lim
n! 1
ba n
Xn
i¼1
f aþr ba n
) ð3
2
xdx¼ lim
n! 1
1 n
Xn
r¼1
f 2þr n
¼ lim
n! 1
1 n
Xn
r¼1
2þr n
" #
) fðxÞ ¼x
f
¼ lim
n! 1
1 n 2Xn
r¼1
1þ1 n
Xn
r¼1
r
" #
¼ lim
n! 1
1 n 2:nþ1
n:nðnþ1Þ 2
¼ lim
n! 1 2þ1 2 1þ1
n
¼2þ1 2¼5
2 Ans:
Method (II):Now, using the second fundamental theorem of integral Calculus:
ðb
a
fðxÞdx¼½ðxÞba¼ðbÞ ðaÞ where
ð
fðxÞdx¼ðxÞ;we get ð3
2
xdx¼ x2 2
3 2
¼32 222
2
¼9 24
2¼5
2 Ans:
Example (2) ExpressÐ2
0ð3xþ5Þdxas the limit of a sum and hence evaluate.
Solution:Divide the interval [0,2] inton equal parts. The length of each subinterval¼2n0¼2n. The partition so formed by the points, is given by
P ¼ 0; 0þ2 n
; 0þ2 2 n
; 0þ3 2 n
;. . .; 0þ ðn1Þ 2 n
; 0þn 2 n
¼ 0; 1 2 n
; 2 2 n
;. . .; ðn1Þ 2 n
;2
Letxr¼xr¼0þr2n¼2rn;½r¼1;2;3;. . .;n(i.e., we choose theright-hand endpoint of each subintervalas the sample pointxr, for computing the sum).
Method (I):Now, we have by definition (i.e., the result “B”):
ðb
a
fðxÞdx¼ lim
n! 1
ba n
Xn
r¼1
f aþr ba n
wherea¼0 andb¼2 so thatððbaÞ=nÞ ¼2=n.
) ð2
0
ð3xþ5Þdx¼ lim
n! 1
2 n
Xn
r¼1
f 0þr2 n
¼ lim
n! 1
2 n
Xn
r¼1
f 2r n
¼ lim
n! 1
2 n
Xn
r¼1
3 2r n þ5
; f ) fðxÞ ¼3xþ5
) ð2
0
ð3xþ5Þdx¼ lim
n! 1
2 n
6 n
Xn
r¼1
rþ5Xn
r¼1
1
" #
¼ lim
n! 1
2 n
6
nnðnþ1Þ
2 þ5n
¼ lim
n! 1
2 n
3
nn2 1þ1 n
þ5n
Note thatðnþ1Þ ¼n 1þ1 n
¼ lim
n! 1 6 1þ1 n
þ10
¼6þ10¼16 Ans:
Method (II):Using the fundamental theorem of integral Calculus:
ð2 0
ð3xþ5Þdx¼ 3x2 2 þ5x
2
0
; where 3x2 2 þ5x
is antiderivative ofð3xþ5Þ
¼ 322 2 þ5ð2Þ
ð3:0þ5:0Þ
¼6þ100¼16 Ans:
Example (3): FindÐ2
0ðx2þ1Þdxas the limit of a sum.
Solution: Divide the interval [0,2] into n equal parts. The length of each subinterval¼ ðð20Þ=nÞ ¼2=n. The partition so formed by the points, is given by
P ¼ 0; 0þ2 n
; 0þ2 2 n
; 0þ3 2 n
;. . .; 0þ ðn1Þ 2 n
; 0þn 2 n
¼ 0; 1 2 n
; 2 2 n
;. . .; ðn1Þ 2 n
;2
Letxr¼xr¼0þr 2n ¼2rn, (i.e., we choose theright-hand endpoint of each subintervalas the sample pointxr, for computing the sum).
RIEMANN SUMS AND THE ANALYTICAL DEFINITION OF THE DEFINITE INTEGRAL 161
Method (I):We have by definition (i.e., the result (B)):
ðb
a
fðxÞdx¼ lim
n! 1
ba n
Xn
r¼1
f aþr ba n
Herea¼0,b¼2, f(x)¼x2þ1,h¼ba n ¼20
n ¼2 n
) ð2
0
ðx2þ1Þdx¼ lim
n! 1
2 n
Xn
r¼1
f 0þr 2 n
¼ lim
n! 1
2 n
Xn
r¼1
f 2r n
" #
¼ lim
n! 1
2 n
Xn
r¼1
2r n
2
þ1
" #
) fðxÞ ¼x2þ1
¼ lim
n! 1
2 n
Xn
r¼1
4r n2
2
þ1
¼ lim
n! 1
2 n
4 n2
Xn
r¼1
r2þXn
r¼1
1
" #
¼ lim
n! 1
8
n3nðnþ1Þð2nþ1Þ
6 þ2
nn
¼ lim
n! 1
8
6n3n3 1þ1 n
2þ1 n
þ2
¼ 4
3ð1þ0Þð2þ0Þ þ2
¼14 3 Ans:
Method (II):Using the second fundamental theorem of integral Calculus.
ð2 0
ðx2þ1Þdx¼ x3 3 þx
2
0
;where x3 3þx
is the antiderivative ofðx2þ1Þ
¼ ð2Þ3 3 þ2
! 0
" #
¼8 3þ2
¼14 3 Ans:
Example (4): EvaluateÐ2
0exdx, using the definition of a definite integral as the limit of a sum.
Solution:Divide the interval [0,2] inton equal partsso that we get each subinterval of the lengthðð20Þ=nÞ ¼2=n. Then, the partition formed by the points is given byP¼f0;ð2=nÞ;
ð4=nÞ;ð6=nÞ;. . .;ðð2ðr1ÞÞ=nÞ;2g.
Method (I):Letxr¼xr¼0þr 2n ¼2rn We have by definition (i.e., the result “B”)
ðb
a
fðxÞdx¼ lim
n! 1
ba n
Xn
r¼1
f aþr ba n
:
Herea¼0,b¼2,f(x)¼ex,ððbaÞ=nÞ ¼2=n )
ð2 0
exdx¼ lim
n! 1
2 n
Xn
r¼1
f 0þr 2 n
¼ lim
n! 1
2 n
Xn
r¼1
f 2r n
¼ lim
n! 1
2 n
Xn
r¼1
e2r=n; ½ ) fðxÞ ¼ex
¼ lim
n! 1
2
n e2=nþe4=nþe6=nþ þe2n=n
h i
The sum in the square bracket is a geometric series with the first term¼e2/nand the common ratioðr0Þ ¼ee4=n2=n¼eð4=n2=nÞ¼e2=n.
)This sumðSnÞ ¼e2=n 1 ðr0Þn 1r0
¼e2=n 1 ðe2=nÞn 1e2=n
)Sn¼e2=nð1e2Þ 1e2=n )
ð2 0
exdx ¼ lim
n! 1
2 n
ð1e2Þ e2=n 1e2=n
¼L ðsayÞ
Put (2/n)¼ton right-hand side and note that asn! 1;t!0 We get
L ¼lim
t!0t ð1e2Þet 1et
¼lim
t!0
ð1e2Þet
ð1etÞ=t; ½Imp step
¼lim
t!0
ðe21Þet ðet1Þ=t
¼ðe21Þe0
logee ¼ ðe21Þ 1
) lim
x!0
ax1
x ¼logea )lim
t!0
et1
t ¼logee¼1 2
66 64
3 77 75 )
ð2 0
exdx¼e21 Ans:
RIEMANN SUMS AND THE ANALYTICAL DEFINITION OF THE DEFINITE INTEGRAL 163
Method (II):Using the fundamental theorem of integral Calculus:
ð2 0
exdx¼½ ex20;where exis the antiderivative of ex
¼e2e0
¼ ðe21Þ Ans:
The four examples are meant to illustrate the theory behind the concept ofdefinite integrals, as the limit of a sum.Also we have seen that by applying the second fundamental theorem of Calculus, we can compute, very easily, the exact values of these definite integrals without using Riemann sums. Hence, we now dispense with the terminology of antiderivatives and antidiffer-entiation and begin to call the expressionÐ
fðxÞdx an indefinite integral—the term derived from the definite integral. Accordingly, the process of evaluating bothan indefinite integralor the definite integralis calledintegration.This is what we had pointed out in Chapter 1 of this volume.
In terms of the symbol for indefinite integrals, we may write the conclusion of the second fundamental theorem as:Ðb
a
fðxÞdx¼ Ð fðxÞdx
b
a.
Note that, for applying the second fundamental theorem,an important requirementisto find the indefinite integralÐ
fðxÞdx, by using any suitable technique that we have learnt in previous chapters.
Note: The distinction between an indefinite integral and the definite integral should be emphasized. The indefinite integralÐ
fðxÞdx represents (jointly) all functions whose derivative is f(x). However,the definite integralÐb
a
fðxÞdxis a number whose value depends on thefunction f and the numbers a and b, and is defined as the limit of a Riemann sum.
Remark: Note that the definition of the definite integral makes no reference to differentiation.
Important Note:Integral Calculus(like the differential Calculus) has important applications in situations where the quantities involved vary. We know that the area of the rectangular region changes if one or both of its dimensions are changed. If we consider the shaded region of Figure 5.3, wherein the height [of the curvey¼f(x)] varies as we travel across the region from left to right, then the area of the regionRchanges continuously. This is the type of situation where integral Calculus comes into play. More complicated situations are considered in Chapter 8a of this book.
6a The Fundamental Theorems of Calculus
6a.1 INTRODUCTION
Until now, the limiting processes of thederivativeanddefinite integralhave been considered as distinct concepts. We shall now bring these fundamental ideas together and establish the relationship that exists between them. As a result, definite integrals can be evaluated more efficiently.
We have defined the definite integralÐb
afðxÞdx, as thelimit of a sumand have had some practice in estimating the integral. Calculating definite integrals this way isalways tedious, usually difficult, and sometimes impossible.(1)
Since evaluation of the definite integral Ðb
afðxÞdx has a great variety of important applications, it is highly desirable to have an easy way to computeÐb
afðxÞdx.
The purpose of this section is to develop ageneral methodfor evaluatingÐb
afðxÞdxthat does not necessitate computing various sums. The method will allow us to evaluate many (but not all) of the definite integrals that arise in applications. It turns out that the exact value of Ðb
afðxÞdx can be easily found if we can compute Ð
fðxÞdx [i.e., if we can find the antiderivativeoff(x)].
6a.2 DEFINITE INTEGRALS
In the previous chapter, we evaluated certain definite integrals using two methods: first-as the limit of a sum(which is based on the definition of definite integral) andsecond- by applying the second fundamental theorem of Calculus, for which it is only necessary (as we will see shortly) that one should be able to computeÐ
fðxÞdxto evaluateÐb
afðxÞdx.
Having experienced the convenience in estimating exact values of definite integrals by the second method, one should appreciate the elegance and beauty of such easy computations.(2)
Introduction to Integral Calculus: Systematic Studies with Engineering Applications for Beginners, First Edition.
Ulrich L. Rohde, G. C. Jain, Ajay K. Poddar, and A. K. Ghosh.
2012 John Wiley & Sons, Inc. Published 2012 by John Wiley & Sons, Inc.
(1)In fact, we have been able to evaluate a few definite integrals directly from the definition (as the limit of a sum) only because we have nice formulas for 1þ2þ3þ þn, 12þ22þ32þ þn2, and so on.
(2)This method is of great practical importance, since it enables us to calculate not only areas, but also volumes, lengths of curves, centers of mass, moments of inertia, and so on, which are capable of being expressed in the form Px¼b
x¼afðxÞDx.
The definite integral 6a-The fundamental theorems of the Calculus and their applications (Differentiation and integration as inverse processes and the MVT for integrals)
165
Now, we shall show why antidifferentiation enables us to find the area under the graph of the function f(x). Thetrickis to considerthe integral functionÐx
a fðxÞdx(to be discussed shortly). Let Ðx
afðxÞdx¼AðxÞ, then we shall show thatA0(x)¼f(x), and this will remove the mystery. If thederivative of the integral functionÐx
afðxÞdxisf(x), then surely to find A(x), we should find an antiderivative of f(x) [i.e., we must evaluate ÐfðxÞdx].
To understand the approach of Newton and Leibniz in developing the two theorems, we use the Integrability Theorem, which states thatif f is continuous on[a,b], thenÐb
afðxÞdx exists.
We begin our development of these theorems by discussingdefinite integrals having a variable upper limit.
Let the functionfbecontinuous on the closed interval[a,b]. Then, thevalue of the definite integralÐb
afðxÞdx depends onlyonfand the numbers “a” and “b”, andnoton the symbolx, used here as the variable of integration. In other words, the definite integralsÐb
afðxÞdx,Ðb
afðtÞdt, andÐb
afðuÞdu, and so on, represent the same (closed) area fromatob(Figure 6a.1). For the present, we will assume thata<b.
Let us nowuse the symbol x to represent a number in the closedinterval [a,b]. Then, becausef iscontinuouson [a,b],it is continuouson [a,x] andÐx
afðxÞdxexists. It represents the area enclosed by the graph offand thex-axis, fromatox(Figure 6a.2).
Furthermore,this definite integral is a unique number whose value depends on x, that is the upper limit of the integral. It is a new function ofx(in the form of a definite integral, with a variable upper limit). We call it the area function and denote it byA(x).
In order to usex as a variable in our discussion, we write it as an upper limit (of a definite integral) and replace the expressionf(x)dxby the expressionf(t)dt. Thus, we get the definite integral Ðx
afðtÞdt [in place ofÐx
afðxÞdx] that clearly indicates the upper limit x, avoiding the confusion with the variable of integration, which is nowt.
y
y = f(x)
x
a b
a
f(x) dx = f(t) dt,etc.
b a
b
FIGURE 6a.1
6a.3 THE AREA OF FUNCTIONA(x) We have defined Ðb
afðxÞdxas the area of the region bounded by the curve y¼f(x), the ordinatesx¼aandx¼b, and thex-axis (Figure 6a.1). Sincexis a point in [a,b],Ðx
afðxÞdx represents the area of the shaded region inFigure 6a.2. Here, it is assumed thatf(x)0 for x2[a,b].
Thus,A(x) definesa function of x which is the variable upper limit of the integral function Ðx
afðxÞdx, whose domain is all numbers in [a,b] and whose function valueat any number x in [a,b] is given by
AðxÞ ¼ ðx
a
fðxÞdx ð1Þ
Note that,f iscontinuouson the interval [a,b] and sincef(x)0,its graph does not fall below the x-axis.
6a.3.1 First Fundamental Theorem of Calculus
From the definition of the area functionA(x), we state its two properties immediately:
(i) AðaÞ ¼Ða
afðxÞdx¼0,since there is no area from a to a.
(ii) AðbÞ ¼Ðb
afðxÞdx,represents the area from a to b.
If x is increased byhunits, then the area of the shaded region¼A(xþh), as shown in Figure 6a.3. Hence, thedifference of areasin Figures 6a.3 and 6a.2 will beA(xþh)A(x), as shown by the area of the shaded region in Figure 6a.4.
y
y = f(x)
x x
a A(x)
b
A(x) = f(t) dt a x
FIGURE 6a.2
THE AREA OF FUNCTIONA(x) 167
The area of the shaded region in Figure 6a.4 is the same asthe area of a rectanglein Figure 6a.5, whose base is h and height is some value y between f(x) and f(xþh).
Thus, the area of this rectangle is, on the one hand, A(xþh)A(x), andon the other hand it is hy.
Therefore, we haveA(xþh)A(x)¼hyorðAðxþhÞ AðxÞÞ=ðhÞ ¼y.
Ash!0, theny approaches the number f(x)[as clear from Figure 6a.5], and so,
h!lim0
AðxþhÞ AðxÞ
h ¼fðxÞ ð2Þ
But the left-hand side of(2)is merely the derivative of A(x). Thus, equation (2) becomes A0(x)¼f(x).
y
x
a x b
FIGURE 6a.3
y
x
x+h
a x b
FIGURE 6a.4
Note that equation (2) can also be expressed as A0ðxÞ ¼ d
dx ðx
a
fðxÞdx¼fðxÞ ðIÞ
This result is a crucial equation and it is so important that it is called the first fundamental theorem of Calculus.
We conclude that the area functionA(x)has the additional property of having its derivative A0(x)asf(x). In other words,A(x) is an antiderivative off(x).
6a.3.2 The Background for the Second Fundamental Theorem
Now, using the first fundamental theorem of Calculus, we try to understand the second fundamental theorem that is useful in evaluating definite integrals. Suppose (x) is any antiderivative of f(x), then we have0(x)¼f(x).
Since bothA(x) and(x) areantiderivatives of the same function, we conclude that they must differ by a constantc.
)AðxÞ ðxÞ ¼c
or AðxÞ ¼ðxÞ þc ð3Þ
Let us apply the property (i) ofA(x)to equation (3).(3)
SinceA(a)¼0, evaluating both sides of equation (3), whenx¼a, we get 0¼ðaÞ þC
) C¼ ðaÞ y
h y f(x+ h)
f(x)
x
FIGURE 6a.5
(3)This equation supports our observation that we made, as property (i) of the area functionA(x), defined on [a,b], that AðaÞ ¼Ða
a
fðxÞdx¼0
THE AREA OF FUNCTIONA(x) 169