02 definite integration (1-25) - vkr...

2DEFINITE

INTEGRATION

C H A P T E R

2.1 INTRODUCTIONThe definite integral is one of the basic concepts ofmathematical analysis and is a powerful research toolin mathematics, physics, mechanics, and otherdisciplines. Calculation of areas bounded by curves,of arc length, volumes, work, velocity, path length,moment of inertia, and so forth reduce to theevaluation of a definite integral.There are various problems leading to the notion ofthe definite integral : determining the area of a planefigure, computing the work of a variable force, findingthe distance travelled by a body with a given velocity,and many other problems.In the previous chapter we dealt with integration asthe inverse process of differentiation. The concept ofintegration first arose in connection with determinationof areas of plane regions bounded by curves and anintegral was recognized as the limit of a certain sum. Itwas only later that Newton and Leibnitz establishedan intimate relationship between the processes ofintegration and differentiation, known now as thefundamental theorem of integral calculus which weshall discuss in this chapter. A definite integral will bedefined as the limit of a sum and it will be shown howa definite integral can be used to define the area ofsome special region.A definite integral may be described as an analyticalsubstitute for an area. In the usual elementary treatmentof the definite integral, defined as the limit of a sum, itis assumed that the function of x considered may be

represented graphically, and the limit in question isthe area between the curve, the axis of x, and the twobounding ordinates, say at x = a and x = b.

The Area ProblemLet us understand the problem of finding the area ofa curvilinear trapezoid. Consider a nonnegativecontinuous function y = f(x), x [a, b].

Y

0 a

B

A

XbThe figure AabB bounded by a segment of the axis ofabscissas, segments of vertical lines x = a andx = b, and the graph of the given function is called acurvilinear trapezoid.In other words, a curvilinear trapezoid is the set ofpoints in the plane whose coordinates x, y satisfy thefollowing conditions: a x b, 0 y f(x).Let us find the area of this curvilinear trapezoid. Tothis end we partition the closed interval [a, b] into nsubintervals of equal length

[a, x1], [x1, x2], ....., [xn–1, b],using for this purpose the points

xi = a + inab , i = 0, 1, ...., n.

2.2 INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED

We then denote by mi and Mi the least and greatestvalues (respectively) of the function f(x) on the interval[xi–1, xi], where i = 1,....., n.

Y

0 a

BA

Xbx 1 x 2 x 3

The curvilinear trapezoid AabB is thus separated inton parts. Obviously, the area of the ith part is not lessthan mi(xi – xi–1) and is not greater than Mi(xi – xi–1).Consequently, the area of the entire curvilineartrapezoid AabB is not less than the sum

mix1 + ....... + mnxn =

n

1iii xm

where xi = xi – xi –1 and is not greater than the sum

M1x1 + ....... + Mnxn = n

i ii 1

M x

Denoting respectively these sums by sn and Sn, wesee that SAabB satisfy the inequalities

sn SAabB Sn,Here sn represents the area of the stepped figurecontained in the given curvilinear trapezoid, and Snthe area of the steplike figure containing the givencurvilinear trapezoid.If the interval [a, b] is divided into sufficiently smallsubintervals, i.e. if n is sufficiently large, then the areasof these figures differ but slightly from each other asalso from the area of the curvilinear trapezoid.Consequently, we may assume that the sequences(sn) and (Sn) have one and the same limit and this limitis equal to the area of the figure AabB.This assertion is obtained in the assumption that thecurvilinear trapezoid under consideration has an area,but the latter notion is not yet defined. And so, theabove reasoning leads us to the following definition.Definition Let there be given a continuousnonnegative function f(x), x [a, b]. Then, if the limitsof the sequences (sn) and (Sn) exist and are equal toeach other, their common value is called the area ofthe cuvilinear trapezoid.

{(x, y) | a x b, 0 y f(x)}.

Example 1. Let us show that, according to thegiven definition, the area of the right angled trianglewith the vertices at points (0, 0), (a, 0) and (a, b) is

equal to 21

ab, that is, it is computed by the knownformula.Solution The given triangle is a curvilinear

trapezoid for the function

f(x) = xab , x [0, a]

b

Y

0 a X

Using the points xi = ina , i = 0, 1,.....,n, we divide the

interval [0, a] into n equal parts of length na . Then

mi = f(xi –1) = )1i(nb

Mi = f(xi) = b in

and therefore

sn =

n11

2ab

2n)1n(

.nab

na)1i(

bn

2

n

1i

Sn =

n11

2ab

2n)1n(

.nab

na)i(

bn

2

n

1iHence, it is obvious that

2abSlimslim n

nn

n

Thus, it has been proved that the area of the given

triangle is equal to ab21 .

Example 2. Find the area of the figure boundedby a portion of the parabola y = x2 and segments ofthe straight lines x = 0 and x = a, where a > 0.Solution Proceeding as in previous example, we

partition the interval [0, a] into n subintervals each

of length na

, using points xi = ina , i = 0, 1,...., n.

DEFINITE INTEGRATION 2.3

Then

b

Y

0 a X

sn =

n

1i

23

3n

1i

21i )1i(

nax

na ,

Sn =

n

1i

23

3n

1i

2i i

nax

na

We have6

)1n2)(1n(ni

n

1i

2

Hence,

Sn =

n211

n11

3a

6)1n2)(1n(n

.na 3

3

3

sn =

n211

n11

3a

6)1n2(n)1n(

.na 3

3

3

and therefore

3aSlimslim

3

nn

nn

Thus, the area of the given figure is equal to 3a

31 .

Here we would like to note several points.

Note:1. Let us return to the curvilinear trapezoid AabB,

and proceeding in the usual manner, partition theinterval [a, b] into n parts of equal length, makinguse of points xi, i = 0, 1, ...., n. On each subinterval[xi–1, xi] we choose arbitrarily some point anddenote it by i .If mi and Mi are respectively, the least and thegreatest values of the function f on thesubinterval [xi -1, xi], then, obviously, mi f(i) Mi,where i = 1, ...., n. Let us now multiply either ofthese inequalities by xi = xi – xi–1 and addtermwise the inequlities thus obtained.Then we get the following inequality :

n

1iii

n

1iii

n

1iii xMx)(fxm

Hence, it follows that the limit

n

1iii

nx)(flim

exists, does not depend on the choice of points xi andis always equal to the area of the figure AabB.

Thus,

n

1iii

nx)(flim = SAabB ...(1)

2. In the previous examples we divided the interval[a, b] into n equal subintervals. It can be provedthat formula (1) remains true also for the casewhen [a, b] is separated into n parts of arbitrarylengths but such that the greatest of these lengthstends to zero as n .

The Definite IntegralConsider the function f(x) defined on the interval [a, b].As before, we divide the interval [a, b] into n equalsubintervals by means of points

xi = a + inab , i = 0, 1, ...., n.

On each of these subintervals [xi – 1, xi], i = 1, ...., n wechoose one point denoting it by i where i [xi–1 , xi]

Then the sum

f(1)x1 + .....+f(n)xn =

n

1iii x)(f

where xi = xi – xi –1 , is called an integral sum of thefunction f.Obviously, this sum depends both on the manner theinterval [a, b] is subdivided and on the choice of points i.

Definition. If the limit

n

1iii

nx)(flim exists and is

independent of the coice of points i , then the functionf is said to be integrable on the interval [a, b] and thelimit is called the definite integral or simply the integralof the function f(x) with respect to x over the interval [a,

b] and is denoted as b

af(x)dx

(read as " the integral of f(x)dx from a to b").The symbol is called the integral sign, the functionf(x) the integrand, x the variable of integration, theexpression f(x)dx the element of integration. Thenumbers a and b are called the lower and upper limitsof integration.Thus, according to the definition,nb

i inai 1

f(x)dx lim f( ) x

2.4 INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCEDThe interval [a, b] is called the interval of integration.The word 'limit' here has nothing to do with the wordlimit as used in differential calculus. It only signifiesthe 'end points' of the interval of integration.We have a theorem that every continuous function isintegrable, but integrability extends to a class offunctions wider than the class of continuous functions.

Note:1. The above definition of a definite integral is a

special case of the more generalized definition asgiven below.Let f(x) be a bounded function defined in theinterval (a, b), and let the interval (a, b) be dividedin any manner into n sub-intervals (equal orunequal) of lengths 1, 2, ......, n. In each sub-interval choose a perfectly arbitrary point (whichmay be within or at either end points of theinterval), and let these points be x = 1, 2, ..... n.

Let

n

1rrrn )(fS .

Now, let n increase indefinitely in such a way thatthe greatest of the lengths 1, 2, ...... , n tends tozero. If, in this case, Sn tends to a definite limitwhich is independent of the way in which theinterval (a, b) is sub-divided and the intermediatepoints 1, 2, ......, n are chosen, then this limit,when it exists, is called the definite integral of f(x)from a to b.

2. The process of forming the definite integral shows

that the symbol b

af(x) dx is a certain number..

Its value only depends on the properties of theintegrand and on the numbers a and bdetermining the interval of integration.

Geometrical interpretation of definiteintegral

In general, b

a f(x) dx represents an algebraic sum ofareas of the region bounded by the curve y = f(x), thex-axis and the ordinates x = a and x = b. Here algebraicsum means that area which is above the x–axis will beadded in this sum with + sign and area which is belowthe x–axis will be added in this sum with – sign.So, value of the definite integral may be positive, zeroor negative.This is because the value of f(x) in the integral sum is

considered without modulus sign. The area above thex-axis enter into this sum with a positive sign, whilethose below the x-axis enter it with a negative sign.

Y

X0 aA1

A2A3

A4 b

b

a f(x) dx = AA1 – A2 + A3 – A4where A1, A2, A3, A4 are the areas of the shaded region.

Hence, the integral b

af(x)dx represents the "net

signed area" of the region bounded by the curvey = f(x), x–axis and the lines x = a, x = b.

Example 3. Evaluate 4

1(2x 3)dx

.

Solution y = 2x – 3 is a straight line, which lie

below the x–axis in 3

1,2

and above in 3

, 42

x = 4 Xx = –1A

Y

D

E (4, 5)

C(3/2, 0)

B(–1, 5)–

Now area of ABC = 1 5

52 2 =

25

4

Area of CDE = 1 5

52 2 =

25

4(using formula from geometry)

So, 4

1(2x 3)dx

= 25 25 04 4 .

Example 4. Evaluate the following integrals byinterpreting each in terms of areas :

(a)1

2

01 x dx (b)

2

1(x 2)dx

(c)3

0(x 1)dx

Solution(a) We sketch the region whose area is represented

by the definite integral, and evaluate the integralusing an appropriate formula from geometry.


Since f(x) = 21 x 0 , we can interpret thisintegral as the area under the cure y = 21 x

from x = 0 to x = 1. We have y2 = 1 – x2, that isx2 + y2 = 1, which shows that the graph of f is thequarter-circle with radius 1.

1

1 X

Y

2x1y -=

0

Therefore, 1

2

01 x dx = area of quarter-circle

21 (1)4 4

.

(b) The graphs of the integrand is the line y = x + 2,so the region is a trapezoid whose base extendsfrom x = –1 to x = 2. Thus,

2

1(x 2)dx

= (area of trapezoid)

= 2

1(1 + 4)(3) =

2

15

0 X

Y

1–1 2

1

234 y=x+2

(c) The graph of y = x – 1 is a line with slope 1 asshown in the figure. We compute the integral asthe difference of the areas of the two triangles :

3

0(x 1) dx = AA1 – A2 =

1

2(2 . 2) –

1

2(1 . 1) = 1.5.

0 3 X

Y

1

–1

y = x – 1

A2

(3, 2)

A1

2.2 DEFINITE INTEGRAL ASA LIMIT OF SUM

Let us use rectangles to estimate the area under theparabola y = x2 from 0 to 1.

0 1

(1, 1)

X

Y

y = x2

S

We first notice that the area of S must be somewherebetween 0 and 1 because S is contained in a squarewith side length 1, but we can certainly do betterthan that. Suppose we divide S into four strips S1, S2,

S3, and S4 by drawing the vertical lines x = 1

4, x =

1

2a n d

x =3

4 as in Figure (a)

0 1

(1, 1)Y

y = x2

S4S3S2

S1

43

21

41 0 1

(1, 1)

X

Y

43

21

41X

(a) (b)We can approximate each strip by a rectangle whosebase is the same as the strip and whose height is thesame as the right edge of the strip [see Figure (b)]. Inother words, the heights of these rectangle are thevalues of the function f(x) = x2 at the right end pointsof the subintervals

1 1 1 1 30, , , , ,

4 4 2 2 4 and

3,1

4 .

Each rectangle has width1

4 and the heights are

2 2 21 1 3, ,

4 2 4

, and 12.

If we let R4 be the sum of the areas of these

2.6 INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCEDapproximating rectangles, we get R4

= 2 2 2

21 1 1 1 1 3 1 15, . . .14 4 4 2 4 4 4 32

= 0.46875

From the Figure (b) we see that the area A of S is lessthan R4, so A < 0.46875Instead of using the rectangles in Figure (b) we coulduse the smaller rectangles in Figure (c) whose heightsare the values of f at the left endpoints of thesubintervals. (The leftmost rectangle has collapsedbecause its height is 0). The sum of the areas of theseapproximating rectangles is

0 1

(1, 1)

X

Y

43

21

41

y = x2

Z

(c)

L4 =1

4 . 02 +

2 2 21 1 1 1 1 3 7. . .

4 4 4 2 4 4 32

= 0.21875We see that the area of S is larger than L4, so we havelower and upper estimates for A :

0.21875 < A < 0.46875We can repeat this procedure with a larger number ofstrips. Figure (d), (e) shows what happens when wedivide the region S into eight strips of equal width.

Y

18

18

1 1 X

Y

(1,1)(1,1)y=x2

0 X

(d) Using left endpoint (e) Using right endpointBy computing the sum of the areas of the smallerrectangles (L8) and the sum of the areas of the largerrectangles (R8), we obtain better lower and upperestimates for A :

0.2734375 < A < 0.3984375So, we can say that the true area of S lies somewherebetween 0.2734375 and 0.3984375.We could obtainbetter estimates by increasing the number of strips.For the region S in previous example, we now showthat the sum of the areas of the upper approximating

rectangles approaches 1

3.

Rn is the sum of the areas of the n rectangles in Figure.Each rectangle has width 1/n and the heights are thevalues of the function f(x) = x2 at the points 1/n, 2/n, 3/n,..., n/n; that is, the heights are (1/n)2, (2/n)2, (3/n)2, ...,(n/n)2.

(1, 1)Y

y = x2

0 1 X1n

Thus, Rn

= 2 2 2 21 1 1 2 1 3 1 n

...n n n n n n n n

= 21 1.n n

(12 + 22 + 32 + .... + n2)

= 31

n(12 + 22 + 32 + ... + n2)

= 2(n +1)(2n +1)

6nThus, we have

nlim Rn = 2n

(n 1)(2n 1)lim

6n

= n

1 n 1 2n 1lim

6 n n

= n

1 1 1lim 1 2

6 n n =

1 1. .1 26 3

It can be shown that the lower approximating sumsalso approach 1/3, that is,

2 22nn n

1 1 1 1 2lim L lim (0)

n n n n n + .....


+ 21 n 1n n= 2n

(n 1(2n 1) 1lim

36n

.F r o m F i g u r e s i t a p e a r s t h a t , a s n i n c r e a s e s , b o t h L nand Rn become better and better approximations tothe area of S. Therefore, we define the area A to bethe limit of the sums of the areas of the approximatingrectangles, that is,

A =nlim Rn = nlim Ln =

1

3.

Definite integral as a limit of sumLet f(x) be a continuous real valued function definedon the closed interval [a, b] which is divided into nparts as shown in the figure.

The points of division on the x-axis area, a + h, a + 2h ..........a + (n – 1)h, a + nh,

where b a

n

= h.

Left end estimationLet Ln denotes the area of these n rectangles.Then, Ln = hf(a) + hf(a + h) + hf(a + 2h) + ........

+ h f(a + (n – 1)h)Clearly, Ln represents an area very close to the area ofthe region bounded by curve y = f(x), x–axis and theordinates x = a, x = b.

Hence, b

af(x) dx = nlim Ln

b

af(x) dx = nlim h [f (a) + f (a + h) + f (a + 2h)

+ ..... + f (a + (n – 1)h)]

= nlim n 1

r 0

h f(a rh)

, where nh = b – a.

= nlim n 1

r 0

b a

n

f

(b a) ra

n

.

If for a function f(x) the limit exists, then we say thefunction is integrable on the interval [a, b].It can be shown that, when f(x) is a continuousfunction, the above limit always exists. Hence, if afunction f(x) is continuous on an interval [a, b], then itis integrable on that interval.

Right end estimationConsidering the sum of areas of rectangles using theheights at the right end points of the subintervals, wehaveRn = hf(a + h) + hf (a + 2h) + .........+ hf(a + nh) and

b

af(x) dx = nlim

n

r 1

b a

n

f

b aa r

n

It follows from theorems that for a continuous function f,

nlim Rn = nlim Ln =

b

af(x)dx .

That is, we may compute the integral using either theleft end estimation or the right end estimation.

Note:1. If a = 0, b = 1, then

1

0f(x) dx = nlim

n 1

r 0

1 rf

n n

2. From the definition of definite integral, we have(x)

b

n ar (x)

1 rlim f f(x)dx

n n

, where(i) is replaced by sign,

(ii)n

r is replaced by x,

(iii)n

1 is replaced by dx,

(iv) To obtain the limits of integration, we use

a = n

)x(limn

and b = n

(x)limn

.

For example, nlim

pn

r 1

1 rf

n n

=

p

0f(x) dx

since n

1lim

n

= 0, n

pnlim

n

= p.

Example 1. Calculate I = 4

1(1 x)dx

as the

limit of sums.


Solution We divide the interval [–1, 4] into n equalparts. On each subinterval

[xi–1, xi] =

ni51,

n)1i(5

1

the continuous function 1+ x attains the least value atthe left endpoint of the interval and the greatest valueat the right end point.Therefore

Ln n

i 1

5(i 1) 5f 1 .n n

=

n

1i

n

1i2

)1i(n25

n5)1i(

n5

Rn n

i 1

5i 5f 1 .n n

=

n

1i

n

1i2

in25

n5.i.

n5

Hence

Rn – Ln =

n

1i

n

1i2

)1i(in25

= 225 25n 0 as n

nn .

This means that the integral I = 4

1(1 x)dx

exists.

To calculate it as the limit of sums, we can considerany of the sequence of sums.Here, we use right end estimation.

I=n4

2 2n n1i 1

25 25n(n 1) 25(1 x)dx lim i lim

2n 2n

Thus, 4

1

25(1 x)dx

2 .


0 x2 dx.Solution The function f to be integrated is defined

by f(x) = x2, and the interval of integration is [0, 2].2

0 x2 dx = nlim RnThe partition {x0,... , xn} which subdivides [0, 2] inton subintervals of equal length is given by

xi = a +b a

n

i = 0 +

2 2ii

n n ,

for each i = 0, ...., n.

Moreover, xi – xi – 1 =b a

n

=

2

n, i = 1, ...., n.

0 1 X

Y

–2 –1 2

y = x2

Since f(xi) = x2i and xi =

2i

n, it follows that

Rn = n n n2 2

22 3 3

i 1 i 1 i 1

4i 2 8i 8i

nn n n

= 3 2

3 2

8 2n 3n n 4 3 12

6 3 nn n

,

and so nlim Rn= 2n

4 3 1 4 8lim 2 .2

3 n 3 3n .

We conclude that 2

2

0

8x dx

3 .


1 (5 – x) dx.Solution The function f, defined by f(x) = 5 – x, is

linear and decreasing on the interval [1, 4]. Its graphis shown in the figure.

X

y = 1

Y

4

y = 5 ̶ x

1

P

The partition {x0, . . ., xn} subdivides the interval

[1, 4] into subintervals of length 4 1 3

n n

, and the

points are given by

xi = 1 +3

n i, i = 0, . . . , n.


In addition, xi – xi – 1 = 3

n, i = 1, ......, n.

We shall compute the integral as4

1(5 x) dx = nlim Ln.

We have xi = 1 +3i

n and f(xi) = 5 – xi, and so

f(xi) = 5 –3i

1n

= 4 –

3i

n.

Since xi – xi – 1 = 3

n, we get

Ln = n

i 1

3i 34

n n

.

= n n

2i 1 i 1

3i 3 12 9i4

n n n n

= n n

2i 1 i 1

12 9i

n n =

n n

2i 1 i 1

12 91 i

n n

= 212 9 n(n 1)

nn 2n

= 12 –

9 11

2 n .

But it is easy to see that

n

9 1 9 1lim 12 1 12 7

2 n 2 2 ,

and we finally conclude that4

1(5 x) dx = nlim Ln = 7

1

2.

This answer can be checked by looking at the figure.The value of the integral is equal to the area of theshaded region P, which is divided by the horizontalline y = 1 into two pieces : a right triangle sitting ontop of a rectangle.

The area of the triangle is 1

2(3 . 3) =

9

2, and that of

the rectangle is 3 . 1 = 3. Hence4

1(5 x) dx = area(P) =

9

2 + 3 = 7

1

2.

The excessive lengths of the computations in theabove examples make it obvious that some powerfultechnique is needed to streamline the process ofevaluating definite integrals.

Example 4. Evaluate 3 3

0(x – 6x) dx using

limit of sum and interpret the result.Solution

y=x– 6x35

Y

X30 A2

A1

3 3

0(x – 6x)

= n n

in n

i 1 i 1

3i 3lim f(x ) x lim f

n n

= 3n

ni 1

3 3i 3ilim 6

n n n

=n

33n

i 1

3 27 18lim i i

n nn

= n n

34 2n

i 1 i 1

81 54lim i i

n n

= 2

4 2n

81 n(n 1) 54 n(n 1)lim

2 2n n

= 2

n

81 1 1lim 1 27 1

4 n n

=81

4 – 27 = –

27

4 = – 6.75

This integral can be interpreted as net signed areabecause f takes on both positive and negative values.It is the difference of areas A1 – A2, where A1 and A2are shown in the figure.

33

0(x 6x)dx = AA1 – A2 = – 6.75

Example 5. Express the following limit as a

definite integral : n

1 1 1lim ...

n 1 n 2 6n

.

Solution n1 1 1

lim ...n 1 n 2 6n


= n1 1 1

lim ...n 1 n 2 n 5n

5n

n r 1

1limn r

=

5n

nr 1

1 1lim

rn 1n

Since the lower limit of r is 1, the lower limit of

integration = n

1lim 0n

Since the upper limit of r is 5n, the upper limit of

integration is n

5nlim 5n .

Hence, the given limit is equivalent to

5

0

1dx

1 x.

Example 6. Use the definition of the integral as

the limit of a sum to evaluate 0

sin xdx

Solution Here a = 0, b = , = nh, f(x) = sinx.

n

n rsin x dx lim ( / n)sin(r / n)

01

nnsin...

n2sin

nsin

nlim

n

n2sin

n2nsin

n.

21n

nsin

nlim

n

n

2( / 2n) 1 1lim . sin 1 sinsin( / 2n) 2 n 2

= 2. 1 sin

2 sin

2 = 2 sin2

2 = 1 – cos.

Example 7. Show that b

a

1 bdx ln

x a (0 < a < b)

using limit of sum where the interval (a, b) is divided inton parts by the points of division a, ar, ar2, ....... , arn–1, arn.Solution We have arn = b, i.e. r = (b/a)1/n.

Evidently as n , r = (b/a)1/n 1, so that each ofthe intervals a(r – 1), ar(r – 1) ..... 0. Now, bydefinition,

nb

ia ni 1

1 dx lim f (x ) xx

= n

k 1nk 1

1limar

(ark – ark–1)

n nlim (r ) lim n(r )

1 1/ n

nlim n[(b / a) ]

1 1

1/n

n

(b / a) 1lim1/ n

blna

.

Example 8. Evaluate the integral 2

1

dx

x.

Solution We subdivide the interval [1, 2] into nparts so that the points of division xi (i = 0, 1, 2, ..., n)form the geometric progression :

x0 = 1; x1 = q; x2 = q2; x3 = q3; ....; xn = qn = 2.Here q = n 2 .The length of the ith subinterval is equal to

xi = qi + 1 – qi = qi(q – 1),and qn – 1(q – 1) 0 as n , i.e. as q 1.Now let us choose the right hand endpoints of thesubintervals as the points xi + 1 = qi + 1.Forming an integral sum :

Rn =

1n

0i1iq

1 qi(q – 1) = qn

(q – 1) = 1n

1n

1 n(2 1)

2

1n

n 1n nn

n(2 1)lim R lim

2

n 2

and so, 2

1

dx

xn 2.

Example 9. Evaluate the integral b

2a

dx

x(a < b).

Solution

Sn=h 2 2 2 21 1 1 1

....a (a h) (a 2h) (a n 1h)

< h1 1 1

(a h)a a(a h) (a h)(a 2h)

1....

(a n 2 h) (a n 1h)


= 1 1 1 1 1 1

(a h) a a (a h) (a h) (a 2h)

1 1....

(a n 2 h) (a n 1h)

Thus Sn < hnha1

ha1

...(1)Similarly, Sn >

h

.......

)h2a)(ha(1

)ha(a1

= 1 1 1 1

a (a h) (a h) (a 2h)

1 1....

(a nh)(a n 1h)

Thus Sn > 1 1

a a nh

...(2)

We have b = a + nhFrom (1) and (2), using Sandwich Theorem, we get

b

2a

dx

x=

nlim Sn = h 0

1 1lim

a h b h

b1

a1

Example 10. Evaluate the integral b

a

dx

x, a > 0 ,

b > 0Solution By definition

Sn = h 1 1 1 1

......a a h a 2h a (n 1)h

We know that r2 < hr + r for sufficientlysmall h > 0.

r21

> rhr1 = h

rhr

rh

> ]rhr[2 Substituting r = a , a + h , a + 2h , ....... we have

ah

> 2 aha hah2a2

hah

h2ah3a2h2a

h

h)1n(anha2h)1n(a

h

On addition,

1 1 1h ....

a a h a n 1h

2 a nh a = aaba2 = ab2

nlim Sn ab2

Similarly, by considering hrrr2 , wecan prove that nlim Sn ab2 .Hence,

b

a

dx

x = nlim Sn = ab2 .

A1. Compute the area of the figure bounded by a

portion of the straight line y = x and segments ofthe straight lines y = 0 and x = 3.

2. Evaluate

0

2

24 x dx ,

3. Find the area of the curvilinear trapezoid definedby the graph of the function y = ex on the interval 0 x 1.

4. First show that 0

sin xdx

=

n

1kn n

ksin

nlim

Use the fact

n

1k n2cot

n

ksin and L'Hospital's

rule to show that finally that 0

sin xdx 2

.

5. Evaluate the following integrals as limit of sums:

(i) 4

2

1(x x)dx (ii)

32

0(x 1)dx

(iii) 2

3

0(2x 5)dx (iv)

12x

0(x e )dx

(v)

/ 2

0cos x dx (vi)

b

asin x dx

6. Let f(x) denote a linear function that isnonnegative on the interval [a, b]. For each valueof x in [a, b], define A(x) to be the area betweenthe graph of f and the interval [a, x].

(a) Prove that A(x) = 2

1[f(a) + f(x)] (x – a).

(b) U s e p a r t ( a ) t o v e r i f y t h a t A (x) = f(x).


7. Use appropriate formulas from geometry toevaluate the integrals.

(a)1

2

0(x 2 1 x )dx (b)

3

1(4 5x)dx

(c)2

2(1 3 | x |)dx

8. Evaluate the integral10

2

010x x dx by

complet ing the square and applyingappropriate formulas from geometry.

9. Prove that the area of the curvilinear trapezoiddefined by a portion of the parabola y = 1 – x2

(0 x 1) is equal to 32 .

10. Determine a region whose area is equal to the

limit

n

1in n4itan

n4lim . Do not evaluate the limit.

11. Express the integral

6

52

xdx

1 xas a limit of

integral sums. Do not evaluate the limit.

12. Evaluate 2n

ni 1

2 2ilim 5

n n

13. Explain why 100

i 1

1 if100 100

is an estimate of1

0f(x) dx .

A14. Prove using the concept of definite integral as

limit of sum that

n

1k

1n

0k n

kacos

n

aasin

n

kacos

n

a, 0 < a <

2

15. Form the integral sum sn by dividing the interval[a, b] into n parts by the points xi = aqi (i = 0, 1, 2,

.., n), where q = n ba

and pass to the limit to

compute the following definite integrals:

(i)b

2

ax dx

(ii)b

a

dx,

x where 0 < a < b

(iii)b

ax dx.

16. Let A denote the area between the graph off(x) = x and the interval [0, 1], and let Bdenote the area between the graph of f(x) = x2and the interval [0, 1]. Explain geometrically why A + B = 1.

17. Let A denote the area between the graph off(x) = 1/x and the interval [1, 2], and let B denotethe area between the graph of f and the interval

1,2

1. Explain geometrically why A = B.

2.3 RULES OF DEFINITEINTEGRATION

We have a number of rules regarding the definite

integral b

af(x)dx .

1. If a = b, then x = 0 and so f(x)a

adx = 0. This is

natural also from the geometric standpoint.Indeed, the base of a curvilinear trapezoid haslength equal to zero. Consequently, its area iszero too.

2. Order of integrationa

bf(x) dx = –

a

bf(x) dx

When we defined the definite integral b

af(x) dx, we

implicitly assumed that a < b. But the definition as alimit of sum makes sense even if a > b.Notice that if we reverse a and b, then x changesfrom (b – a)/n to (a – b)/n.

Therefore, a

bf(x) dx = –

b

af(x) dx.


Example 1. Evaluate the definite integrals :

(a) sin x

dx (b) 1

4(5 x) dx

Solution(a) Because the sine function is defined at x = , and

the upper and lower limits of integration are

equal, we can write sin x

dx = 0.

(b) This integral is the same as 4

1(5 x) dx except

that the upper and lower limits are interchanged.Because the integral has a value of 7.5, we can

write 1

4(5 x) dx = –

4

1(5 x) dx = –7.5.

3. Dummy variable

The definite integral b

af(x) dx is a number which

depends only on the form of the function f(x) and thelimits of integration, and not on the variable of inte-gration, which may be denoted by any letter. In fact,we could use any letter in place of x without changingthe value of the integral

b

af(x) dx =

b

af(t) f(t) dt =

b

af(u) du.

Because the variable of integration in a definiteintegral plays no role in the end result, it is oftenreferred to as a dummy variable.Whenever you find it convenient to change theletter used for the variable of integration in adefinite integral, you can do so without changingthe value of the interval.This result should not be surprising, since the areaunder the graph of the curve y = f(x) over an interval[a, b] on the x-axis is the same as the area under thegraph of the curve y = f(t) over the interval [a, b]on the t-axis (See figure).

A

a b X

Y

y=f(x)

A

a b t

Y

y=f(t)

4. Homogeneous Propertyb

acf(x) dx = c

b

af(x) dx, where c is any

constant.This property says that the integral of a constanttimes a function is the constant times the integral ofthe function. In other words, a constant (independentof x) can be taken in front of an integral sign.

For example, b

a2f(x) dx = 2

b

af (x) dx

Y

Xa b

y = f(x)

Y

Xa b

y = 2f(x)

5. Additivity Propertyb

a [f(x) + g(x)] dx = b

a f(x)dx + b

a g(x) dx

and b

a [f(x) – g(x)] dx = b

a f(x) dx – b

a g(x) dxThis property says that the integral of a sum is thesum of the integrals. It says that the net signed areaunder f + g is the area under f plus the area under g.The figure helps us understand why this is true inview of how graphical addition works.


Y

X

f + g

0 a b

g

f

+=+b

a

b

a

b

adx)x(gdx)x(fdx)x(g)x(f[

In general, this property follows from the fact thatthe limit of a sum is the sum of the limits :

nb

i ia n

i 1

[f(x) g(x)]dx lim [f(x ) g(x )] x

= n n

i in

i 1 i 1

lim f(x ) x g(x ) x

= n n

i in n

i 1 i 1

lim f(x ) x lim g(x ) x

= b

a f(x) dx + b

a g(x) dx.The above two properties can be combined intoone formula known as the linearity property.

6. Linearity PropertyFor every real c1 and c2, we have

b

a [c1f(x) + c2g(x)] dx = c1b

a f(x) dx + c2b

a g(x) dx.Example 2. Use the properties of integrals to

evaluate1

0 (4 + 3x2) dx.Solution Using linearity property of integrals, we

have1

0 (4 + 3x2) dx = 1

0 4 dx + 1

0 3x2 dx=

1

0 4 dx + 3 1

0 x2 dxWe have

1

0 4 dx = 4(1 – 0) = 4 and we found in one

of the previous examples that 1

0 x2 dx = 31

. So,1

0 (4 + 3x2) dx = 1

0 4 dx + 3 1

0 x2 dx

= 4 + 3 . 31

= 5.

The next property tells us how to combine integrals ofthe same function over adjacent intervals :

7. Additivity with respect to the intervalof integration

c

a f(x) dx + b

c f(x) dx = b

a f(x) dxThis theorem reflects the additive property of area.If an interval is decomposed into two intervals, thesum of the areas of the two parts is equal to thearea of the whole.Note that c may or maynot lie between a and b,Property 2 allows us to conclude that this propertyis valid not only when c is between a and b but forany arrangement of the points a, b, c, but thefunction must be integrable in the desired intervals.It is easy to prove for the case where f(x) 0 and a < c <b. This can be seen from the geometric interpretationof the following figure

0 Xa b

Y

c

y = f(x)

The area under y = f(x) from a to c plus the area fromc to b is equal to the total area from a to b.

Example 3. Given that 10

0 f(x) dx = 17 and8

0 f(x) dx = 12, find 10

8 f(x) dx.Solution By the rule of additivity, we have

10

8 f(x) dx + 10

8 f(x) dx = 10

0 f(x) dx

So,10

8 f(x) dx = 10

0 f(x) dx – 8

0 f(x) dx = 17 – 12 = 5

Example 4. If for every integer n, n 1

nf(x)

dx = n2, then find the value of

4

2f(x)dx.

Solution We have n 1

nf(x)

dx = n2

DEFINITE INTEGRATION 2.15Putting n = 2, 1, 0, 1, 2, 3 we get

1

2f(x) = 4 ,

0

1f(x) dx = 1 ,

1

0f(x) = 0,

2

1f(x) dx = 1,

3

2f(x) dx = 4 ,

4

3f(x) dx = 9

Hence, 4

1f(x) dx = 4 + 1 + 0 + 1 + 4 + 9 = 39.

Comparision properties of integralNext, we have a comparison theorem which tells usthat if one function has larger values than anotherthroughout [a, b], its integral over this interval isalso larger.

1. If f(x) 0 for a x b, then b

a f(x) dx 0.2. Domination Law

If f(x) g(x) for for every x in [a, b], then

b

a f(x) dx b

a g(x) dx.3. If f(x) < g(x) for every x in [a, b], then

b b

a af(x)dx g(x)dx .

4. Max-Min InequalityIf m f(x) M for a x b, then

m (b – a) b

a f(x) dx M(b – a)

If f(x) 0, then b

a f(x) dx represents the area underthe graph of f, so the geometric interpretation ofProperty 1 is simply that areas above the x-axis arepositive.Property 2 says that a bigger function has a biggerintegral. It follows from Property 1 because g – f 0.

0 Xa b

Y

y = f(x)

m

M

Max-Min Inequality is illustrated in the figure forthe case where f(x) 0. If f is continuous we couldtake m and M to be the absolute minimum and

maximum values of f on the interval [a, b]. In this casethe inequality says that the area under the graph of fis greater than the area of the rectangle with height mand less than the area of the rectangle with height M.Since m f(x) M, Property 2 gives

b

a m dx b

a f(x) dx b

a M dxEvaluating the integrals on the left and right sides,we obtain

m(b – a) b

a f(x) dx M(b – a)This inequality is useful when we want to find a roughestimate of the value of an integral.Example 5. Use Max-Min Inequality to estimate

21 x

0e dx.

Solution Because f(x) = 2xe is a decreasing

function on [0, 1], its absolute maximum value isM = f(0) = 1 and its absolute minimum value is m = f(1)= e–1. Thus by Max-Min Inequality,

e–1(1 – 0) 21 x

0e dx 1(1 – 0)

or e–1 21 x

0e dx 1

Since e–1 0.3679, we can write

0.367 21 x

0e dx 1

The result is illustrated in the figure.

0 X1

Y y = 11

y = 1/e

2xey –=

The integral is greater than the area of the lowerrectangle and less than the area of the square.

Example 6. Prove that 4 3

3

13 x dx 2 30

Solution Since the function f(x) = 3x3 increaseson the interval [1, 3]. M = maximum value of 3x3

= 333 = 30


m = minimum value of 3x3= 33 1 = 2

m = 2, M = 30 , b – a = 2

Hence, 2.2 3

3

13 x dx 2 30

4 3

3

13 x dx 2 30

Mean Value Theorem for IntegralsIf f is continuous on the interval [a, b], there is atleastone number c between a and b such that

b

a f(x) dx = f(c) (b – a)Proof : Suppose M and m are the largest and smallestvalues of f, respectively, on [a, b]. This means that

m f(x) M when a x bb

a m dx b

a f(x) dx b

a M dx[ Domination Law ]

m(b – a) b

a f(x) dx M(b – a)

m b

a

1

b a f(x) dx MBecause f is continuous on the closed interval [a, b]and because the number

I = b

a

1

b a f(x) dxlies between m and M, the Intermediate Value Theoremsays there exists a number c between a and b forwhich f(c) = I, that is,

b

a

1

b a f(x) dx = f(c)

b

a f(x) dx = f(c) (b – a).The Mean Value Theorem for Integrals does notspecify how to determine c. It simply guarantees theexistence of atleast one number c in the interval.Since f(x) = 1 + x2 is continuous on the interval[– 1, 2], the Mean Value Theorem for Integrals saysthere is a number c in [–1, 2] such that

2

1 (1 + x2) dx = f(c)[2 – (–1)] ...(1)In this particular case we can find c explicitly.

From definite integral as limit of sum, we can evaluate2

1 (1 + x2) dx to be equal to 6.Placing this value in (1), we get f(c) = 2.Therefore, 1 + c2 = 2 so c

2 = 1.Thus, in this case there happen to be two numbersc = ± 1 in the interval [–1, 2] that work in the MeanValue Theorem for Integrals.

0 X

Y

f = 2avg

–1 1 2

(–1, 2)

(2, 5)y=1 + x2

For a nonnegative function the Mean Value Theoremfor Integrals has a simple geometrical interpretation.It asserts that the area of the curvilinear trapezoidcorresponding to the function f is equal to the area ofthe rectangle whose base is equal to the base of thetrapezoid, and the altitude to one of the values of theintegrable function.

Note: The formula in the theorem holds truenot only for integrals in which the lower limit ofintegration is less than the upper one, but also forthose in which the lower limit exceeds the upper one.Example 7. Let f be the function defined by

1 11 x, when 0 x , x 12 2f (x)

10, when x2

Show that 1

0

3f (x)dx2

but there is no point c in [0, 1]such that

1

0f (x)dx f (c) .

Solution Here f is bounded and has only one point of

discontinuity, i.e. x = 21 .

To find 1

0f (x)dx , we consider the partition

1 2 n0, , , ... ,n n n

obtained by dissecting [0, 1]into n equal parts.

DEFINITE INTEGRATION 2.17n1

0 nr 1

1 rf (x)dx lim 1n n

nn1....

n11

n1lim

n

n

1 n(n 1)lim nn 2

n

3n 1 3lim2n 2

.

But there is no point in [0, 1] at which f takes thisvalue. The only likely candidate for the point c is thepoint x = 1/2 but f(1/2) = 0. This does not contradictthe Mean Value Theorem for Integrals as the givenfunction is discontinuous.

Average value of a functionIf f is integrable on the interval [a, b], the averagevalue of f on this interval is given by the integral

f avg = = b

a

1

b a f(x) dx.

From the Mean Value Theorem for Integrals weconclude that this average value is definitely attainedby continuous functions at some c [a, b].Hence, f(c) = Example 8. Find the average value of the

function f(x) = 1 + x2 on the interval [–1, 2].Solution With a = –1 and b = 2 we have

favg = b 2

2

a 1

1 1f(x)dx (1 x )dx

b a 2 ( 1)

=1

63 = 2.

Another frequently seen averaging process is theroot mean square (r.m.s.) of f over [a, b], defined asfollows :

fr.m.s = ab

dx)]x(f[b

a

2

One can show that favg fr.m.s., the equality holdingonly for constant functions.

B1. Given that

12

0

1x dx

3. Use this fact and the

properties of integrals to evaluate 1

2

0(5 6x )dx

2. Write as a single integral in the form

b

af(x)

2 5 1

2 2 2f(x)dx f(x)dx f(x)dx

3. Find 2

1[f(x) 2g(x)]dx

if

2

1f(x)dx 5

and

2

1g(x)dx 3

4. If the function f is integrable in a closed intervalcontaining a, b, c and d, prove that

b c d d

a b c af(x)dx f(x)dx f(x)dx f(x)dx .

5. The graph of g consists of two straight lines anda semicircle. Use it to evaluate each integral.

(a) 2

0g(x)dx (b)

6

2g(x)dx

(c) 7

0g(x)dx

0 4 7

1

4

y = g(x)

62

Y

X

6. Replace the symbol * by either or so thatthe resulting expressions are correct. Give yourreasons.

(a)1 1

2 3

0 0x dx * x dx

(b)1 1

2 3

1 1x dx * x dx

(c)3 3

2 3

1 1x dx * x dx .

2.18 INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED10. Assume that f is integrable and nonnegative

on [a, b]. If b

af(x)dx = 0, prove that f(c) = 0 at

each point of continuity of f.11. If favg really is a typical value of the integrable

function f(x) on [a, b], then the number favgshould have the same integral over [a, b] that fdoes. Does it ? That is, does

b

avga

f dx = b

af(x)dx ?

12. It would be nice if average values of integrablefunctions obeyed the following rules on aninterval [a, b] :(a) (f + g)avg = favg + gavg(b) (kf)avg = k favg , (any number k)(c) favg gavg if f(x) g(x) on [a, b].Do these rules ever hold ? Give reasons foryour answers.

7. Find out which integral is greater :

(i)2 31 1x x

0 02 dx or 2 dx ?

(ii)2 32 2x x

1 12 dx or 2 dx?

(iii)2 2

2

1 1ln xdx or (ln x) dx?

(iv)4 4

2

3 3ln xdx or (ln x) dx?

8. (a) If 1

07f(x) dx = 7, does

1

0f(x) dx = 1 ?

(b) If 1

0f(x) dx = 4 and f(x) 0, does

1

0f(x)dx = 4 = 2 ?

9. If f is continuous on [a, b], f(x) 0 on [a, b],and f(x0) > 0 for some x0 in [a, b], prove that

b

af(x)dx 0 .

B

13. If 9

0f(x)dx = 37 and

9

0g(x)dx = 16, find

9

0[2f(x) 3g(x)]dx

14. Suppose 2

2f(x)dx

= 4, 5

2f(x)dx = 3,

5

2g(x)dx

= 2. Which, if any, of the followingstatements are true ?

(a)2

5f(x)dx = – 3

(b)

5

2(f(x) g(x))dx = 9

(c) f(x) g(x) on the interval –2 x 515. The graph of f is shown. Evaluate each integralby interpreting it in terms of areas.

(a) 2

0f(x)dx (b)

5

0f(x)dx

(c) 7

5f(x)dx (d)

9

0f(x)dx

0 2 4 6 8

2

Y

X

y = f(x)3

1

3 5

‒3‒3‒2‒

7 9

16. Evaluate the integral

0

2

3(1 9 x )dx by

interpreting it in terms of areas.

17. Let f(x) = 2x 1 if 3 x 0

1 x if 0 x 1

Evaluate

1

3f(x)dx by interpreting the integral as a

difference of areas.18. Determine whether the value of the integral is

positive or negative.

(i)

1

3

4

x3

xdx (ii)

34

2

x

| x | 1 dx

DEFINITE INTEGRATION 2.1924. Find the average value of

f(x) =

2x1,2x

1x4,4x on [–4, 2], using

graph of f (without integrating).25. Suppose that f and g are continuous on [a, b],

a b, and that b

a(f(x) – g(x)) dx = 0. Show that

f(x) = g(x) atleast once in [a, b].26. The inequality sec x 1 + (x2/2) holds on

(–/2, /2). Use it to find a lower bound for the

value of 1

0sec x dx .

27. Let f be a function that is differentiable on[a, b]. In the chapter of derivatives, we definedthe average rate of change of f over [a, b] to be

ab

)a(f)b(f

and the instantaneous rate ofchange of f at x to be f(x). Here we defined theaverage value of a function. For the newdefinition of average to be consistent with the

old one, we should have ab

)a(f)b(f

= averagevalue of f on [a, b] Is this the case ?

28. Is it true that the average value of an integrablefunction over an interval of length 2 is half thefunction's integral over the interval ?

19. Draw the graph of the functionf(x) = x(x – 2) (x – 4) = x3 – 6x2 + 8x, and indicate theregion P+ defined by the inequalities 0 x 3 and0 y f(x), and the region P– defined by 0 x 3and f(x) y 0. Let P = P+ P–, and suppose that

2

0f(x) dx 4 and

3

0

1f(x) dx 2

4 . Find area

(P+), area (P–), and area (P).20. Use the properties of integrals to verify the

inequality without evaluating the integrals.2 2

1 15 x dx x 1dx

21. Show that the value of 1

2

0sin(x ) dx cannot

possibly be 2.22. Given that, when x > 0, the function f(x) is positive

and is strictly decreasing, prove that

f(n + 1) n 1

nf(x)dx f(n)

23. Find the maximum and minimum values of

3x 2 for 0 x 3, and use these values tofind bounds on the value of the integrals

33

0x 2dx

2.4 FIRST FUNDAMENTALTHEOREM OF CALCULUS

The Fundamental Theorem of Calculus is appropriatelynamed because it establishes a connection betweenthe two branches of calculus : differential calculus andintegral calculus. Differential calculus arose from thetangent problem, whereas integral calculus arose froma seemingly unrelated problem, the area problem. TheFundamental Theorem of Calculus gives the preciseinverse relationship between the derivative and theintegral. It was Newtonand Leibnitz who exploited this relationship and used itto develop calculus into a systematic mathematicalmethod. In particular, they saw that the FundamentalTheorem enabled them to compute areas and integralsvery easily without having to compute them as limits ofsums as we did before.

Area functionThe First Fundamental Theorem deals with functionsdefined by an equation of the form

g(x) = x

a f(t) dt ...(1)

where f is a function defined on [a, b] and x variesbetween a and b. Observe that g depends only on x,which appears as the variable upper limit in the integral.

If x is a fixed number, then the integral x

a f(t) dt is adefinite number.

If we then let x vary, the number x

a f(t) dt also variesand defines a function of x denoted by g(x). Ifhappens to be a positive function then g(x) can beinterpreted as the area under the graph of f from a tox, where x can vary from a to b. Think of g as the “areaso far” function (See figure).

0 ta b

Y

x

y = f(t)

area = g(x)


x

adt)t(f = algebraic sum of areas

a x

f(x)

Let us discuss the question of notation. theindependent variable in the upper limit is usuallydenoted by the same letter (say x) as the variable ofintegration. thus, we write

x

ag(x) f(x)dx

however the letter x in the element of integration onlyserves to designate the auxiliary variable (the variableof integration) which runs over the values rangingfrom the lower limit a to the upper limit x as the integralis formed. If it is necessary to evaluate a particularvalue of the function g(x), for instance, for x = b, i.e.g(b), we substitute b for x in the upper limit of theintegral but do not replace by b the variable ofintegration. Therefore, it is more convenient to write

g(x) = x

af(t)dt

denoting the variable of integration by some otherletter (in this case by t).However, for simplicity, we shall often denote by thesame letter both the variable of integration and theindependent variable in the upper limit bearing in mindthat in the upper limit and under the integral sign theyhave different meanings.Now consider the graph of a bounded piecewisecontinuous function f with a point of discontinuity c.Let us take an arbitrary value x [a, b]. We shall beagain interested in the definite integral of f on [a, x].Let us denote it by g(x).

Y

Xx+hx ba0

A B

Hence, g(x) = x

af(t)dt

The number g(x) for the given x is represented in thefigure by the area of the figure ABxa. g(x) changes asx varies on [a,b].Theorem. If a function f is integrable on a closed

interval [a, b], then the function g(x) = x

af(t)dt is

continuous at any point x [a, b].Proof : Let us take an arbitrary point x and assign anincrement h to it (shown in figure is a positive h).We have

|g(x + h) – g(x)| =

x h x

a af(t)dt – f(t)dt

= x h

xf(t)dt

M | h |(M | f(t) |, t [a, b]).We have obtained the inequality| g(x + h) – g(x) | M | h |,it follows that

0hlim [g (x + h) – g(x)] = 0,

i.e. g is continuous at the point x.It should be underlined that x may turn out to be eithera point of continuity, or a point of discontinuity of f,but all the same, the function g(x) is continuous at thispoint.Example 1. If f is the function whose graph is

shown in the figure and g(x) = x

a f(t) dt, find thevalues of g(0), g(1), g(2), g(3), g(4) and g(5). Thensketch a rough graph of g.

f(t)

t1 2 3 4 5

2

1

0–1

–2

Solution First we notice that g(0) = 0

0f(t) dt = 0.

From Figure we see that g(1) is the area of a triangle :

g(1) = 1

0 f(t) dt = 21

(1 . 2) = 1

To find g(2) we add to g(1) the area of a rectangle :


g(2) = 2

0 f(t) dt = 1

0 f(t)dt + 2

1 f(t) dt= 1 + (1 . 2) = 3

g(3) = g(2) + 3

2 f(t) dt = 3 + 21

(1 . 2) = 4For t > 3, f(t) is negative and so we start subtractingareas :

g(4) = g(3) + 4

3 f(t) = 4 – 21

(2 × 1) = 3

g(5) = g(4) + 5

4 f(t) dt = 3 – 21

× 2 × 1 = 2We use these values to sketch the graph of g.

0 X1

Y

2 3 4 5

1

234

g

=x

0dx)t(f)x(g

Notice that, because f(t) is positive for t < 3, we keepadding area for t < 3 and so g is increasing up to x = 3,where it attains a maximum value.For x > 3, g decreases because f(t) is negative.If we take f(t) = t and a = 0, then, we have

g(x) = x

0 t dt = 2x

2Notice that g(x) = x, that is , g = f. In other words, if gis defined as the integral of f, then g turns out to bean antiderivative of f, atleast in this case.And if we sketch the derivative of the function gshown in the figure by estimating slopes of tangents,we get a graph like that of f.

First Fundamental Theorem of CalculusIf f is continuous on [a, b], then the function g defined

by g(x) = x

af(t) dt a x b ...(1)

is continuous on [a, b] and differentiable on (a, b),and g(x) = f(x).Proof : If x and x + h are in (a, b), then

g(x + h) – g(x) =

x h

xf(t) dt –

x

af(t) dt

= x h a

a xf(t)dt f(t)dt

=

x h

xf(t) dt

and so, for h 0,x h

x

g(x h) g(x) 1

h h

f(t) dt ...(2)

0 ta b

Y

x

x + h

f(x)

h

For now let us assume that h > 0. Since f is continuousin [x, x + h], the Extreme Value Theorem says thatthere are numbers u and v in [x, x + h] such thatf(u) = m and f(v) = M, where m and M are the absoluteminimum and maximum values of f on [x, x+ h]. ByMax-Min Inequality, we have

mh x h

x

f(t) dt Mh

that is, f(u)h x h

x

f(t) dt f(v) hSince h >0, we can divide this inequality by h:

f(u) x h

x

1

h

f(t) dt f(v)Now we use (2) to replace the middle part of thisinequality :

f(u) g(x h) g(x)

h f(v) ...(3)

Inequality 3 can be proved in similar manner for thecase h < 0.Now we let h 0. Then u x and v x, since u andv lie between x and x + h.Therefore

h 0lim f(u) = u xlim f(u) = f(x) and

h 0lim f(v) = v xlim f(u) = f(x)

because f is continuous at x. We conclude, from (3)and the Sandwich Theorem, that

g(x) =h 0lim

g(x h) g(x)

h

= f(x) ...(4)

2.22 INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCEDIf x = a or b, then equation 4 can be interpreted as aone-sided limit.Using Leibnitz notation for derivatives, we can writethe First Fundamental Theorem (FTC1) as

x

a

d

dx f(t) dt = f(x) ...(5)when f is continuous. Roughly speaking, equation 5says that if we first integrate f and then differentiatethe result, we get back to the original function f.This theorem is called the theorem on differentiatingthe definite integral with respect to its upper limit.For a continuous function the derivative of theintegral with respect to its upper limit is equal to thefunction itself.

The antiderivative of a continuous functionIt follows from the First Fundamental Theorem thatany continuous function has an antiderivative(primitive) which is definite integral with variable upperlimit of the given function.Theorem The function f(x) continuous on theclosed interval [a, b] has an antiderivative on thisinterval. One of the antiderivatives is a function

F(x) = xa

f(t)dt ...(1)For example,

F(x) = x 2t

0e dt is an antiderivative of f(x) = 2xe ,

since f(x) is continuous and FTC1 ensures that F(x) = f(x).

Note: An integral with a variable upper limitis defined for any function f(x) integrable on [a, b].However, for the function F(x) of form (1) to be anantiderivative for f(x), it is essential that the functionf(x) be continuous. Thus, out of the definite integral wehave constructed a function, which we call an indefiniteintegral of the integrand. Loosely speaking, an indefiniteintegral is the definite integral with a varying upper endpoint. The theorem is stated generally for indefiniteintegrals based at an arbitrary point x0 in [a, b], namely

x

x0f(t)dt . Of course, for a given x0 the two indefinite

integrals x

x0f(t)dt and

x

af(t)dt just differ by a

constant, namely ± x0

af(t)dt .

Example 2. Find the derivative of the function

g(x) = x

t dt 20 1 .Solution Since f(t) = 21 t is continuous, the

First Fundamental Theorem of Calculus givesg(x) = 21 x

Example 3. If F(t) =t

20

1

x 1 dx, find F(1),F(2), and F(x).Solution The integrand in this example is a

continuous function f defined by f(x) = 2

1

x 1.

By the First Fundamental Theorem,

F(x) = f(x) = 21

x 1.

In particular, F(1) = 21 1

21 1,

F(2) = 2

1 1

52 1.

Example 4. Find 4x

1

d

dx sec t dt.Solution Here we neet to use the Chain Rule in

conjunction with FTC1.Let u = x4. Then

4x

1

d

dx sec t dt = u

1

d

dx sec t dt

= u1d dusec t dtdu dx (by the Chain Rule)= sec u

du

dx (by FTC1)

= sec (x4) . 4x3.

Example 5. Find the derivative dy

dx of the

implicit function

x y

2

/ 2 03 2sin zdz cos tdt 0 .

Solution Differentiate the left side of the equationwith respect to x,

x

2

/ 2

d3 2sin zdz

dx+

y

0

d dycos tdt 0

dy dx


2dy

3 2sin x cosy 0dx

2dy 3 2sin x

dx cosy.

Example 6. Let

f(x) = x

2 2 4 2

0{(a 1)(t t 1) (a 1)(t t 1)}dt

Find the value of ‘a’ for which f (x) = 0 has two distinctreal roots.Solution Differentiating the given equation, we get

f(x) = (a – 1) (x2 + x + 1)2

– (a + 1) (x2 + x + 1) (x2 – x + 1).Now, f(x) = 0

(a – 1) (x2 + x + 1)– (a + 1) (x2 – x + 1) = 0

x2 – ax + 1 = 0.For distinct real roots D > 0 i.e. a2 – 4 > 0

a2 > 4 a (–, – 2) (2, )

Example 7. If 1

sin x t2 f(t) dt = 1 – sin x, where x

,2

,0

then find the value of f

3

1 .

Solution We have sin x

1 t2 f(t) dt = 1 – sin xDifferentiating both sides, we get

– sin2x f(sin x) cos x = – cos x

f(sin x) = cosec2x = xsin

12 ,

2,0x

f(z) = ,z12

z (0, 1).

f

3

1 = 3.

Example 8. Find

x

2

0

x 0

cos(t )dtlim

x

Solution The limit is in indeterminate form 0/0.The integral with a variable upper limit

x

2

0cos(t )dt has derivative

x2

0cos(t )dt = cos(x2).

Therefore, applying L'Hospital's rule, we obtain

x

22

0

x 0 x 0

cos(t) dt cos(x )lim lim 1

x 1Note that an antiderivative of cos(x2) is not an

elementary function, i.e. x

2

0cos(t) dt cannot be

expressed in terms of elementary functions. Thishowever, has not prevented us from calculating therequired limit.

Example 9. Evaluate

x

x 3 3

x sin tlim dt

x 3 t .

Solution

x

x 3 3

x sin tlim dt

x 3 t

= 3

x

3

x 3

sin tdt

tlimx 3

= 3

x 3

F(x) F(3)lim

x 3[ applying L'Hospital's Rule and FTC1]

= 3F(3) = 3 sin3

3 = sin 3.

Example 10. Findx

2

0

x

(ln(t 1 t ) ln(1 t)dtlim

x 1

Solution The limit is in (/) form. Therefore,applying L'Hospital's Rule and FTC1, we get

2xlim ln x 1 x ln(1 x)

2

x

x 1 xlim ln

1 x

2

x

11 1

xlim ln ln 21

1x

.

Example 11. Find the critical points of thefunction f(x) if

(i) f(x) = 1 + x + x

1 (ln2 z + 2 ln z) dz

(ii) f(x) = x – ln x +x

1

12 2cos4t

t dt


Solution

(i) f(x) = 1 + x +x

2 (ln2 z + 2 ln z) dz f(x) = 0 + 1 + ln2 x + 2 ln x

For critical points f(x) = 0 1 + ln2 x + 2 ln x = 0

(ln x + 1)2— = 0 ln x = –1

x = e–1 = 1

e.

(ii) f(x) = x – ln x +x

1

12 2cos4t

t dt

f(x) = 1 – 1 1

x x – 2 – 2 cos 4x – 0

= – 1 – 2 cos 4xFor critical points f(x) = 0

cos 4x = –1

2 = cos

23

4x = 2n ± 23

or x = n

2 6

x =

n

,6 2 6

, n N ( x > 0)

Example 12. If f(x) = x

0(t 1) (et – 1) (t–2) (t + 4)

dt, then find the points of local minima of f(x).

Solution Here f (x) = (x + 1) (ex –1) (x – 2)(x + 4)Sign scheme for f(x) :

– –+ ++–4 –1 0 2

Clearly x = –1 and x = 2 are the points oflocal minima.

Example 13. Find the points of maxima/minima

of 2 2x

t0

t 5t 4

2 e

dt.

Solution Let f(x) = 2 2x

t0

t 5t 4

2 e

dt

f(x) =

2

4 2

x

x 5x 4

2 e.2x – 0

= xe2x2).2x)(2x)(1x)(1x(

– + – + – +

–2 –1 0 1 2From the sign scheme of f(x), it is clear that f(x) haspoints of maxima at x = –1, 1(as sign changes from+ve to –ve) and points of minima at –2, 0, 2 (as signchanges from –ve to + ve) at x = 1.

Example 14. Let f(x) =

x

4 23

dt

t 3t 13If g (x) is

the inverse of f(x) then find g'(0).

Solutiondydx

= 4 2

1

x 3x 13 when y = f (x)

g'(y) = 1

dy dx = 13x3x 24 When y = 0 then x = 3.Hence, g'(0) = 132734 = 121 = 11.

Example 15. If f(x) is a continuous function such

thatx

0 f(t) dt as x , show that every line

y = mx intersects the curve y2 + x

0 f(t) dt = awhere a R+.Solution We have to show that there exists some x

such that m2x2 + x

0 f(t) dt = a (a R+) ...(1)Consider the function

g(x) = m2x2 + x

0 f(t) dtSince f is a continuous function, therefore g is also acontinuous function. Also, g(0) = 0 and g(x) asx . Thus, by intermediate value theorem, theremust be some x (0, ), such that g(x) = a (a R+).Hence, for every real m, there exists some a (a R+)that satisfies equation (1).Example 16. Consider the function

f(x) = cos x – x

0(x – t) f(t) dt.

Show that f(x) + f(x) = – cos x.


Solution We have f(x) = cos x – x

0(x – t) f(t) dt

= cos x – xx

0 f(t) dt + x

0 t f(t) dtDifferentiating w.r.t. x, we have

f(x) = – sin x – xf(x) – x

0 f(t) dt + xf(x)= – sin x –

x

0 f(t) dtDifferentiating again w.r.t. x, we have

f(x) = – cos x – f(x)i.e. f(x) + f(x) = – cos x. Hence proved.

Example 17. Show [assuming (f(t) to becontinuous for all values of t considered in the problem]that, when p and k are constants,

y = x

k

1f(t)sin p(x t)dt

p is a solution of the

differential equation 22

dx

yd + p2y = f(x).

Solution y = x

k

1f(t)

p (sin px cos pt – cos px sin pt) dt

= p1

x xk ksin px f(t)cospt dt cospx f(t)sin pt dt dx

dy = p

1 sin px . f(x) . cos px + cos px

x

kf(t)cospt dt

– p

1 cos px . f(x) sin px + sin px

x

kf(t)sin pt dt

= cos px . x

kf(t) cos pt dt + sin px

x

k)t(f sin pt dt

On differentiating again,

2

2

dx

yd = cos2 px . f(x) – p sin px

x

k)t(f cos pt dt

+ sin2 px f(x) + p cos px x

k)t(f sin pt dt

= f(x) – p x

k)t(f (sin px cos pt – sin pt cos px) dt

= f(x) – p x

k)t(f sin p(x – t) dt = f(x) – p2 y..

C1. Prove that

g(x)

a

d

dx f(t) dt = f[g(x)]g(x).2. Suppose

x

1f(t)dt = x2 – 2x + 1. Find f(x).

3. Let f be continuous on [a, b] and x

a f(t) dt = 0for all x [a, b], then prove that f(x) = 0 for all x [a, b].

4. Find the following derivatives :

(i) b

2

a

dsin(x )dx

dx(ii)

2x2

a

dsin(x )dx

da

(iii) 2x

2

0

d1 x dx

dx(iv)

3x

2 2x

d dt

dx 1 t

(v)

3x

2 2x

d dx

dx 1 x(vi)

3x

2 2 4t

d dt

dx x t

5. Find the derivative of y =

31

21 3X

udu

1 u.

6. If F(x) = x

1f(t)dt where f(t) =

2 4t

1

1 udu

u

find F" (2).

7. Find 22

dx

yd if y =

313

x

t sin 2t dt1 3t .

8. Let f ' be continuous in [a, b]. State under what

conditions

x

a

x

adt)t(f

dtddt)t(f

dxd .

9. Find an antiderivative F of f(x) = x2sin(x2) suchthat F(1) = 0.

10. Let f(t) be a function that is continuous and

satisfies f(t) 0 on the interval

0,2 . Suppose

it is known that for any number x between 0 and

2

, the region under the graph of f on [0, x] has

area A(x) = tan x.

(a) Explain whyx

0f (t) dt tan x for 0 x

2

(b) Differentiate both sides of the equation in

part (a) and deduce the formula of f.

02 definite integration (1-25) - vkr...

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