06.Integral Calculus

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Complete Study Guide & Notes On INTEGRAL CALCULUS

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IMPORTANT TERMS, DEFINITIONS & RESULTS

01. Meaning of integral of a function:

If differentiation of a function F x is f x i.e. if F d

x f xdx

, then we say that one integral or

primitive or anti-derivative of f x is F x and in symbols, we write

F f x dx x . Therefore, we can say that integration is the inverse process of differentiation.

02. List of formulae for Integral Calculus:

FORMULAE FOR INDEFINITE INTEGRALS

A. 1

, 11

n

n xx dx k nn

B. 1

logdx x kx

C. 1

log

x xa dx a ka

D. 1ax axe dx e ka

E. 1

sin cosax dx ax ka

F. 1

cos sinax dx ax ka

G. tan log secxdx x k OR log cos x k

H. cot log sinxdx x k OR log cosec x k

I. sec log sec tanxdx x x k

OR log tan4 2

xk

J. cosec log cosec cot xdx x x k

OR log tan2

xk

K. 2sec tanxdx x k

L. 2cosec cotxdx x k

M. sec . tan sec x xdx x c

N. cosec .cot cosec x xdx x k

O. 2

11 sec1

dx x kx x

P. 2 2

11 1 tanx

dx ka x a a

Q. 2 2

1 1log

2

a xdx k

a x a a x

R. 2 2

1 1log

2

x adx k

x a a x a

S. 2 2

2 2

1logdx x x a k

x a

T. 2 2

2 2

1logdx x x a k

x a

U. 2 2

11 sinx

dx kaa x

V. 1 1

log

dx ax b kax b a

, where a is any constant (and obviously, k is the integral constant)

W. dx x k , where is a constant (and, k is the integral constant)

A Formulae Guide By OP Gupta (Indira Award Winner)

Integral Calculus By OP Gupta (INDIRA AWARD Winner, Elect. & Comm. Engineering)


X. 2

2 2 2 2 2 2log2 2

x a

x a dx x a x x a k

Y. 2

2 2 2 2 2 2log2 2

x ax a dx x a x x a k

Z. 2

2 2 2 2 1sin2 2

x a xa x dx a x k

a

03. Introduction to the terms and symbols used in Integration:

Terms / Symbols Meaning

f x dx Integral of f with respect to x

f x in f x dx Integrand

x in f x dx Variable of integration

Integrate / Evaluate Find the integral

Integral value of f A function F such that

F x f x

Constant of Integration Any real number denoted by C or k

04. Methods of Integration:

Though there is no general method for finding the integral of a function, yet here we have considered the following methods based on observations for evaluating the integral of a function:

a) Integration by Substitution Method

In this method we change the integral f x dx , where independent variable is x, to another integral in which independent variable is t (say) different

from x such that x and t are related by x g t .

Let u f x dx then, du

f xdx

.

Again as x g t so, we have dx

g tdt

.

Now . . du du dx

f x g tdt dx dt

On integrating both sides w.r.t. t, we get

dudt f x g t dt

dt

u f g t g t dt

i.e., f x dx f g t g t dt where x g t .

So, it is clear that substituting x g t in

f x dx will give us the same result as obtained by putting g t in place of x and g t dt in place of dx .

b) Integration by Partial Fractions

Consider

f x

g x defines a rational polynomial

function.

If the degree of numerator i.e. f x is greater than or equal to the degree of

denominator i.e. g x then, this type of rational function is called an improper rational

function. And if degree of f x is smaller than

the degree of denominator i.e. g x then, this type of rational function is called a proper rational function.

In rational polynomial functions if the degree (i.e. highest power of the variable) of numerator (Nr) is greater than or equal to the degree of denominator (Dr), then (without any doubt) always perform the division i.e., divide the Nr by Dr before doing anything and thereafter use the following:

Numerator RemainderQuotient

Denominator Denominator .

On doing this, the rational function is resolved into partial fractions. The table on next page lists the types of simpler partial fractions that are to be associated with various kinds of rational functions which will be dealt in our current study:

A Complete Formulae Guide Compiled By OP Gupta (M.+91-9650350480 | +91-9718240480)


TABLE DEMONSTRATING PARTIAL FRACTIONS OF VARIOUS FORMS

Form of the Rational Function Form of the Partial Fraction

px q

x a x b, a b

A B

x a x b

2

px q

x a

2

A B

x a x a

2

px qx r

x a x b x c

A B C

x a x b x c

2

2

px qx r

x a x b

2

A B C

x a x bx a

2

2

px qx r

x a x bx c

where 2 x bx c cant be factorized further.

2

A Bx C

x a x bx c

c) Integral By Parts

If and V be two functions of x then,

I II.

dV dx Vdx Vdx dx

dx

In finding integrals by this method, proper choice of functions and V is crucial. Though there is no fixed rule for taking and V (their choice is possible by practice) yet, following rule is found to be quite helpful in deciding the functions and V :

If and V are of different types, consider that function as which comes first in the word ILATE .

Here I stands for Inverse trigonometrical function, L stands for Logarithmic function, A stands for Algebraic function, T stands for Trigonometrical function and E stands for the Exponential function.

If both the functions are trigonometrical, take that function as V whose integral is easier.

If both the functions are algebraic, take that function as whose differentiation is easier.

Some integrands are such that they are not product of two functions. Their integrals may be found by integrals by parts taking 1 as the second function. Logarithmic and Inverse Trigonometric functions are examples of such functions.

Follwoing result can be directly applied in case of the objective type questions :

(i) ( ) ( ) ( ) x xe f x f x dx e f x k (ii) log log | |x dx x x x k (iii) ( 1)

x xxe dx x e k

05. Making the Perfect Square:

STEP1 Consider the expression 2 ax bx c .

STEP2 Make the coefficient of 2x as unity i.e., 1 by taking a common, after doing so the original

expression will look like, 2

b ca x x

a a.

STEP3 Add and subtract

2

2

b

a to the expression obtained in STEP2 as depicted here i.e.,

2 2

2

2 2

b c b ba x x

a a a a.

STEP4 The perfect square of 2 ax bx c will be

2 2

2 2

b c ba x

a a a.



06. Various Integral forms:

Integrals of the form 2

px q

dxax bx c

, 2

px qdx

ax bx c, 2 px q ax bx c dx :

Express the numerator px q as shown here, i.e., 2 dpx q A ax bx c Bdx

. Then on, obtain the values

of A and B by equating the coefficients of like powers of x and constants terms on both the sides. Then,

integrate it after replacing px q by 2 dA ax bx c Bdx

using the values of A and B.

Integrals of the form sin cos

sin cos

a x b x

dxc x d x

:

Express Numerator Denominator Denominator d

A Bdx

. Then obtain the values of A and B by equating the

coefficients of sin x and cos x on both the sides and proceed.


sin cos

a x b x cdx

p x q x r

:

Note that the previous integral form can be considered as a special case of this form. Express

Numerator Denominator Denominator d

A B Cdx

. Then obtain the values of unknowns i.e., A, B and C

by equating the coefficients of sin x , cosx and the constant terms on both the sides and hence proceed.

Integrals of the form 2 2sin cos

dx

a x b x,

2sindx

a b x,

2cosdx

a b x,

2sin cosdx

a x b x and

2 2sin cos dx

a b x c x: Divide the Nr and Dr both by

2cos x . Replace 2sec x , if any, in Dr by 21 tan x and

then put tan tx and proceed.


dx

a x b x,

sindx

a b x,

cosdx

a b x and

sin cosdx

a x b x:

Use 2

2 tan2sin

1 tan2

x

xx

and/ or

2

2

1 tan2cos

1 tan2

x

xx

. Replace 21 tan2

x

in the Nr by 2sec2

x and then put

tan t2

x and then after proceed.

Integrals of the form 1

M Ndx where M and N are linear or quadratic expressions in x :

M N Substitutions

Linear Linear 2t N

Quadratic Linear 2t N

Linear Quadratic 1tM

Quadratic Quadratic 2 N 1t or tM

x

Before we go for the questions, here are a few useful Quickies:

a)

1

, 11

n

n f xf x f x dx k n

n b)

log

f x

dx f x kf x

c)

1

1

n

n

f xf xdx k

nf x d)

1

1

1

nn ax b

ax b dx ka n



e) [ ]f x g x dx f x dx g x dx f ) dx x k

g) , where is any real constantf x dx f x dx .

DEFINITE INTEGRALS

01. Meaning of Definite integral of a function:

If F f x dx x i.e. F x be an integral of f x , then F Fb a is called the definite integral

of f x between the limits a and b and in symbols it is written as b

a

f x dx or, F b

ax .

Moreover, the definite integral gives a unique and definite value (numeric value) of anti-derivative of the function between the given intervals. It acts as a substitute for evaluating the area analytically.

02. Formulae & properties of Definite Integrals:

P.01 F F F b

b

aa

f x dx x b a

P.02 b a

a b

f x dx f x dx

P.03 b b

a a

f x dx f t dt

P.04 , b m b

a a m

f x dx f x dx f x dx a m b

P.05 b b

a a

f x dx f a b x dx

P.06

0

2 , if is even function . .,

0, if is odd function . .,

a

aa f x dx f x i e f x f x

f x dx

f x i e f x f x

P.07 0 0

2 2

a a a

a

f x dx f x dx f x dx

0 0

2a a

f x dx f a x dx

P.08

0

/2

0

2 , if

0, if

mm f x dx f m x f x

f x dx

f m x f x

03. Proof of the properties of Definite Integrals:

P.01 F F F b

b

aa

f x dx x b a .

PROOF Let function f x be such that F d

x k f xdx

.

Then, F f x dx x k . That is F x k represents the anti-derivative of a continuous function f x defined on ,a b .



So, by the definition of definite integral, the definite integral value of f x over ,a b is denoted by:

F b

b

aa

f x dx x k .

Now, we proceed as follow to evaluate the value of b

a

f x dx :

F b

b

aa

f x dx x k

F F b k a k

F F b

a

f x dx b a . [H.P.]

As the constant of integration k, is cancelled out, so we intentionally leave it while evaluating the definite integral for a function.

P.02 b a

a b

f x dx f x dx .

PROOF We know that F F F b

b

aa

f x dx x b a (i)

Also, consider F F a

b

f x dx a b (ii)

By (i) and (ii), we can easily deduce that F F b

a

f x dx a b

a

b

f x dx . [H.P.]

P.03 b b

a a

f x dx f t dt i.e., value of definite integral is independent of the change of variable.

PROOF Consider F b

b

aa

f x dx x

F F b a (i)

Also, F F F b

b

aa

f t dt t b a (ii)

By (i) and (ii), we observe that b b

a a

f x dx f t dt . [H.P.]

P.04 , b m b

a a m

f x dx f x dx f x dx a m b .

PROOF We know, F F F b

b

aa

f x dx x b a (i)

Consider F F F m

m

aa

f x dx x m a (ii)

And F F F b

b

mm

f x dx x b m (iii)

Adding the equations (ii) and (iii), we get



F F m b

a m

f x dx f x dx b a

b

a

f x dx . [ By (i)

Hence, , b m b

a a m

f x dx f x dx f x dx a m b . [H.P.]

P.05 b b

a a

f x dx f a b x dx .

PROOF Consider b

a

f x dx .

Let x a b t dx dt . Also when x a t b and, when x b t a .

So, b a

a b

f x dx f a b t dt

b

a

f a b t dt [By using P.02

b

a

f a b t dt

b a

a b

f x dx f a b x dx [Replacing t by x, P.03

Hence, b b

a a

f x dx f a b x dx . [H.P.]

SPECIAL CASE OF P.05 Take 0a and b a .

Then, 0 0

a a

f x dx f a x dx .

The proof for the special case is same as is for the P.05, so it has been left as an exercise for you!

P.06

0

2 , . .,

0, . .,

a

a

a

f x dx if f x is even function i e f x f xf x dx

if f x is odd function i e f x f x

.

PROOF We know that 0

0

a a

a a

f x dx f x dx f x dx (i) [By using P.04

Consider 0

a

f x dx .

Let x t dx dt . Also when x a t a and when 0 0 x t .

So, 0 0

a a

f x dx f t dt

0

a

f t dt [By using P.02

0

a

f t dt



0

0

a

a

f x dx f x dt [Replacing t by x, P.03

Therefore equation (i) becomes,

0 0

a a a

a


0

a a

a

f x dx f x f x dx

0

2 ,

0,

a

a

a

f x dx if f x f xf x dx

if f x f x

i.e.,.

0

2 , is even function

0, is odd function

a

a

a

f x dx if f xf x dx

if f x

[H.P.]

P.07 0 0

2 2

a a a

a


0 0

2a a

f x dx f a x dx .

PROOF We know 0 0

2 2

a a a

a

f x dx f x dx f x dx (i) [By using P.04

Consider 2

a

a

f x dx .

Let 2 x a t dx dt . Also when x a t a and when 2 0 x a t .

So, 2 0

2 a

a a

f x dx f a t dt

0

2

a

f a t dt [By using P.02

0

2 a

f a t dt

2

0

2 a a

a

f x dx f a x dx [Replacing t by x, P.03

So equation (i) becomes,

0 0

2

0

2 a a a

f x dx f x dx f a x dx . [H.P]

P.08

00

2 2 , 2

0, 2

aa f x dx if f a x f x

f x dx

if f a x f x

.

PROOF We know 0 0

2 2

a a a

a

f x dx f x dx f x dx (i)

Consider 2

a

a

f x dx .



Let 2 x a t dx dt . Also when x a t a and when 2 0 x a t .

So, 2 0

2 a

a a

f x dx f a t dt

0

2

a

f a t dt [By using P.02

0

2 a

f a t dt

2

0

2 a a

a

f x dx f a x dx [Replacing t by x, P.03

So equation (i) becomes,

0 0

2

0

2 a a a

f x dx f x dx f a x dx

0

2 a

f x f a x dx

Hence,

00

2 2 , 2

0, 2

aa f x dx if f a x f x

f x dx

if f a x f x

. [H.P.]

04. Definite integral as the limit of a sum (First Principle of Integrals):

Take that function whose integral value is to be calculated as f x and then use the given relation,

0

lim 2 ... ( 1)

b

ha

f x dx h f a f a h f a h f a n h

or, 0

lim 2 3 ...

b

ha

f x dx h f a h f a h f a h f a nh

i.e., 1

0

limb n

nra

f x dx h f a rh

, such that as ,n 0h and nh b a

or, 1

limb n

nra

f x dx h f a rh

, such that as ,n 0h and nh b a .

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06.Integral Calculus