Calculus.pdf

Chapter 4

Mathematics

CHAPTER 4 Calculus Indeterminate form ,

,

,

,

,

,

, -

, 0.

Example: Plot y= [x] Here [x]

greatest integer not greater than x -2 x < -1 , y = -2 -1 x < 0 , y = - 1 0 x < 1 , y =0 1 x < 2 , y = 2

Various Plots

y

y

Y =ax 0
x

Y =ax 0
y y Y

x

=ex

Y =In x x

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Chapter 4

Mathematics

Limit of a function Let y = f(x) Then i = i.e, “ f x |< , | |< 0<|

as x a” i p ies for any (>0), (>0) such that whenever

Some standard expansions ......... ......... =1+x+

+

og

=x

og

=

......... +

x

Sin x = x Cos x = 1

......... ......... .........

+

.........

Sinh x = x

.........

Cosh x = 1

+

.........

Some important limits i i

=1 (

) =e

i

=e

i

= og

i

=1

i i

=1 =n THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com Page 92

Chapter 4

Mathematics

L – Hospita ’s Ru e 

When functions are of limit.

or

form differentiate Numerator & Denominator and then apply

Existence of limits and Continuity  

f(x) is defined at a, i.e, f(a) exists. If i f(x) = i f(x) = L , Then, the i



If i

f(x) exists and equal to L.

f x

i

f x = f(a) then the function f(x) is said to be continuous.

Properties of continuity   

If f and g are two continuous functions at a; then (f+g), (f.g), (f-g) are continuous at a is continuous at a, provided g(a) 0



|f| or |g| is continuous at a

Example i Solution Using the formula above, i = i = i

. i

=

.

=

Example i Solution i

= i = .

. i =1


Chapter 4

Mathematics

Example Evaluate i

*

+

Solution i

[

] =

i

*

=

i

*

=

.

+ +

=

Example Find the limit i

|=?

|

if possible

Solution x < 0,

i

= x

x>0, i

= x

= 0

Hence i

f(x) = 0

=0

Example f(x) =

when x

= 0

when x =0

Solution i

f(x) = i =

i

f(x) = i

= i

=

= 0 = i

= = 1 Here limits does not exists also discontinuous at x =0


Chapter 4

Ro e’s theore If (i) f(x) is continuous in closed interval [a,b]. ii f’ x exists for every va ue of x in open interva a,b . (ii) f(a) = f(b). then there exists at east one point c between a, b such that f’ c

Mathematics

.

Geometrically There exists at least one point c between (a, b) such that tangent at c is parallel to x axis.

C

a

C2

C1

b

Lagrange mean value theorem If (i) f(x) is continuous in the closed interval [a,b] and ii f’ x exists in the open interva a,b , then at east one va ue c of x exist in a,b such that f’ c . Geometrically, it means that at point c, tangent is parallel to the chord line.

Cauchy mean value theorem If, (i) f(x) is continuous in the closed interval [a,a+h] and ii f’ x exists in the open interva a,a h), then there is at least one number (0< <1) such that f(a+h) = f(a) + h f(a+ h) Let f1 and f2 be two functions: THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com Page 95

Chapter 4

Mathematics

f1,f2 both are continuous in [a,b] f1, f2 both are differentiable in (a,b) f ’

0 in (a,b)

then, for a = Example f(x) = cosx

, a=

, b=

; find c fro

Ro e’s theore

Solution f(x) = cos x f’ x f(

-sin x ) = cos (- ) = 0

f( ) = cos Hence f’ c

- sinx = 0

c =0 Example Find c using Lagrange’s Mean value theorem from f(x) = 3

+ 5x + 7 in interval [ , ]

Solution f(1) = 15 f(3) = 49 f’ x

6x 5

a = 1 , b =3 f (c) = 6c + 5 = 6c + 5 = c =

= 17

= 2


Chapter 4

Mathematics

Example f(x) = ln x

in interval [1 ,e]

f’ x f’ c = = c = e-1

Derivative f’ x

i

Provided the i it exists f’ x is ca ed the rate of change of f at x. Algebra of derivative:i

f g ’

f’

(ii) (f-g ’

g’

f’–g’

iii

f.g ’

f ’. g

iv

f/g ’

f.g’

.

.

Derivative of a function of function: chain rule =

.

Homogenous function Any function f(x, y) which can be expressed in from xn ( ) is called homogenous function of order n in x and y. (Every term is of nth degree. f(x,y) = a0xn + a1xn-1y + a2xn- y f(x,y) = xn

………… an yn

( )

Eu er’s theore

on ho ogenous function

If u be a homogenous function of order n in x and y then,  x 

+y

= nu

+ 2xy

+


Chapter 4

Mathematics

Total derivative If u f x,y ,x φ t y Ψ t =

.

+

=

.

x+

Example If z = log( ez= x

+xy+ +y

x.ez x

+xy +

+y

=2

ho ogenous,

=2.f

+ y.ez +y

), show that x

=2ez

=2

Monotonicity of a function f(x) 1.

f(x) is increasing function if for

, f

Necessary and sufficient condition, f’ (x) > 0 2.

f(x) is decreasing function

if for

, f

Necessary and sufficient condition, f x

Maxima-Minima Two Types

a) Global

b) Local

Rule for finding maxima & minima  

If maximum or minimum value of f(x) is to be found, let y = f(x) Find dy/dx and equate it to zero and fro this find the va ues of x, say x is α, β, … ca ed the critical points).



Find

at x

α,

If

, y has a minimum value

If

,y has a maximum value

If

, proceed further and find at x

α,


Chapter 4

If

, y has neither

But If

axi u

nor

ini u

, proceed further and find

at x

If

, y has minimum value

If

, y has maximum value

If

, proceed further

Mathematics

va ue at x

α

α,

Note Greatest / least value exists either at critical point or at the end point of interval. Point of Inflexion If at a point, the following conditions are met, then such point is called point of inflexion

(i)

,

(ii)

0,

(iii)

 Neither minima nor maxima exists

Taylor series f a

h

f a

h f’ a

f” a

Point of inflexion

.........

Maclaurian Series f x

f

x f’

f

h

f

Partial differentiation Taylor series f(x,y) = f(a, b) + [(x-a)fx(a, b) + (y-b) fy(a, b)] + fyy a,b ] ………..

[

fxx(a, b) + 2(x-a)(y-b) fxy(a, b) +

Error & approximation f=

x+

approxi ate y


Chapter 4

Mathematics

Maxima & minima (Two variables) r=

,s=

1)

= 0,

2)

(i) if rt(ii) if rt(iii) if rt(iv) if rt-

, t= so ve these equation. Let the so ution be a, b , c, d … and r axi u at a, b and r ini u at a, b < 0 at (a, b), f(a,b) is not an extreme value i.e, f(a, b) is saddle point. > 0 at (a, b), It is doubtful, need further investigation.

Example Find max and min value of

5

6

Solution 6 5

6 6

x = 2, 3

|

= - 6 < 0  maximum

|

= + 6 > 0  minimum

Maximum value

@x=2

= 16 – 60 +72 +11 = 39

Minimum value

@ x =3

= 54 – 135 +108 +11 =173 – 135 = 38

Example Show that maximum value of

is less than its minimum value.

Solution y= =

,

x=±1 =0+


Chapter 4

At x = 1,

, =2

At x = -1,

Mathematics

minimum

= -2

maximum

Maximum value @ x = -1 = -1 +

= -2

Minimum value @ x= +1 =

= 2

1+

Example Find the maxima and minima of 5

5

Solution 5

5

5

5

x =0, 1, 3 6 |

= - 10

Maximum value = 1 – 5 + 5 – 1 = 0 |

= 90 > 0

Minimum value = |

+ 5 . 34 + 5 . 33 – 1 = -28

=0 6 |

= 30

Hence neither maxima, nor minima at point x = 0

Example Find the maxima and minima of 6

, in interval [-1, 1]

Solution 6

6

, √

x= Hence

=

= 1 +i : Monotonous function


Chapter 4

Mathematics

Integration Reverse process of differentiation or the process of summation ∫ any continuous function.

defines the integral of

Standard Integral results 1. ∫ x dx

, n

2. ∫ dx

og x

3. ∫ e dx = e 4. ∫ a dx =

(prove it )

5. ∫ cos x dx

sin x

6. ∫ sin x dx

cos x

7. ∫ sec x dx

tan x

8. ∫ cosec x dx

cot x

9. ∫ sec x tan x dx

sec x

10. ∫ cosec x cot x dx 11. ∫ √ 12. ∫ 13. ∫

√ √

dx dx dx

cosec x

sin sec sec

x

14. ∫ cosh x dx

sinh x

15. ∫ sinh x dx

cosh x

16. ∫ sech x dx

tanh x

17. ∫ cosech dx

coth x

18. ∫ sech x tanh x dx

sech x

19. ∫ cosech x cot h x dx

cosech x

20. ∫ tan x dx

og sec x

21. ∫ cot x dx

og sin x

22. ∫ sec x dx

og sec x

23. ∫ cosec x dx

tan x = og tan ⁄

og cosec x

⁄

cot x = log tan

24. ∫ √

dx

og x

√x

a

= cosh

25. ∫ √

dx

og x

√x

a

= sinh


Chapter 4 √

26. ∫ √a

x dx

sin

27. ∫ √a

x dx

√x

a

og x

√x

a

28. ∫ √x

a dx

√x

a

og x

√x

a

29. ∫

dx = tan

30. ∫

dx =

31. ∫

dx =

og

where x
og

32. ∫ sin x dx

where x > a

sin x

33. ∫ cos x dx

sin x

34. ∫ tan x dx

tan x

35. ∫ cot x dx

x

cot x

36. ∫ n x dx

x nx

x x

37. ∫ e

sin bx dx

a sin bx

b cos bx

38. ∫ e

cos bx dx

a cos bx

b sin bx

39. ∫ e [f x

Mathematics

f x ]dx

e f x

Method of finding Integrals: (A) (B) (C) (D)

Integration by INSPECTION Integration by TRANSFORMATION Integration by SUBSTITUTION Integration by PARTS

Integration by parts:

∫ u v dx

u. ∫ v dx

∫

∫ v dx dx

I L A T E E

Selection of U & V Inverse circular (e.g. tan

x) Logarithmic

Exponential Algebraic Trigonometric

Note: Take that function as “u” which comes first in “ILATE” THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com Page 103

Chapter 4

Mathematics

Some Other Important Formulae Area = ∫ y dx = Vo u e

∫ r d

π ∫ y dx=

∫ r sin d

cos x

sin x

cos x

cos x

sin x

sin x

sin x

cos x

cos x

cos x

Example ∫ sec x tan x dx = ? Solution = x =

x

∫ sec x tan x dx

∫ sec

tan

=

=

sec

Example ∫ sin

x dx

∫ . sin

= ∫ sin x . dx = sin x, x ∫ √ = x sin

x

∫

= x sin

x

+ .

= x sin

x

x dx

. x dx /

√

. x dx

(z = 1 – x , dz = -2x dx )

/

/

x

/

Rules for definite integral 1)∫

=∫

2) ∫

=∫

+∫

a

Chapter 4

Mathematics

=∫

∫ 3) ∫

/

=∫

+∫

= ∫

∫

/

/

if f(a-x)=f(x)

=0

if f(a-x)=-f(x)

4) ∫

=2 ∫

if f(-x) = f(x), even function

=0

if f(x) = -f(x), odd function

Example (

∫

)

=?

Solution =∫

(

)

Also, = ∫

=∫

(

)

)

+∫

(

)

=∫

------ --------------(

)

------------------------(2)

Add (1) & (2) 2 =∫

(

(

)

=∫

(

.

)

=∫

=0

Example =∫

√ √

-----------------------(1)

√ √

=∫

=∫ √

√

√ √

√

----------------------(2)

Add (1) & (2) 2 =∫

√ √

√ √

=∫

.

= | =

=


Chapter 4

Mathematics

Improper integral Those integrals for which limit is infinite or integrand is infinite in a , then it is called as improper integral.

x

b in case of ∫

Types of improper integral i)

The interval increases without limit

(a) let f(x) be bounded and integral is a or to exist if i

B,

∫

(c) Let f(x) be bounded and integral in A B>a. Then ∫ i ∫

b for every A
exists finitely and

∫

= i

∫

is said to converge

∫

(b) let f(x) be bounded and integral in A x or to exist if i

B >a. then ∫

exists finitely and

∫

= i

∫

x

x

a for every A
is said to be convergent if i

and

∫

exists finitely and we write

∫ =∫

+∫

= i

∫

+ i

∫

ii)

(a) f(x) has infinite discontinuity only at the left end point, then ∫

i

∫

0<
(b) f(x) has infinite discontinuity at the right end point then ∫

∫

0<
(c) f(x) has infinite discontinuity at x=c, a
∫

+

∫

If either of a or b does not exist then integral does not exist. Note on convergence  ∫ is said to be convergent if the value of the integral is finite.  for all x (ii) ∫ converges , then ∫

also

converges 

for all x

(ii) ∫

diverges, then ∫

also diverges


Chapter 4



Assume both proper integrals ∫ and g(x) > 0 x a. If i

and ∫

Mathematics

exist for each b a where f(x)

=c where c 0, then both integrals ∫

and ∫

0

converge or

both diverge. 

∫



∫

is converges when p and ∫

and diverges when p is converges for any constant p



∫

is convergent iff p



∫

is convergent iff p

and diverges for p

Example =?

∫ Solution i

∫

= i

[tan

=i

[tan

=i

tan

] ] =

Example =∫

∫

+∫

Solution = i

∫

= i

[

= i

[

=

+

+ i ] + i ]+ i

∫ [

] * .

+

=

Thus integral diverges Example THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com Page 107

Chapter 4

Evaluate ∫

Mathematics

if it converges

√

Solution The integral becomes infinite at x=3 (second type improper integral) ∫

√

= sin

As t 0 =sin

( )|

( ) = sin

=sin

(

)

=

Vector calculus Scalar point function If corresponding to each point P of region R there is a corresponding scalar (P) is said to be a scalar point function for the region R. (P)= (x,y,z)

Vector point function If corresponding to each point P of region R, there corresponds a vector defined by F(P) then F is called a vector point function for region R. F(P) = F(x,y,z) = f1(x,y,z) +f2 x,y,z ĵ f3(x,y,z) ̂

Vector differential operator or Del operator:

=(

ĵ

̂

)

Directional derivative The directional derivative of f in a direction ⃗ is the resolved part of . ⃗ =|

in direction ⃗ .

|cosα

Where ⃗ is a unit vector in a particular direction. Direction cosine: Where, l=cos , m=cos

, , n=cos ,


Chapter 4

Mathematics

Gradient The vector function as grad f. Grad f =  

=î

is defined as the gradient of the scalar point function f(x,y,z) and written

+̂

ĵ

is vector function If f(x,y,z) = 0 is any surface, then is a vector normal to the surface f and has a magnitude equal to rate of change of f along this normal. Directional derivative of f(x,y,z) is maximum along | |.



Divergence The divergence of a continuously differentiable vector point function F is denoted by div. F and is defined by the equation. div. F = . F=f + ĵ div.F= .

=(

=  

Ψ̂

+

̂

ĵ

) .(f + ĵ

Ψ ̂)

+

. is scalar . is Laplacian operator

Curl The curl of a continuously differentiable vector point function F is denoted by curl F and is defined by the equation. ĵ Curl F =



̂

=|

|

is vector function

Solenoidal vector function If .A = 0 , then A is called as solenoidal vector function. Irrotational vector function If

=0, then A is said to be irrotational otherwise rotational. THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com Page 109

Chapter 4

Mathematics

DEL applied twice to point functions 1) div .grad f = 2) 3) 4) 5)

f=

+

.

=0 =0 F = =

curl grad f = div.curl F = . curl curl F = grad div F =

+

---------- this is Laplace equation

.

+

F F

Vector Identities f, g are scalar functions & F, G are Vector functions 1) 2) 3) 4) 5) 6) 7) 8) 9)

f g = f+ g . F G = .F .G F G = F G fg = f g + g f . fG = f. G f . G fG = f G f G F. G F G G F . F G = G.( F F. G F G = F( . G G .F

Also note 1) (f/g)= (g f – f g)/ 2) F.G ’ F ‘ .G F . G’ 3) (F G ‘ F‘ G + F G’ 4) (fg) = g f + 2 f. g + f

g

Vector product 1) Dot product of A

B with C is called scalar triplet product and denoted as [ABC]

Rule: for evaluating the scalar triplet product (i) Independent of position of dot and cross (ii) Dependent on the cyclic order of the vector [ABC] = A

A 2) (

. = A. B

=B

. = B.C

=C

. = C.A

. = -(B ⃗

. )

⃗ = (extreme

adjacent)

Outer

= (Outer. extreme) adjacent - (Outer. adjacent) extreme THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com Page 110

Chapter 4



(⃗⃗⃗



⃗



(⃗

⃗⃗⃗⃗

Mathematics

⃗ = (⃗ . ⃗ ) ⃗ - (⃗ . ⃗ ) ⃗ ⃗ ) = (⃗ . ⃗ ) ⃗ - (⃗ . ⃗ ) ⃗

(⃗ ⃗ )

⃗

⃗

⃗ )

(⃗

Example Find unit vector normal to the surface x

= 4 at point ( -1 , -1 ,2 )

Solution ⃗

Normal vector = = grad(x

= ⃗

= At point (-1,-1,2) ⃗

= -4 Vector = =

⃗ √

⃗)

(

√

Example Find directional derivatives of f(x , y , z) =x ⃗⃗⃗ ⃗⃗⃗⃗ vector

at the point ( 2 , -1 , 1 )in the direction of

Solution ⃗

f ⃗⃗⃗⃗ at ( 2 , -1 , 1 )

⃗⃗⃗

=

Directional derivation of = (

⃗⃗⃗

⃗⃗⃗⃗⃗⃗⃗

in the direction of

+⃗⃗⃗⃗

⃗⃗⃗⃗ ⃗⃗⃗⃗⃗⃗⃗ √

= = -


Chapter 4

Mathematics

Line integral, Surface Integral & Volume Integral 

Line integral = ∫ F R dR If F R dR

ĵ (x,y,z) + ̂ Ψ x,y,z

f x,y,z dx

ĵ dy

̂ dz

= ∫

∫

)



⃗⃗⃗⃗ or ∫ F ⃗ . ds ⃗ .N ⃗⃗ ds, Surface integral : ∫ F



Volume integral : ∫

If F(R ) = f(x,y,z)î +

x,y,z ĵ

= î∫ ∫ ∫

∫

Where N is unit outward normal to Surface.

Ψ x,y,z ̂ and

ĵ ∫∫∫

=

, then

+̂ ∫ ∫ ∫

Example If F=3xy î -y ĵ , evaluate ∫ F .dR Where c is the curve in the xy plane y = 2x from (0,0) to (1,2). Solution Since the partic e ∫ F. dr =∫

xydx

oves in the xy p ane z

, we ta e R x

yĵ .Then,

y dy

c is y=2x , x goes from 0 to 1 ∫ 6x

6x dx = - 7/6

Green’s theore If R be a closed region in the xy plane bounded by a simple closed curve c and if P and Q are continuous functions of x and y having continuous derivative in R, then according to Green’s theorem. ∮ P dx

dy = ∫ ∫ (

) dxdy

Sto e’s theorem If F be continuously differentiable vector function in R, then ∮ F. dr = ∫

F .N ds

Gauss divergence theorem The normal surface integral of a vector point function F which is continuously differentiable over the boundary of a closed region is equal to the ∫

.N.ds =∫ THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com Page 112

Chapter 4

Mathematics

Problem on surface integral Example ∬ .⃗

where

⃗

=

S is surface of a cube bounded by x = 0, x = 2 y = 0, y = 2 z = 0, z = 2 Solution By divergence theorem ∬ .⃗.

∬

.

= ∫ = ∫ = ∭

=∫ ∫ ∫ =∬

. |

= ∬ 8 = ∫

.

. .

∫ 8

|

6

= 12 x 2 = 24

Even Function f (-x) = f (x) ex. Cos x , x2 etc

x2 cos x

Odd function f (-x) = -f(x) Example : THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com Page 113

Chapter 4

Mathematics

Sin x, x3 etc x3 Sin x

Important Points If the function is even, then for any -a x ∫ t x d

a the function is even then

∫ f x dx

Example ∫

/ /

osxdx

∫

0

-π/2

/

π/2

For a given a

-

cosx dx

x

-a if the function is odd then ∫ f x dx

Example ∫

/ /

sinx dx y

-π/2 π/2

x

Important point for continuity and Differentiability -

If the function is discontinuous at any point it will be non differentiable at that point.

-

If the function is non differentiable at any point it does not mean that it is discontinuous at that point.

|x| = max (x. –x) x -x x x { x x x |x| is non differentiable at x = 0 but it is continuous at x = 0 |x|


Chapter 4

Mathematics

Maxima and Minima For finding maxima and minima of any function -

We first find If

,

at x

critical points than we check

at x

,

Example

1

2

Basically 2. If

at x

2

for minimum value , y for minimum value

1

For the above Diagram

, that is rate of change of slope is –ve so it is giving maximum


Calculus.pdf

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