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 x Y =ax 0 < a <1 y x y Y =In x x y Y =ex Y =ax 0 < a <1 y
Limit of a function Let y = f(x)
Then i = i.e, “ f x as x a” i p ies for any (>0), (>0) such that whenever
0<| |< , | |< 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 =1 i ( ) = e i =e i = og i =1 i =1 i = n
L – Hospita ’s Ru e
When functions are of or form differentiate Numerator & Denominator and then apply limit.
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
f(x) exists and equal to L.
If i 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
Example Evaluate i * + Solution i [ ] = i * + = i * + = . = Example Find the limit i | | = ? if possible Solution x < 0, i = x = 0 x > 0 , i = x = 0 Hence i f(x) = 0 Example f(x) = when x = 0 when x =0 Solution i f(x) = i = i = = = 0 i f(x) = i = i = = 1
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 .
Geometrically
There exists at least one point c between (a, b) such that tangent at c is parallel to x axis.
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:
C
C1
C2
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 -sin x f( ) = 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 = = 17 c = = 2
Example f(x) = ln x in interval [1 ,e] f’ x f’ c = = c = e-1
Derivative
f’ x iProvided the i it exists f’ x is ca ed the rate of change of f at x. Algebra of derivative:-
i f g ’ f ’ g ’ (ii) (f-g ’ f ’ – g ’ iii f.g ’ f ’. g f . g ’ iv 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 ………… an yn f(x,y) = xn ( )
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
Total derivative If u f x,y ,x φ t y Ψ t = . + . = x + Example
If z = log( +xy + ), show that x
+ y =2 ez= +xy+ ho ogenous, x +y =2.f x.ez + y.e z =2e z x + y =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
If
, y has neither axi u nor ini u va ue at x α
But If
, proceed further and find at x α,
If
, y has minimum value
If
, y has maximum value
If
, proceed further
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 . . . 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)] +
[ fxx(a, b) + 2(x-a)(y-b) fxy(a, b) +
fyy a,b ] ………..
Error & approximation f = x + approxi ate y Point of inflexion
Maxima & minima (Two variables) r =
, s = , t = 1) = 0,
so ve these equation. Let the so ution be a, b , c, d …
2) (i) if rt- and r axi u at a, b (ii) if rt- and r ini u at a, b
(iii) if rt- < 0 at (a, b), f(a,b) is not an extreme value i.e, f(a, b) is saddle point. (iv) if rt- > 0 at (a, b), It is doubtful, need further investigation.
Example
Find max and min value of 5 6 Solution 6 6 5 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 +
At x = 1, , =2 minimum At x = -1, = -2 maximum Maximum value @ x = -1 = -1 + = -2 Minimum value @ x= +1 = 1 + = 2 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 = √ = = 1 +i
Integration
Reverse process of differentiation or the process of summation ∫ defines the integral of any continuous function.
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 cosec x 11. ∫ √ dx sin 12. ∫ √ dx sec 13. ∫ √ dx 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 tan x = og tan ⁄ ⁄ 23. ∫ cosec x dx og cosec x cot x = log tan 24. ∫ √ dx og x √x a = cosh 25. ∫ √ dx og x √x a = sinh
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 = og where x <a 31. ∫ dx = og where x > a 32. ∫ sin x dx sin x 33. ∫ cos x dx sin x 34. ∫ tan x dx tan x x 35. ∫ cot x dx cot x x 36. ∫ n x dx x n 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 f x ]dx e f x
Method of finding Integrals: (A) Integration by INSPECTION
(B) Integration by TRANSFORMATION (C) Integration by SUBSTITUTION (D) Integration by PARTS Integration by parts: Selection of U & V I L A T E E Logarithmic Inverse circular (e.g.
tan
x)
Algebraic Trigonometric ExponentialNote: Take that function as “u” which comes first in “ILATE”
∫ u v dx u. ∫ v dx ∫
Some Other Important Formulae Area = ∫ y dx = ∫ r d
Vo u e π ∫ y dx= ∫ r sin d
sin x cos 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 x dx = ∫ sin x . dx = sin x, x ∫ √ . x dx = x sin x ∫ / √ . x dx (z = 1 – x , dz = -2x dx ) = x sin x + . / / = x sin x x /
Rules for definite integral
1)∫ =∫ +∫ a<c<b 2) ∫ =∫
∫ =∫ 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 =∫ √ √ √ √ =∫ . = | = =
Improper integral
Those integrals for which limit is infinite or integrand is infinite in a x b in case of ∫ , then it is called as improper integral.
Types of improper integral
i) The interval increases without limit
(a)
let f(x) be bounded and integral is a x B, B >a. then ∫ is said to converge or to exist if i ∫ exists finitely and∫ = i ∫
(b)
let f(x) be bounded and integral in A x b for every A<b.Then f(x) is said to converge or to exist if i ∫ exists finitely and∫ = i ∫
(c)
Let f(x) be bounded and integral in A x a for every A<a and in a x B for every B>a. Then ∫ is said to be convergent if i ∫ andi ∫ exists finitely and we write
∫ = ∫ + ∫ = i ∫ + i ∫
ii)
(a)
f(x) has infinite discontinuity only at the left end point, then ∫ i ∫ 0< <b-a(b)
f(x) has infinite discontinuity at the right end point then ∫ ∫
0< <b-a
(c)
f(x) has infinite discontinuity at x=c, a<c<b then ∫ ∫
+ ∫
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
Assume both proper integrals ∫ and ∫ exist for each b a where f(x) 0 and g(x) > 0 x a.
If i
=c where c 0, then both integrals ∫ and ∫ converge or
both diverge.
∫ is converges when p and diverges when p
∫ and ∫ is converges for any constant p and diverges for p ∫ is convergent iff p ∫ is convergent iff p Example ∫ =? Solution i ∫ = i [tan ] = i [tan ] = i tan = Example ∫ = ∫ + ∫ Solution = i ∫ + i ∫ = i [ ] + i [ ] = i [ ] + i * . + = + =
Thus integral diverges Example
Evaluate ∫
√ if it converges
Solution
The integral becomes infinite at x=3 (second type improper integral)
∫ √ = sin ( )| =sin ( ) As t 0 =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: ,
Gradient
The vector function is defined as the gradient of the scalar point function f(x,y,z) and written as grad f. Grad f = = î 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 Ψ ̂ div.F= . = ( 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.
Curl F = = | ĵ ̂ | is vector function
Solenoidal vector function
If .A = 0 , then A is called as solenoidal vector function. Irrotational vector function
DEL applied twice to point functions 1) div .grad f = f=
+ + --- this is Laplace equation
2) curl grad f = = 0 3) div.curl F = . =0
4) curl curl F = F = . - F 5) grad div F = . = + F
Vector Identities
f, g are scalar functions & F, G are Vector functions 1) f g = f + g 2) . F G = . F . G 3) F G = F G 4) fg = f g + g f 5) . fG = f. G f . G 6) fG = f G f G 7) F. G F G G F 8) . F G = G.( F F. G 9) 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. B
= B . = B.C = C . = C.A A . = -(B . )
2) ( ⃗⃗ ⃗ = (extreme adjacent) Outer
( ⃗⃗⃗ ⃗⃗⃗⃗ ⃗ = ( ⃗ . ⃗⃗ ) ⃗⃗ - ( ⃗ . ⃗⃗ ) ⃗⃗ ⃗⃗ ( ⃗⃗ ⃗ ) = ( ⃗⃗ . ⃗ ) ⃗⃗ - ( ⃗⃗ . ⃗⃗ ) ⃗ ( ⃗⃗ ⃗⃗ ) ⃗ ⃗⃗ ( ⃗⃗ ⃗ )
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 at the point ( 2 , -1 , 1 )in the direction of vector ⃗⃗⃗ ⃗⃗⃗⃗
Solution
f ⃗ = ⃗⃗⃗ ⃗⃗⃗⃗ at ( 2 , -1 , 1 )
Directional derivation of in the direction of + ⃗⃗⃗⃗ = ( ⃗⃗⃗ ⃗⃗⃗⃗⃗⃗⃗ ⃗⃗⃗⃗ ⃗⃗⃗⃗⃗⃗⃗
√
= = -
Line integral, Surface Integral & Volume Integral Line integral = ∫ F R dR
If F R f x,y,z ĵ (x,y,z) + ̂ Ψ x,y,z dR dx ĵ dy ̂ dz
∫ = ∫ )
Surface integral : ∫ F⃗ . ds⃗⃗⃗⃗ or ∫ F⃗ . N⃗⃗ ds, Where N is unit outward normal to Surface. Volume integral : ∫
If F(R ) = f(x,y,z)î + x,y,z ĵ Ψ x,y,z ̂ and = , then ∫ = î∫ ∫ ∫ ĵ ∫ ∫ ∫ + ̂ ∫ ∫ ∫
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 oves in the xy p ane z , we ta e R x yĵ .Then, ∫ F.dr =∫ xydx 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
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
Odd function f (-x) = -f(x) Example :
x2
Sin x, x3 etc
Important Points
If the function is even, then for any -a x a the function is even then ∫ t x d ∫ f x dx
Example
∫ / / osxdx ∫ / cosx dx
- For a given a x -a if the function is odd then ∫ f x dx Example
∫ / / sinx dx
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| is non differentiable at x = 0 but it is continuous at x = 0 -x x π/2 -π/2 x y π/2 -π/2 0 x3 Sin x
Maxima and Minima
For finding maxima and minima of any function
- We first find , critical points than we check
at x
If
at x ,
Example
Basically for minimum value 2. If
at x , y for minimum value
For the above Diagram
, that is rate of change of slope is –ve so it is giving maximum
1 2 1 2
