1. Proper Univariate and Multivariate Integrals
Rajandra Chadra Bhowmik
Lecturer
Dept. of Mathematics
Pabna Science and Technology University

2. Let’s start with some classical problems
What is the area of a circle?
How is it = pi*(radius)2
???
What is the volume of a sphere?
How is it = 4*pi*radius3/3
???
What is the mass of a wire?
if density is given ???

3. Lecture contents Pre-history and Archimedes
Invention of Newton and Leibnitz
Definite integrals and Riemann
Geometric Interpretation and Applications of Definite Integrals
Double Integrals and Triple Integrals
Applications of Double Integrals and Triple Integrals
Line Integrals
Applications of Line Integrals

4. Pre-history and Archimedes
Integration can be traced as far back as ancient Egypt, circa 1800 BC, with the Moscow Mathematical Papyrus demonstrating knowledge of a formula for the volume of a pyramidal frustum.
The first documented systematic technique capable of determining integrals is the method of exhaustion of Eudoxus (circa 370 BC), which sought to find areas and volumes by breaking them up into an infinite number of shapes for which the area or volume was known.
This method was further developed and employed by Archimedes (287 BC BC) and used to calculate areas for parabolas and an approximation to the area of a circle.
Archimedes devised a heuristic method based on statistics to do private calculations that would be classified today as integral calculus, but then presented rigorous geometric proofs for his results. To what extent Archimedes’ version of integral calculus was correct is debatable.

5. Invention of Newton and Leibnitz
The major advance in integration came in the 17th century with the independent discovery of the fundamental theorem of calculus by Newton ( ) and Leibnitz ( ).
The theorem demonstrates a connection between integration and differentiation. This connection, combined with the comparative ease of differentiation, can be exploited to calculate integrals.
In particular, the fundamental theorem of calculus allows one to solve a much broader class of problems. Equal in importance is the comprehensive mathematical framework that both Newton and Leibnitz developed. Given the name infinitesimal calculus, it allowed for precise analysis of functions within continuous domains. This framework eventually became modern calculus, whose notation for integrals is drawn directly from the work of Leibnitz.

6. Drawbacks of Newton and Leibnitz
While Newton and Leibnitz provided a systematic approach to integration, their work lacked a degree of rigor.
Bishop Berkeley memorably attacked infinitesimals as "the ghosts of departed quantities".
Calculus acquired a firmer footing with the development of limits and was given a suitable foundation by Cauchy in the first half of the 19th century.

7. Definition of Riemann integral
Let [a, b] be a compact (closed and bounded) interval in R.
Let p={a=x0,x1,x2,..,xn=b}, with xk-1<=xk for k=1,2,…,n, be a partition of [a, b].
Let f be a bounded real function defined on [a, b]
Integration was first rigorously formalized, using limits, by German Mathematician Riemann ( )

8. Graphical Representationsf
a=x0 x1 ξ2 x2 ξk xk ξn ξ1
Xk-1
Xn-1
Xn=b
∆xk
∆xn
∆x1
∆x2

9. Definition of Riemann integral
Write ∆xk=xk-xk-1 for k = 1,2,…,n.
Set δ = max {∆xk : k = 1,2,…,n}
For each k = 1,2,…,n, Choose point
Then the Riemann integral of f with respect to x is defined by
Provided the limits exist.

10. Graphical Representations
f(ξn)∆xn
f(ξn)
f(ξk)∆xk f
f(ξ2)∆x2
f(ξk)
f(ξ1)∆x1
f(ξ2)
f(ξ1)
x0=a x1 ξ2 x2 ξk xk ξn ξ1
Xk-1
Xn-1
Xn=b
∆xk
∆xn
∆x1
∆x2

11. Riemann Integral (Rigorous definition )
Let [a, b] be a compact (closed and bounded) interval in R.
Let p={a=x0,x1,x2,..,xn=b}, with xk-1<=xk for k=1,2,…,n, be a partition of [a, b].
Write ∆xk=xk-xk-1 for k = 1,2,…,n.
Let f be a bounded real function defined on [a, b]

12. Corresponding to each partition P, and k=1,2,…n, put
And then

13. Lower sum L(P, f) f xk x0=a x1 x2 x3 mn ∆xn mk ∆xk m2 ∆x2 m3 ∆x3
Xn=b
∆xk
∆xn
∆x1
∆x2
∆x3

14. Upper sum U(P, f) f xk x0=a x1 x2 x3 Mn ∆xn Mk ∆xk M2 ∆x2 M3 ∆x3
Xn=b
∆xk
∆xn
∆x1
∆x2
∆x3

15. Let be the set of all partitions of [a, b]
Let be the set of all partitions of [a, b]. Then, finally taking infimum and supremum over all partitions of [a, b] to define the upper Riemann integral and lower Riemann integral respectively as

16. If the lower Riemann integral and the upper Riemann integral are equal, then we say that the function is Riemann integrable on [a, b] and write
In this case we write the integral as

17. variable of integration
upper limit of integration
Integration
Symbol
integrand
variable of integration
(dummy variable)
lower limit of integration
It is called a dummy variable because the answer does not depend on the variable chosen.

18. The exact value from integrationxk
x0=a x1 x2 x3
Xk-1
Xn-1
Xn=b
∆xk
∆xn
∆x1
∆x2
∆x3

19. Some results For every P,
For any refinement P*(a super set of P) of P,

20. <= M (b-a) f xk x0=a x1 x2 x3 M (b-a) Xk-1 Xn-1 Xn=b ∆xk ∆xn ∆x1

21. <= U (P1, f) <= M (b-a)
M1 ∆x1
M2 ∆x2 f
x0=a X1
X2=b
∆x1
∆x2

22. <=U (P2, f)<= U (P1, f) <= M (b-a)
M3 ∆x3
M1 ∆x1
M2 ∆x2 f
x0=a X1 X2
X2=b
∆x1
∆x2
∆x3

23. m (b-a) <= f m (b-a) xk x0=a x1 x2 x3 Xk-1 Xn-1 Xn=b ∆xk ∆xn ∆x1

24. m (b-a)<= L (P1, f)<=
m2 ∆x2
m1 ∆x1
x0=a X1
X2=b
∆x1
∆x2

25. m (b-a) <= L (P1, f) <= L (P2, f) <=
m3 ∆x3 f
m2 ∆x2
m1 ∆x1
x0=a X1 X2
X2=b
∆x1
∆x2
∆x3

26. Necessary and sufficient conditions
f is continuous on [a, b] implies f is Riemann integrable on [a, b].
f is monotonic on [a, b] implies f is Riemann integrable on [a, b].
f is bounded on [a, b] and f has only finitely many points of discontinuity on [a, b] implies f is Riemann integrable on [a, b].
A bounded function f on a compact interval is integrable iff its set of discontinuities has measure zero(0).
Proof of the Theorem is out of scope here (a matter of Lebesgue Integration in Measure Theory). For instant, we only say that all countable sets, including finite sets, and the Cantor set have measure 0, so that functions that are discontinuous only on these sets are integrable.

27. Dirichlet function and Thomae function
Both functions are bounded. First one is discontinuous everywhere in R. i.e. the set of discontinuities has a nonzero measure. On the other hand, second one is discontinuous only at the rationals. i.e. set of discontinuities has measure 0.
So, Dirichlet function is not integrable but Thomae function is.

28. Properties of Riemann integral

29. Fundamental Theorem of Calculus (Part 1)

30. Proof:
Let f be a real Riemann integrable function on [a, b]. Then for any arbitrary ε >0, we must have a partition P={a=x0, x1,x2,…,xn=b} with U(P, f)-L(P, f) < ε
Now for each k=1,2,….,n, applying Mean Value Theorem in the interval [xk-1, xk], we must have tk in [xk-1, xk] such that
F(xk) - F(xk-1) = ∆xk F/(tk) = f(tk) ∆xk

32. Fundamental Theorem of Calculus (Part 2)

33. Proof:

34. Continue:

35. Continue:

36. Drawbacks of Riemann Integral
Although all bounded piecewise continuous functions are Riemann integrable on a bounded interval, subsequently more general functions were considered, to which Riemann's definition does not apply.
Stieltje formulated an integral called Stietjes integral, to which Riemann integral is a special case.
Lebesgue formulated a different definition of integral, founded in measure theory (a subfield of real analysis).
Other definitions of integral, extending Riemann's and Lebesgue's approaches, were proposed.

37. Drawbacks of Riemann Integral
A major limitation towards more widespread implementation of Bayesian approaches is that obtaining the posterior distribution often requires the integration of high-dimensional functions. This can be computationally very difficult, among the several approaches short of direct integration Markov Chain Monte Carlo (MCMC) methods is one which attempt to simulate direct draws from some complex distribution of interest.
MCMC approaches are so-named because one uses the previous sample values to randomly generate the next sample value, generating a Markov chain (as the transition probabilities between sample values are only a function of the most recent sample value).

38. Drawbacks of Riemann Integral
Markov chain Monte Carlo integration, or MCMC, is a term used to cover a broad range of methods for numerically computing probabilities, or for optimization. They are simulation methods, mostly used in complex stochastic systems where exact computation and even simple simulation are not computationally feasible.
Methods that fall under this heading include Metropolis sampling, Hastings sampling and Gibbs sampling which are for integration and simulated annealing and sometimes genetic algorithms which are optimization techniques.
Although these methods are mainly used for complex systems it can be used to find the exact p-value for a test of association between the rows and columns of a contingency table.

39. Geometrical meaning of Definite Integrals
Let f be a real non-negative continuous function defined in the closed interval [a, b]. Then

40. Area and Definite integral
Curve y= f (x)
Vertical line x=b R
Vertical line x=a
x0=a
Xn=b
X-axis

41. Geometrical meaning of Definite Integrals
Let f be a real continuous function (non negativity is not assured) defined in the closed interval [a, b]. Then

42. Area and Definite integral
x0=a
Xn=b
R 2 c

43. Applications of Definite Integrals (1) (Area of a region)
Let f and g are continuous functions on the interval [a, b] with
f (x) >= g (x) in [a, b]. Then the area A of the region
bounded above by the curve y = f (x)
below by the curve y = g (x)
between the vertical line x =a and x = b is given by

44. Area of the region R bounded by y = sqrt (x) and y = x.^2 (filled red)

46. Application of Definite Integrals (2) (Volume of solid revolution)
Theorem: Let S be a solid bounded by two parallel planes perpendicular to the x-axis at x = a and x = b [[ y-axis at y = c and y = d ]], If, for each x in [a, b] [[y in [c, d] ]], the cross sectional area of S perpendicular to the x-axis is A (x) [[y-axis is B(y) ]], then the volume of the solid is

47. Solid Revolution
Let f be continuous and non negative in [a, b] and let R be the region bounded by y = f (x) above x-axis between x = a and x = b. Then the solid generated by revolving the region R about the x-axis is such as require in the above theorem [each cross sectional area is a circular disk with radius f (x)]. Thus volume of the solid revolution is

48. Volume of a Cone (r, h)

49. Volume of a Cone (r, h)

50. On the other hand, we know volume of a cone of radius r (=3/2) and height h (=3) is

51. In fact if someone take y = xr/h as a function in between x = 0 and x = h, for some positive real numbers r and h, then the solid generated by revolving the region, bounded by the line y = xr/h above x-axis between x = 0 to x = h, would be a cone of radius r and height h. And thus its volume is

52. Application of Definite Integrals (3 & 4) (Arc length and Surface area)
As before Definite integral also applied to find arc length of a curve in 2-space and surface area of a solid revolution in 3-space by the following respective formulas :

53. Double Integrals Let R be a closed region in the xy- plane.
Subdivide R into n-sub-regoins ∆Rk of area ∆Ak, for k = 1,2,..,n.
Let f (x, y) be a bounded real function of double variables defined on R .

54. Double Integrals And then For each k = 1,2,…,n, Define
Also for each k =1,2,…,n, choose

55. Geometrical interpretation of Double Integrals
diam (∆Rk) R
Elementary sub-region ∆R k of area ∆Ak

56. Geometrical interpretation of Double Integrals

57. Double Integrals
Then the Double integral of f over the region R is defined by
Provided the limits exist.

58. Triple Integrals
Let V be a closed bounded three dimensional region in the space.
Subdivide V into n-subregoins ∆Vk of volume ∆Vk, for k = 1,2,..,n.
Let f (x, y, z) be a bounded real function of triple variables defined on V .

59. Triple Integrals And then For each k = 1,2,…,n, Define
Also for each k =1,2,…,n, choose

60. Triple Integrals
Then the Triple integral of f over the volume V is defined by
Provided the limits exist.

61. Sufficient condition for Double and Triple integrals
In both the cases of Double and Triple, we have mentioned that the multiple integral exists if the corresponding limit exists.
And the limits exist if the functions are continuous on the corresponding regions.
And all the properties of single integral hold here likely.

62. Technique to solve Double integrals (Fubini’s Theorem)
Let f be a real continuous function defined in the closed and bounded region
R = {(x, y): a<=x<=b, f1(x)<=y<=f2(x)} . Then

63. Technique to solve Double integrals (Fubini’s Theorem)
Let f be a real continuous function defined in the closed and bounded region
R = {(x, y): f1(y)<=x<=f2(y), c<=y<=d} . Then

64. Geometrical meaning of Double Integrals
Let f be a real non-negative continuous function defined in the closed and bounded region R . Then

65. Geometrical meaning of Double Integrals
Let f be a real continuous function (non-negativity is not assured) defined in the closed and bounded region R . Then

66. Application of Double Integrals (1) (Area of a region)
Let R be a closed and bounded region in the xy-plane. Then

69. Application of Double integrals (2) (Volume of a solid )
z-axis h
z = h
x2+y2 = r2 R r
y-axis
xy-plane
x-axis

70. It coincide with our well known formula of volume of a cylinder with radius r and height h is πr2h

71. Application of Double Integrals (3) (Mass of a lamina)
If a lamina with continuous density function σ(x, y) occupies a region R in the xy-plane, then the mass M of the lamina is

72. Application of Double Integrals (4) (Center of gravity or Centroid of a lamina)
If a lamina with continuous density function σ(x,y) occupies a region R in the horizontal xy-plane then the Center of gravity (u, v) of the lamina is
For uniform density, Center of gravity is Centroid. And this time σ(x, y)=1 is to be considered.

73. Solving Techniques for Triple Integrals
Let G be a solid bounded above by the surface z = g1(x,y) and below by the surface z=g2(x,y), and R be the projection of G on the xy-plane. Also, say g1 and g2 are continuous on R , and f (x, y, z) is a function continuous on G, then

74. Transformation of co-ordinates and Jacobian
Let T be a transformation from the uv-plane to the xy-plane defined by the equations x=x(u,v) and y=y(u,v), then the Jacobian J of T is

75. Transformation of co-ordinates and Jacobian
If the transformation x=x(u,v) and y=y(u,v), maps the region S in the uv-plane to the region R in the xy-plane and if the Jacobian J(u,v) is non-zero and does not change sign on S, then with appropriate restrictions on the transformation and the region it follows that
Analogous for triple integrals

76. Applications of Triple Integrals (1) (Volume of a solid)
Let G be a solid bounded above by the surface z = g1(x,y) and below by the surface z = g2(x,y), and R be the projection of G on the xy-plane. Also, say g1 and g2 are continuous on R , then volume of the solid is

77. Volume of a sphere with radius r

78. Applications of Triple Integrals (2) (Mass of a solid)
Let a solid with continuous density function σ (x, y, z) occupies a region G in the space, then mass M of the solid is

79. Application of Triple Integrals (3) (Center of gravity and centroid of a solid)
If a solid with continuous density function σ(x, y, z) occupies a region G in the space, then the center of gravity (u, v, w) of the solid is
For uniform density, center of gravity is centroid. And this time σ(x, y, z)=1 is to be considered.

80. Line Integrals (for Scalar functions)
Let C be a smooth (having smooth parametrization x = x (t), y = y (t), z = z (t) for a<=t<=b) curve in the 3-space joining the points P and Q.
Further, Let f (x, y, z) be a real function defined on C.

81. Line Integrals (for Scalar functions)
Subdivide C into n very small sections using a succession of distinct points P=P0,P1,P2,..,Pn=Q.
Write ∆sk= length(Pk-1,Pk) for k = 1,2,…,n.
Set ||∆|| = max {∆sk : k = 1,2,…,n}

82. Line Integrals (for Scalar functions)
For each k = 1,2,…,n, Choose point
Then the Line integral of f with respect to s along C is defined as
Provided the limits exist and does not depend on the choice of partition and points.

83. Geometric interpretation
Pn=Q
Pn*
Pn-1 C
P4*
Pk* P3 Pk P4
Pk-1
P5*
∆sk P5
P1*(ξk, ζk, ηk)
P3* P2
P2* P1
P=P0

84. Evaluating technique of Line integrals

85. Applications of Line Integrals (1) (Mass of wire)
If C is a smooth curve in 3-space that models a thin wire and if f (x, y, z) is the linear density function of the wire, then the mass M of the wire is given by the line integral

86. Mass of a helix like wire of density σ = 15-2xy

87. Applications of Line Integrals (2) (Length of wire)
If C is a smooth curve in 2-space. Then the length L of the curve is given by the line integral

88. Applications of Line Integrals (3) (Area of sheet)60 40 20 y 4 2 -2 -4 x
If C is a smooth curve in 2-space and if f (x, y) is a nonnegative function defined on C, then the area A of the of the sheet (swept out by a vertical line segment that extends upward from the point (x, y) to height f (x, y) and moves along C from one end-point P to the other Q) is given by the line integral. 60 40 20 y 4 2 -2 -4 x

89. Applications of Line Integrals (4) (Moments of Inertia)
Moments of inertia about the x-axis
Moments of inertia about the y-axis
Moments of inertia about the z-axis

90. Line Integrals (for Vector fields)
If C is a smooth oriented curve parameterized by r=r(t)=x(t)i+y(t)j+z(t)k, and F=F1(x(t),y(t),z(t))i+F2(x(t),y(t),z(t))j+ F3(x(t),y(t),z(t))k is a continuous vector field over C, then the line integral of F over C is

91. F is conservative if and only if
Theorem
If F is conservative i.e. F = f for some f , then every line integral of F will be independent of path and where (x(a),y(a),z(a)) is the beginning of the curve and (x(b),y(b),z(b)) is the end of the path.
If every line integral of the form is independent of path, then F is conservative.
Corollary
F is conservative if and only if
for any closed path lying inside the domain of F.

92. Applications of Line Integrals (4) (Work done)
Suppose that under the influence of a continuous force field F a particle moves along a smooth curve C and that C is oriented in the direction of motion of the particle. Then the work performed by the force field on the particle is
If the integral is positive, then the particle will gain energy (and usually moves faster). Otherwise the particle will slow down.
In the special case that the force is always perpendicular to the path, the total work done will be 0.

93. Find the work done moving once around an ellipse C in the xy-plane having center at the origin and semi major and semi minor axes 4 and 3 respectively with in a force field
F=(3x-4y+2z)i+(4x-2y-3z2)j+(2xz-4y2+z3)k
Soln: In the xy-plane z=0. Thus rewrite
F=(3x-4y)i+(4x-2y)j+(2xz-4y2)k and dr=dx i + dy j
And parameterized C as:
x=4 cos(t), y=3 sin(t), 0<=t<=2*pi.

94. Thus the Work done is given by the line integral

95. Ampere's Law
The line integral of a magnetic field around a closed path C is equal to the total current flowing through the area bounded by the contour C. This is expressed by the formula
Where is the vacuum permeability constant, equal to H/m.

96. Faraday's Law
The electromotive force ε induced around a closed loop C is equal to the rate of the change of magnetic flux ψ passing through the loop.

97. Green’s Theorem
Let R be a simply connected plane region bounded by a simple closed piecewise smooth counterclock-wise oriented curve C. If F=F1(x,y)+F2(x,y) be a function with component functions are continuous and having continuous partial derivatives on some region containing R , then

98. References
D V Widder (1961), Advanced Calculus, PHL Learning Private Ltd.
H Anton, I Bivens, S Davis (2008), Calculus, 8th edition, John Willey and Sons.
MR Spigel, Theory and Problems of Advanced Calculus, SI (Metric) Edition, Schaum’s Outline Series..
W Rudin (1976), Principles of Mathematical Analysis, 3rd Edition, Mcgraw-Hill International

99. THANKS EVERYBODY THANKS EVERYBODY
QUESTIONS???
QUESTIONS???