Newton, Leibniz, and the Invention of Calculus:
Sir Isac Newton (1642-1727) and Gottfried Wilhelm Leibniz (1646-1716) invented calculus.
Basic ideas of integration were investigated 2500 years ago by ancient Greeks-Eudoxus and
Archimedes
And methods for finding tangents were pioneered by Pierre de Fermat, Isaac Barrow (1630-
1677), and others.
Barrow, Newton’s teacher at Cambridge, was the first to understand the inverse relationship
between differentiation and integration.
What Newton and Leibniz did was to use this relationship, in the form of Fundamental
theorem of calculus.
In this sense, Newton and Leibnitz are credited with the invention of calculus.
Two view points of integration:
There are two distinct view-points from which the process of integration can be considered.
We may consider integration as the inverse of differentiation and make this as our starting
point.
We also may consider integration as a certain summation (area). Continuing with this
summation the result is identical with the reversal of a differentiation.
Antiderivative or Integral:
A function F ( x) is called an antiderivative or integral of a function f ( x) on a given open
d
interval if F ( x) f ( x) i.e. [ F ( x)] f ( x) for all x in the interval.
dx
Example-1: The function F ( x) sin x is the antiderivative of f ( x) cos x on the interval
(, ) because for each x in this interval
d
F ( x) [sin x] cos x f ( x)
dx
Here, sin x is the antiderivative of cos x .
d
Symbolically, if
dx
[ F ( x)] f ( x) f ( x)dx F ( x) C .
d
[ F ( x)] f ( x) is recast using integral notation f ( x)dx F ( x) C . It is important to note
dx
that they are just different notations to express the same fact.
Two mathematical expressions are same; actually they are in disguise of two different aspects.
It is like the same person with different dress.
Theorem: If F ( x) is any antiderivative of f ( x) on an open interval, then for any constant C ,
the function F ( x) C is also an antiderivative on that interval.
Note that if we differentiate an antiderivative of f ( x) , we obtain f ( x) back again. Thus
d d [ F ( x)] f ( x)
dx f ( x ) dx dx
Indefinite Integral:
The expression f ( x) dx is called an indefinite integral. The adjective “indefinite”
emphasizes that the result of antidifferentiation is a “generic” function, described only up to a
constant term.
x2
Example-2: The function F ( x) is the antiderivative of f ( x) x on the interval (, )
2
because for each x in this interval
d x2
F ( x) x f ( x)
dx 2
x2
Here, is the antiderivative of x . Therefore,
2
d x2 x2
dx 2
x x dx 2
d
Symbolically, if
dx
[ F ( x)] f ( x) f ( x)dx F ( x) .
x2
However, F ( x) is not the only antiderivative of f ( x) x on this interval. If we add any
2
x2 x2
arbitrary constant C to F ( x) , then the function G( x) C is also an antiverivative
2 2
of f ( x) x on this interval. Since
d x2
G( x) C [ x 0] f ( x)
dx 2
which implies that
x2
x dx 2
C
This is the reason why an arbitrary constant is added after the integration.
x2 x2 x2 x2
1, 2, 2, 9 all are antiderivatives of x .
2 2 2 2
Graphs of antiderivatives of a function f are called integral curves.
d x3 x3
C
2 2
dx 3
x x dx
3
d x3
0 x 2
dx 3
d x 3
1 x 2
dx 3
d x3
2 x 2
dx 3
d x 3
1 x 2
dx 3
d x3
2 x 2
dx 3
Integral, Integrand, Integration, Integrable, Integral curves:
I F ( x) f ( x ) dx , I is called the integral, f ( x) is called integrand, the process of finding
the antiderivative F ( x) is called the integration. If the value of the integral exists, then the
integral is integrable. Graphs of antiderivatives of a function f are called integral curves.
elongated S is used to denote integration. Leibnitz devised this symbol.
For simplicity, the dx is sometimes absorbed into the integrand. For example,
1dx can be written as dx
1 dx
x 2
dx can be written as x 2
The Area Problem:
Find the area of the region S that lies under the curve y f ( x) from a to b . This means that
S is bounded by the graph of a continuous function f (where f ( x) 0 ), the vertical lines
x a and x b , and the x axis.
In trying to solve the area problem, question may arise-what is the meaning of the word
“area”. The answer of this question is easy to answer for those regions whose boundaries are
straight line.
For example, the area of a rectangle is defined as the product of the length and the width.
The area of a triangle is half the base times the height.
The area of a polygon is found by dividing it into triangles and adding the areas of the
triangles.
Boundaries of all the shapes are straight line.
Formulas for the areas of polygons, such as squares, rectangles, triangles, and trapezoids, are
well known in many early civilizations. However, the problem of finding explicit formulas for
regions with curved boundaries (a circle being the simplest case) caused difficulties for early
mathematicians. It is not easy to find the area of a region with curved sides.
The first real progress in dealing with the general area problem was made by the Greek
mathematician Archimedes, who obtained areas of regions bounded by circular arcs,
parabolas, spirals, and various other curves using an ingenious procedure that was called later
“the method of exhaustion”.
Two ways to finding the area of region: The antiderivative method and the rectangle method
Rectangle method:
Divide the interval [a, b] into n equal subintervals, and over each subinterval construct a
rectangle that extends from the x-axis to any point on the curve y f ( x) .
For each n , the total area of the rectangles can be viewed as an approximation to the exact
area under the curve over the interval [a, b] . Moreover, it is evident intuitively that as n
increases, these approximations will get better and better and will approach the exact area as a
limit.
n
If A denotes the exact area then A lim An lim f ( xi* )x
n n
i 1
The sum
n
f ( x )x
i 1
*
i
that occurs in the definition of integral is called Riemann sum after the German
mathematician Bernhard Riemann (1826-1866).
The antiderivative method for finding areas:
Though, the rectangle method is appealing intuitively, the limits that result can only be
evaluated in certain cases.
For this reason, progress on the area problem remained at a rudimentary level until the latter
part of the seventeenth century when Isaac Newton and Leibnitz independently discovered a
fundamental relationship between area and derivative. They showed that if f is a nonnegative
continuous function on the interval [a, b] , and if A( x) denotes the are under the graph of f
over the interval [a, x] , where x is any point in the interval [a, b] , then
d
A( x) f ( x) i.e. [ A( x)] f ( x)
dx
The following example confirms this relationship in some cases where a formula for A( x) can
be found using elementary geometry.
Example: For each of the functions f , find the area A( x) between the graph of f and the
interval [a, x] [1, x] , and find the derivative A( x) of this area function
(a) f ( x) 2 (b) f ( x) x 1 (c) f ( x) 2 x 3
Solution of (a):
From figure A( x) 2( x (1)) 2( x 1) 2 x 2
Which is the are of a rectangle of height 2 and base x 1 . For this area fucntion,
A( x) 2 f ( x)
d
A( x) f ( x)
dx
f ( x)dx A( x)
A( x) is the antiderivative of f ( x) . Therefore, above example corroborates, integration of a
function f ( x) , yields (results, thereby, produces, outputs) area. This is the reason integration
means “area”.
1 x2 1
Solution (b): From the figure area A( x) ( x 1)( x 1) x , which is the area of an
2 2 2
isosceles right triangle with base and height equal to x 1 . For this the area function
A( x) x 1 f ( x)
1
Solution (c): The area of a trapezoid is (b b)h , where b and b denote the lengths of the
2
parallel sides of the trapezoid, and the altitude h denotes the distance between the parallel
1
sides. From the figure A( x) ((2 x 3) 1)( x (1)) x 2 3x 2 , which is the area of a
2
trapezoid with parallel sides of lengths 1 and 2 x 3 and with altitude x (1) x 1 . The are
function is
A( x) 2 x 3 f ( x) .
Example: Sketch the region whose area is represented by the definite integral, and evaluate
the integral using an appropriate formula from geometry.
4 2 1
(a) 2 dx (b) ( x 2) dx (c) 1 x 2 dx
1 1 0
Solution (a): The graph of the integrand is the horizontal line y 2 , so the rectangle of height
2 extending over the interval from 1 to 4. Therefore
4
1
2 dx (area of rectangle) 2.3 6
Solution (b): The graph of the integrand is the line y x 2 , so the region is a trapezoid
whose base extends from x 1 to x 2 . Therefore
1 15
2
( x 2) dx (area of trapezoid) (1 4)(3)
1 2 2
Solution (c): The graph of y 1 x 2 is the upper semicircle of radius 1, centered at the
origin, so the region is the right quarter-circle extending from x 0 to x 1 .
1
1
1 x2 dx (area of quarter circle) (11 )
0 4 4
Net signed area:
Net signed area: The area of the region that is above the interval [a, b] and below the curve
y f ( x) minus the area of the region that is below the interval [a, b] and above the curve
y f ( x) . For example the net signed area in the figure is
( AI AIII ) AII [area above [a, b]] [area below[a, b]]
2 1
Example: Evaluate ( x 1)dx and ( x 1)dx
0 0
Solution: The graph of y x 1 is shown in the figure. Applying the formula of plane
1
geometry, the shaded both the triangular regions AI and AII have area . Over the interval
2
[0,2], the net signed area is
AI AII 0
1
And over the interval [0,1], the net signed area is AII .
2
Example: (a) Find the area under the curve y cos x over the interval [0, 2]
(b) Make a conjecture about the value of integral 0
cos x dx .
Solution (a): The area A under the curve is
/2 /2
A cos x dx sin x 0
sin sin 0 1
0 2
Solution (b): The graph suggests that over the interval [0, ] , the portion of area above the -
x-axis is the same as the portion of area below the x-axis. The conjectured signed area is zero.
0
cos x dx sin x sin sin 0 0
0
Properties of the definite integral:
a
(a) If a is in the domain of f , then a
f ( x) dx 0 .
b a
(b) If f is integrable on [a,b], then a
f ( x) dx f ( x) dx
b
(a) x 2 dx 0 (b)
1 0
1 x 2 dx 1 x 2 dx
1
1 1 0 4
(c) If f is integrable on a closed interval containing the three points a, b and c , then
b c b
a
f ( x) dx f ( x) dx f ( x) dx
a c
Definition of Area: We now turn to the problem of giving a precise definition of what is
meant by the “area” under the curve.
Rectangle Method:
ba
1. Divide the interval [a, b] into n equal subintervals of width x , equally spaced
n
points x1 , x2 , , xn .
2. Over each subinterval construct a rectangle whose height is the value of f at a selected
point in the subinterval. Thus if x1* , x2* , x3* , ... , xn* denote the n selected points, then the
rectangles will have heights f ( x1* ), f ( x2* ), f ( x3* ), , f ( xn* ) and areas
f ( x1* )x, f ( x2* )x, , f ( xn* )x
3. The union of the rectangles forms a region Rn whose area can be regarded as an
approximation to the area A of R ; that is
n
A area( R) area( Rn ) f ( x1* )x f ( x2* )x f ( xn* )x f ( xk* )x
k 1
4. Repeat the process using more and more subdivisions, and define the area of R to be the
“limit”
n
A lim
n
f ( x )x
k 1
*
k
Definition of area under a curve:
If the function f is continuous on [a, b] and if f ( x) 0 for all x in [a, b] , then the area A
under the curve y f ( x) over the interval [a, b] is defined by
n
A lim
n
f ( x )x
k 1
*
k
Riemann sums and the definite integral:
In the rectangle method, we assume that for each positive number n , the interval [a, b] was
subdivided into n subintervals of equal length for approximating rectangles.
For some functions it may be more convenient to use rectangles with different widths. It is
important to note that the successive subdivisions be constructed in such a way that the widths
of all the rectangles approach zero as n increases.
The interval [a, b] will be divided with a collection of points
a x0 x1 x2 xn1 xn b
that divides [a, b] into subintervals of lengths
x1 x1 x0 , x2 x2 x1 , x3 x3 x2 , , xn xn xn 1
When this is done, the sum
n n
f ( x )x
k 1
*
k is replaced by f ( x )x
k 1
*
k k
We also need to replace the expression n by an expression that guarantees us that the
lengths of all subintervals approach zero. We will use the expression max xk 0 . Therefore
n
A lim
max xk 0
f ( x )x
k 1
*
k k
This fundamental concept of integral calculus forms the basis for the following definition:
Integrable:
A function f is said to be integrable on a finite closed interval [a, b] , if the limit
n
lim
max xk 0
f ( x )x
k 1
*
k k
exists. We denote this limit by the symbol
n
f ( x )x
b
f ( x)dx lim *
k k
a max xk 0
k 1
which is called the definite integral of f from a to b . The numbers a and b are called lower
limit of integration and the upper limit of integration respectively.
The notation used for the definite integral deserves some comment. Historically, the
expression “ f ( x)dx ” is interpreted to the “infinitesimal area” of a rectangle with height f ( x)
and “infinitesimal ” width dx . By “summing” these infinitesimal areas, the entire area under
the curve is obtained. The integral symbol “ ” is an “elongated S” that is used to indicate
this summation.
b
The sum lim
max xk 0
f ( xk* )xk is called the Riemann sum and the integral a
f ( x)dx is called the
Riemann integral in honor of the German mathematician Bernhard Riemann who formulated
many of the concepts of integral calculus.
n n
f ( xk* )x x f ( xk* ) x[ f ( x1* ) f ( x2* ) f ( xk* )]
k 1 k 1
Example: Find the left endpoint, right endpoint, and midpoint approximations of the area
under the curve y 9 x 2 over the interval [0,3] with n 10, n 15, n 20 subintervals.
Example: Use rectangles to estimate the area under the parabola y x 2 from 0 to 1. The
parabolic region S illustrated in figure.
bydra
Solution: Divide the region S into four strips S1, S2 , S3 , S4 by drawing the vertical lines
x 1/ 4, x 1/ 2, and x 3 / 4 as in figure shown. Find the area of the each rectangle and
added all. The heights of the each rectangle are the values of the function f ( x) x 2 at the
1 1 1 1 3 3
right endpoints of the subintervals 0, , , , , , , 1 .
4 4 2 2 4 4
2 2 2
1 1 1 3
Each rectangle has width and the heights are , , , and 12 .
4 4 2 4
2 2 2
1 1 1 1 1 3 1 15
R4 . . . .12 0.46875
4 4 4 2 4 4 4 32
Left endpoints
2 2 2
1 2 1 1 1 1 1 3 7
L4 . 0 . . . 0.21875
4 4 4 4 2 4 4 32
0.21875 A 0.46875
Example: From the same example prove that the sum of the areas of the upper approximating
1 1
rectangles approaches i.e. lim Rn .
3 n 3
Solution: Rn is the sum of the areas of the n rectangles. Each rectangle has width 1/ n and the
heights are the values of the function f ( x) x 2 at the points 1/ n, 2 / n, 3 / n, , n / n and the
heights are (1/ n) 2 , (2 / n) 2 , (3 / n) 2 , , ( n / n) 2 .
Thus
2 2 2 2
11 12 13 1n
Rn
nn nn nn nn
1 1 2
. 2 (1 22 32 n 2 )
n n
1 2
3
(1 22 32 n2 )
n
1 n(n 1)(2n 1)
.
n3 6
n(n 1)(2n 1)
6n 2
n(n 1)(n 2)
lim Rn lim
n n 6n 2
1 1 1
lim 1 2
n 6 n n
1
3
