|
| Integral
calculus |
|
|
|
| |
|
The
definite integral |
The area between the graph
of a function and the x-axis over a closed interval
|
|
The upper sum and the lower sum |
|
The definition of the definite integral |
|
Riemann
sum |
|
Calculating
a definite integral from the definition |
|
Calculating
a definite integral from the definition, examples |
|
Physical
applications to the definite integral |
|
Describing
motion of objects using velocity - time graphs |
|
Evaluating
the area under the graph of a function using the definition of the
definite integral, examples |
|
|
|
|
|
|
|
| The area between the graph
of a function and the x-axis over a closed interval |
| The
problem of finding the area enclosed by the graph of a function and the x-axis
over a closed interval is |
| solved
by use of the quadrature. The quadrature is the method of the calculation of
planar areas that consists |
| of
the
construction of a square with the same area as a given figure or
surface. |
| Let
y = f (x)
be a continuous positive function defined on the closed interval [a,
b]. |
| Define
the area A
of a region of the coordinate plane bounded above by the graph
of f, below by the interval
|
| [a, b]
on the x-axis,
and bounded to the left by the vertical line x
= a and on the right by the
vertical line |
| x = b. |
|
|
The upper sum and the lower sum |
| Divide
the given interval [a,
b] into n
subintervals by a finite sequence of points such that |
| a
= x0
< x1 < x2 < . . . < xn
-
1 < xn
= b. |
| The
obtained subintervals are [a,
x1], [x1,
x2],
. . . , [xi
-
1, xi],
. . . , [xn
-
1, b]
|
| of
lengths Dx1,
Dx2,
. . . , Dxi
,
. . . ,
Dxn
where, Dxi
= xi -
xi
-
1
and ( i = 1, 2 , . . . , n ). |
| Since,
by the assumption f
is continuous on [a,
b]
then there exist the highest value M and
the lowest value |
| m
of the function inside the interval [a,
b]
and similarly, inside the subinterval Dx1
it will have some highest |
| value
M1
and a lowest m1,
in Dx2
values M2
and m2,
. . . and in Dxn
a function values Mn
and mn. |
| Such
that always |
| mi
< Mi,
and
Mi
< M
as mi
> m |
|
( i = 1, 2 , . . . , n ),
because the highest value in a |
| subinterval
cannot exceed the highest value of the |
| function in the whole interval [a,
b],
as the lowest |
| value in
a
subinterval cannot be lower than that in the |
| whole interval, as is shown
in the right figure. |
| Now
let write |
|
S
= M1
Dx1
+ M2
Dx2
+
·
· ·
+
Mn
Dxn |
|
 |
|
|
and s
= m1
Dx1
+ m2
Dx2
+
·
· ·
+
mn
Dxn,
where
S
represents the sum of areas of the circumscribed |
| rectangles, each base
of witch is one of the subintervals Dxi
in the sequence, with height of the highest |
| function value
Mi
in that subinterval. Therefore, S
denotes the upper sum of f
over the interval [a,
b]. |
| While,
the sum s
denotes the lower sum of areas of the
inscribed rectangles with the same bases and with |
| heights
that equal the lowest function value mi
in each subinterval, as shows the figure above. |
| Since
mi
< Mi
( i = 1, 2 , . . . , n ),
then s
< S |
| and
since, m
< mi
and
Mi
< M
then, |
|
m
(b -
a)
= m
(Dx1
+
·
· ·
+ Dxn)
< s
and S
< M
(Dx1
+
·
· ·
+ Dxn)
= M
(b -
a), |
| that
is, the lower sum does not exceed the upper sum, both sums are bounded from
above and from below. |
| If we now repeat the process and continue to divide each subinterval
Dx into smaller intervals, so that the |
| number of subintervals increases to infinity
(n
®
oo), that is, bases of the rectangles become smaller and |
| smaller while the number of rectangles increases to infinity. |
| By passing to the infinitely small (infinitesimal)
interval Dx, each of two infinite sequences of numbers
S and
s |
| converge to a fixed
limit value. |
| That is, the sequence of areas of circumscribed rectangles will decrease toward the fixed limit value, as the |
| number of intervals
increases, and the sequence of areas of inscribed rectangles will increase toward
that |
| fixed limit
value. |
| Thus,
if the sequence of circumscribed rectangles S
and the sequence of inscribed rectangles
s
tends to the |
| same limit
value I independent of the partition of the interval [a,
b],
provided the length of each subinterval |
| tends
to zero while number of the partition
points tend to infinity, then the limit value I
equals the area A. |
| Therefore,
using the above notation we can write |
 |
|
|
The definition of the
definite integral |
| Finally,
we say a function
f
is integrable on an interval [a,
b] if there exists a
unique number A
such that |
 |
| for
any partition of [a,
b]. |
|
If
y = f (x)
is integrable on [a,
b]
then we call A
the definite integral of f
on [a,
b] and write |
 |
| where
∫ symbolize the sum, the function
f (x) is called
the integrand, the differential dx
shows that x
is the |
| variable
of integration and, the numbers a
and b
are called the limits of integration (a
is the lower limit and b |
| is
the upper limit). |
|
| Riemann
sum |
| Until
now, in the definition of the sums, S
and s
we've used the maximum and the minimum values, |
| Mi
and mi
of a given continuous function
f, so that mi
< f (x)
< Mi for x
in [xi
-
1, xi],
i = 1, 2 , . . . , n. |
| Now,
if we arbitrarily choose a point xi'
in every interval Dxi
and make products f (xi' )
Dxi,
then |
| mi
Dxi
<
f (xi' )
Dx
< Mi
Dxi,
i = 1, 2 , . . . , n |
| and
by adding up |
 |
| The
left-hand and right-hand sides of the above inequality are the sums, s
and S
respectively that, because of |
| continuity of
f,
tend to the same limit value I
when the number
of subintervals n
increases to infinity, such |
| that the length of
every interval Dxi
tends to zero, for any partition of the interval [a,
b]
and arbitrarily |
| chosen
points xi'
in the subintervals [xi
-
1, xi].
Hence,
the middle term |
 |
| called
a Rieman sum, will tend to the same limit value.
|
| Therefore,
if f
is a positive continuous function on the interval [a,
b] then, the definite
integral of the function |
| from a
to b
is defined to be the limit
|
 |
| Note
that the limit value of the sum changes as the number
of subintervals n
increases to infinity while the |
| length
(Dxi)
of each tends to zero, for any partition of the interval [a,
b].
|
|
| Calculating
a definite integral from the definition |
|
As the sequence of inscribed rectangles s
tends to the definite
integral increasingly while the sequence of |
| circumscribed rectangles S
tends to the same value decreasingly then |
 |
| Thus,
we
can approximate the area under the graph of a function over the interval
[a,
b]
to any desired level of |
| accuracy using
the Riemann sum of inscribed or circumscribed rectangles. |
| The
area of the ith
rectangle f (xi' )
Dxi,
denoted as height times base, represents the ith
term of Riemann |
| sum
and is called the element of area. |
| When
we use the partition of the interval [a,
b] into n
equal subintervals (regular partition) then |
 |
| be
the length of the intervals [xi
-
1, xi],
i = 1, 2 , . . . , n. |
|
| Calculating
a definite integral from the definition, examples |
| Example:
Evaluate |
 |
where f (x)
=
1, using the
definition of the definite integral. |
|
| Solution: Since
the graph of the constant f (x)
=
1
is the line passing through the point (0,
1)
parallel to |
| the
x-axis,
the region under the graph is the rectangle of
the base
=
(b
-
a)
with the height h
=
1. |
| Thus,
the area |
| A
=
(b
-
a) · 1
=
b
-
a, |
| as
shows the right figure. Therefore, |
 |
|
 |
|
|
| Example:
Evaluate |
 |
where f (x)
=
x, using the
definition of the definite integral. |
|
| Solution: Since
the graph of f (x)
=
x
is the line through the origin, coordinates of every its point y
=
x, |
| so
the region under the graph is the trapezium
with the height
b
-
a
and whose parallel sides are
a
and b. |
| Let's
use the partition of the interval [a,
b] into n
equal subintervals,
so that Dx
=
(b
-
a) / n
and calculate |
| the
lower sum s
of inscribed rectangles, as
shows the right |
| figure.
If
we choose the point xi'
to be the left-hand end of |
| each
subinterval,
then |
| x1'
=
a
and
f (x1'
)
=
a |
| x2'
=
a
+
Dx,
f (x2'
)
=
a
+
Dx |
|
x3'
=
a
+
2Dx,
f (x3'
)
=
a
+
2Dx |
|
·
· ·
·
· · |
| xn'
=
a
+
(n
-
1)Dx,
f (xn'
)
=
a
+
(n
-
1)Dx. |
|
 |
|
| We
use the Riemann sum to calculate the sum of inscribed rectangles with
bases of the same length, |
 |
|
therefore, s
=
a
Dx
+
(a
+
Dx)Dx
+
(a
+
2Dx)Dx
+
·
· ·
+
[a
+
(n
-
1)Dx]Dx |
|
s
=
Dx[n
a +
Dx(1
+
2
+
·
· ·
+
(n
-
1))] |
| to
calculate the sum of natural numbers inside of square brackets we use
the formula |
| Sn
= [2a1 + (n -
1)d] for the sum of the
arithmetic sequence whose first term a1
= 1 and difference d
= 1, |
| so
we get the sum equals
(n
-
1)n
/ 2, and since Dx
=
(b
-
a) / n
then, |
 |
| Thus
the area under the graph of f (x)
=
x
over the interval [a,
b] |
 |
|
| Physical
applications to the definite integral |
| Describing
motion of objects using velocity - time graphs |
| In
kinematics motion is defined as change in position relative to some fixed point or
object. |
| The
distance the object travels in a period of time is its speed. Velocity
is speed in a given direction. |
| The
displacement is the vector form of the distance involving direction and
magnitude of the change in position |
| of
the object, meaning it is the straight line distance between the initial
and final positions of the body. |
| Acceleration
is the increase in speed or velocity over a period of time. |
|
| Example:
When
an object travels with a constant velocity v
over a period of time t,
the displacement d,
for |
| the object
we
calculate using the formula d
= v · t. |
| Then,
the
constant velocity v
is represented in the coordinate system (t,
v)
as
a line parallel to the t-axis |
| passing
through the point (0,
v)
while the displacement
d
is given by the area of the rectangle with the base t |
| and
height v,
as is shown in the left figure below. |
|
|
|
| Example:
If
an object travels with constant acceleration a
then its velocity v
is changing by a constant |
| amount
each unit of time t. |
| Thus,
the velocity v
= a · t
is represented as the line through the origin with slope equal to the
acceleration, |
| as
shows the velocity-time graph in the right picture above. |
| When
an object travels with constant acceleration a
then
displacement
d
traveled in a certain amount of time |
| t
is the same as object travels with constant velocity v
= (a · t)/2, that
is, equals
the half of the final velocity |
| at
the end of the time interval. |
| Therefore,
the length of the path an object travels with constant acceleration
in a given time interval is |
| represented
in the velocity-time graph by the area of the rectangle of the base t
and height (at)/2 |
 |
| that
correspond to the area of the triangle of the base t
with height at,
shown in the above figure. |
|
| Evaluating
the area under the graph of a function using the definition of the
definite integral, examples |
| Example:
Evaluate
the area under the graph of the quadratic function over the interval [0,
x],
where
x
is any |
| value
of the argument, using the
definition of the definite integral. |
| Solution: Let's
use the partition of the interval [0,
x] into n
equal subintervals,
so that Dx
=
x /
n
and |
| calculate the
lower sum s
of inscribed rectangles, as
is shown in the left figure below. According to
definition |
 |
| The
area under the graph of
f(x) = x2
over the interval [0,
x] amounts 2/3
of the area of the triangle OPxP. |
|
|
|
| Example:
Evaluate
the area under the graph of the rectangular or equilateral hyperbola f(x)
= 1/x over the |
| interval [1,
x],
using the
definition of the definite integral. |
| Solution:
Let's divide the interval [1,
x]
into n
subintervals of different lengths by partition points that form |
| the
geometric sequence, 1,
q, q2, . . . , qn -
1, qn = x (q > 1) |
| and
calculate the lower and upper sums, s
and S,
as is shown in the right figure above. |
