Consider a real-valued, bounded function f (x) defined on the closed and bounded interval[a, b ], a < b.

**Definite Integral
as the Limit of a Sum**

**Riemann Integral**

Consider
a real-valued, bounded function *f (x)*
defined on the closed and bounded interval[*a*,
*b* ], *a* < *b*.
The function *f *(* x*) need not have the same sign on [*a* , *b*] ; that is, *f* ( *x*)
may have positive as well as negative values on [*a* , *b*] . See Fig 9.2.
Partition the interval [*a* , *b*] into *n* subintervals [*x*_{0}
, *x*_{1} ],[*x*_{1} , *x*_{2} ], â€¦ ,[*x _{n}*

*a *=* x*_{0}* *<* x*_{1}* *<* x*_{2}* *< <* x _{n} *

In each
subinterval [*x _{i}*

Consider
the sum âˆ‘^{n}_{i=1} *f*
(*Î¾** _{i}* )(

The sum in
(1) is called a **Riemann
sum** of *f* ( *x*) corresponding to the partition [*x*_{0}* *,* x*_{1}* *],[*x*_{1}* *,*
x*_{2}* *], . . . ,[*x _{n} *

**Note**

In the
present chapter, we consider bounded functions *f* ( *x*) that are
continuous on[*a* , *b*] . However, the **Riemann integral** of *f* ( *x*)
on [*a* , *b*] also exists for bounded functions *f* ( *x*) that are piece-wise
continuous on[a , b] .We have used the same symbol âˆ« both for definite integral
and anti-derivative (indefinite integral). The reason will be clear after we
state the Fundamental Theorems of Integral Calculus. The variable* x *is **dummy** in the sense that it is
selected at our choice only. So we can write ^{b}âˆ«_{a} *f (x)dx* as ^{b}âˆ«_{a}* f *(u)*du *. So, we have ^{b}âˆ«_{a} *f (x)dx* = ^{b}âˆ«_{a}* f *(u)*du *. As max ( *x*_{i} âˆ’ *x*_{iâˆ’1}) â†’ 0, all the three points *x _{i} *

Equation
(2) is known as the **left-end rule** for evaluating the Riemann
integral.

By
choosing *Î¾** _{i}* =

Equation
(3) is known as the **right-end rule** for evaluating the Riemann
integral.

Equation
(4) is known as the **mid-point rule** for evaluating the Riemann
integral.

**Remarks**

(1) If
the Riemann integral ^{b}âˆ«_{a} *f* ( *x*)*dx* exists, then the Riemann integral ^{x}âˆ«_{a}
*f* (*u*)*du* is a

well-defined
real number for every *x* âˆˆ[*a*,
*b*] ** _{.}** So, we can define a function

(2) If *f *(*x*)* *â‰¥* *0 for all *x *âˆˆ[*a*,*
b*]* *, then the Riemann integral ^{b}âˆ«_{a}*f *(*
x*)*dx *is equal to the geometric
area of the region bounded by the graph of *y*
=
*f* ( *x*) , the *x*-axis, the
lines *x* =
*a* and *x* = *b* . See Fig. 9.3.

(3) If *f *(*
x*)* *â‰¤* *0* *for all* x *âˆˆ[*a*,*
b*]* *, then the Riemann integral ^{b}âˆ«_{a} *f* ( *x*)*dx* is
equal to the negative of the geometric area of the region bounded by the graph
of *y *=* f *(* x*)*
*, the* x*-axis, the lines *x* = *a*
and *x* = *b*
. See Fig. 9.4. In this case, the geometric area of the region bounded by the
graph of *y* =
*f* ( *x*) , the *x*-axis, the
lines *x* =
*a* and *x* = *b* is given by

(4) If *f (x)* takes positive as well as negative
values on [a , b] , then the interval [a , b] can be divided into subintervals
[a , c_{1} ] , [c_{1} , c_{2} ] ,. . . , [c_{k} , b] such that *f (x)* has the same sign throughout each
of subintervals. So, the Riemann integral ^{b}âˆ«_{a} *f (x)dx* is given by

In this
case, the geometric area of the region bounded by the graph of *y *=* f *(*
x*)* *, the* x-*axis, the lines *x* =
*a* and *x* = *b* is given by

For
instance, consider the following graph of a function *f* ( *x*), *x* âˆˆ[*a*,
*b*] . See Fig. 9.5. Here, *A*_{1} , *A*_{2} and, *A*_{3}* *denote geometric areas of the
individual parts.

Then,
the definite integral ^{b}âˆ«_{a} *f* ( *x*)*dx* is given by

= *A*_{1}âˆ’*A*_{2}+*A*_{3}.

The
geometric area of the region bounded by the graph of *y* = *f* ( *x*) , the *x *âˆ’* *axis, the lines *x* =* a *and* x *=* b *is given by* A*_{1}* *+* A*_{2}* *+* A*_{3}* *. In view of the above discussion,
it is clear that a Riemann integral need not represent geometrical area.

**Note**

Even if
we do not mention explicitly, it is always understood that the areas are
measured in square units and volumes are measured in cubic units.

** **

**Example 9.1**

Estimate
the value of âˆ«_{0}^{0.5} *x*^{2}*dx* using the Riemann sums corresponding
to 5 subintervals of equal width and applying (i) left-end rule (ii) right-end
rule (iii) the mid-point rule.

**Solution**

Here *a* = 0, *b* = 0.5, *n* =
5, *f* (*x*) = *x*^{2}

So, the width
of each subinterval is

* h*
=* *Î”*x *=* b*âˆ’a / *n* = 0.5âˆ’0 / 5 = 0.1.

The
partition of the interval is given by the points

*x*_{0 }= 0,

*x*_{1 }= *x*_{0} + *h* = 0 + 0.1 = 0.1

*x*_{2 }= *x*_{1} + *h* = 0.1+ 0.1 = 0.2

*x*_{3}= *x*_{2} + *h* = 0.2 + 0.1 = 0.3

*x*_{4}= *x*_{3} + *h* = 0.3 + 0.1 = 0.4

*x*_{5}= *x*_{4} + *h* = 0.4 + 0.1 = 0.5

(i) The
left-end rule for Riemann sum with equal width Î”*x* is

S = [ *f*(*x*_{0}) +* f *(*x*_{1}) + . .
. +* f *(* x *_{nâˆ’1} )Î”*x *.

S = [*f*
( 0) +* f *( 0.1) +* f *( 0.2) +* f *( 0.3) +* f *( 0.4) ] (0.1)

= [ 0.00 + 0.01+ 0.04 +
0.09 +
0.16] (0.1)
=
0.03

âˆ´ âˆ«_{0}^{0.5} *x*^{2} *dx* is approximately 0.03 .

(ii) The
right-end rule for Riemann sum with equal width D*x* is

S = [ *f*(*x*_{1})
+* f *(*x*_{2}) + . . . +* f *(* x *_{n} )Î”*x *.

* S = *[
*f *(* *0.1)* *+* f *(* *0.2)* *+* f *(* *0.3)* *+* f *(* *0.4)* *+* f *(* *0.5) ]* *(0.1)

= [ 0.01+ 0.04 + 0.09 + 0.16 + 0.25](0.1) = 0.055 .

âˆ´ âˆ«_{0}^{0.5} *x*^{2} *dx* is approximately 0.055 .

(iii) The
mid-point rule for Riemann sum with equal width Î”*x* is

*S = *[ *f *(* *0.05)* *+* f *(* *0.15)* *+* f *(* *0.25)* *+* f *(* *0.35)* *+* f *(* *0.45) ]* *(0.1)

= [ 0.0025 + 0.0225 + 0.0625 + 0.1225 + 0.2025](0.1)

= 0.04125 .

âˆ´ âˆ«_{0}^{0.5} *x*^{2}*dx* is approximately 0.04125 .

Tags : Applications of Integration | Mathematics , 12th Maths : UNIT 9 : Applications of Integration

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12th Maths : UNIT 9 : Applications of Integration : Definite Integral as the Limit of a Sum | Applications of Integration | Mathematics

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