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Chapter: 12th Maths : UNIT 9 : Applications of Integration

Definite Integral as the Limit of a Sum

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 [x0 , x1 ],[x1 , x2 ], … ,[xn 2 , xn1 ],[xn 1 , xn ] such that

a = x0 < x1 < x2 < < xn 1 < xn = b.


In each subinterval [xi 1 , xi ], i = 1, 2, …. , n, choose a real number ξi arbitrarily such that xi 1 ξi xi.

Consider the sum ni=1 f (ξi )( xi xi 1 ) = f (ξ1 )(x1 x0 ) + f (ξ2 )(x2 x1 ) + … + f (ξn )(xn xn1 ) ….(1)

The sum in (1) is called a Riemann sum of f ( x) corresponding to the partition [x0 , x1 ],[x1 , x2 ], . . . ,[xn 1 , xn ] of [a , b]. Since there are infinitely many values ξi satisfying the condition xi 1 ξi xi , there are infinitely many Riemann sums of f ( x) corresponding to the same partition [x0 , x1 ],[x1 , x2 ], . . . ,[xn 1 , xn ] of [a , b]. If, under the limiting process n → ∞ and max ( xi xi 1)→ 0, the sum in (1) tends to a finite value, say A, then the value A is called the definite integral of f ( x) with respect to x on [a , b] . It is also called the Riemann integral of f ( x) on [a , b] and is denoted by ba f ( x)dx and is read as the integral of f ( x) with respect to x from a to b . If a = b, then we have aa f ( x)dx = 0.

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 ∫ 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 ba f (x)dx as ba f (u)du . So, we have ba f (x)dx = ba f (u)du . As max ( xixi−1) → 0, all the three points xi 1 , ξi , and xi of each subinterval [ xi 1 , xi ] are dragged into a single point. We have already indicated that there are infinitely many ways of choosing the evaluation point ξi in the subinterval [ xi 1 , xi ] , i = 1, 2, . . . , n . By choosing ξi =xi 1 , i = 1, 2, , n , we have


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

By choosing ξi =xi , i = 1, 2,. . . , n , we have


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 ba f ( x)dx exists, then the Riemann integral xa f (u)du is a

well-defined real number for every x [a, b] . So, we can define a function F ( x) on [a , b] such that F ( x) = xa f (u)du, x [a, b] .

(2) If f (x) 0 for all x [a, b] , then the Riemann integral baf ( 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) 0 for all x [a, b] , then the Riemann integral ba 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 , c1 ] , [c1 , c2 ] ,. . .  , [ck , b] such that f (x) has the same sign throughout each of subintervals. So, the Riemann integral ba 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 [a, b] . See Fig. 9.5. Here, A1 , A2 and, A3 denote geometric areas of the individual parts.

Then, the definite integral ba f ( x)dx is given by


= A1A2+A3.

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 A1 + A2 + A3 . 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 00.5 x2dx 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) = x2

So, the width of each subinterval is

 h = Δx = b−a / n  = 0.5−0 / 5 = 0.1.

The partition of the interval is given by the points

x0 = 0,

x1 = x0 + h = 0 + 0.1 = 0.1

x2 = x1 + h = 0.1+ 0.1 = 0.2

x3= x2 + h = 0.2 + 0.1 = 0.3

x4= x3 + h = 0.3 + 0.1 = 0.4

x5= x4 + h = 0.4 + 0.1 = 0.5

(i) The left-end rule for Riemann sum with equal width Δx is

S = [ f(x0) + f (x1) + . . . + f ( x n−1x .

 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

 ∴00.5 x2 dx is approximately 0.03 .

(ii) The right-end rule for Riemann sum with equal width Dx is

 S = [ f(x1) + f (x2) + . . . + f ( x nx .

 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 .

00.5 x2 dx is approximately 0.055 .

(iii) The mid-point rule for Riemann sum with equal width Δx 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 .

00.5 x2dx is approximately 0.04125 .

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