Definite Integral as the Limit of a Sum

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 ], … ,[x_{n − 2} , x_n−1 ],[x_{n −1} , x_n ] such that

a = x₀ < x₁ < x₂ < < x_{n −1} < x_n = b.

In each subinterval [x_{i −1} , x_i ], i = 1, 2, …. , n, choose a real number ξ_i arbitrarily such that x_{i −1} ≤ ξ_i ≤ x_i.

Consider the sum ∑ⁿ_i=1 f (ξ_i )( x_i − x_{i −1} ) = f (ξ₁ )(x₁ − x₀ ) + f (ξ₂ )(x₂ − x₁ ) + … + f (ξ_n )(x_n − x_n−1 ) ….(1)

The sum in (1) is called a Riemann sum of f ( x) corresponding to the partition [x₀ , x₁ ],[x₁ , x₂ ], . . . ,[x_{n −1} , x_n ] of [a , b]. Since there are infinitely many values ξ_i satisfying the condition x_{i −1} ≤ ξ_i ≤ x_i , there are infinitely many Riemann sums of f ( x) corresponding to the same partition [x₀ , x₁ ],[x₁ , x₂ ], . . . ,[x_{n −1} , x_n ] of [a , b]. If, under the limiting process n → ∞ and max ( x_i − x_{i −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 ^b∫_a 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 ^a∫_a 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 ^b∫_a f (x)dx as ^b∫_a f (u)du . So, we have ^b∫_a f (x)dx = ^b∫_a f (u)du . As max ( x_i − x_i−1) → 0, all the three points x_{i −1} , ξ_i , and x_i of each subinterval [ x_{i −1} , x_i ] are dragged into a single point. We have already indicated that there are infinitely many ways of choosing the evaluation point ξ_i in the subinterval [ x_{i −1} , x_i ] , i = 1, 2, . . . , n . By choosing ξ_i =x_{i −1} , i = 1, 2, , n , we have

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

By choosing ξ_i =x_i , 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 ^b∫_a f ( x)dx exists, then the Riemann integral ^x∫_a 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) = ^x∫_a f (u)du, x ∈[a, b] .

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

Then, the definite integral ^b∫_a f ( x)dx is given by

= A₁−A₂+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 ∫₀^0.5 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−1 )Δx .

 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´ ∫₀^0.5 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 _n )Δx .

 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 .

∴ ∫₀^0.5 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 .

∴ ∫₀^0.5 x²dx is approximately 0.04125 .