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let [a; b] be a non-degenerate closed interval in r, and let f : [a; b] ! r. for ant x 2 [a; b] and any t 2 [a; b] s.t. t 6

Sagot :

Final answer :

The non - degenerated closed interval in r is calculated as [tex]\int\limits^t_s {f(t) - f(s)}[/tex]

What is the degenerated interval ?

Any set that only contains one real number is known as a degenerate interval (i.e., an interval of the form [a,a]). The empty set is included in this definition by some authors. A real interval is considered to be appropriate if it has an infinite number of elements and is neither empty nor degenerate.

When an interval is enclosed in square brackets, it means that it contains all of its limit points. For instance, [0,1] denotes values larger or equal to 0 and lower or equal to 1. The endpoints are not included in an open interval and are denoted by parentheses. For instance, the interval (0,1) is described as being higher than 0 and smaller than 1. The endpoints are included and are indicated by square brackets in a closed interval.

Explanation :

[tex]f(x_{r}) - f(x_{r-1}) = F^{'} ( e_{r}) (x_{r} - x_{r-1})\\[/tex]

By summating the equation it becomes,

[tex]f(t) - f(s)[/tex]

[tex]\int\limits^t_s {f} < =f(t) - f(S) ---- > 1[/tex]

[tex]\int\limits^-t_s < ={f(t)- f(s) --- > 2[/tex]

From 1 and 2

[tex]\int\limits^t_-s < ={f(t) - f(s) < = \int\limits^t_d {f}[/tex]

So, it finally becomes

[tex]\int\limits^t_s {f(t) - f(s)}[/tex]

To learn more about degenerated intervals, visit

brainly.com/question/15875011

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