IDNLearn.com: Your reliable source for finding precise answers. Ask your questions and receive comprehensive, trustworthy responses from our dedicated team of experts.
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
#SPJ4
Your presence in our community is highly appreciated. Keep sharing your insights and solutions. Together, we can build a rich and valuable knowledge resource for everyone. Your search for solutions ends at IDNLearn.com. Thank you for visiting, and we look forward to helping you again.
