That's true. If a function is continuous, it's (Riemman) integrable, but the converse is not true.
No, a non-continuous function cannot be differentiable at the points of discontinuity. Differentiability requires the existence of a well-defined tangent line at a point, which necessitates continuity at that point. However, a function can be differentiable on intervals where it is continuous, even if it has discontinuities elsewhere.
Υou show that it is continuous in every element of it's domain.
No. Not all functions are continuous. For example, the function f(x) = 1/x is undefined at x = 0.
continuous
The way I understand it, a continuos function is said not to be "uniformly continuous" if for a given small difference in "x", the corresponding difference in the function value can be arbitrarily large. For more information, check the article "Uniform continuity" in the Wikipedia, especially the examples.
yes, every continuous function is integrable.
An antiderivative, F, is normally defined as the indefinite integral of a function f. F is differentiable and its derivative is f.If you do not assume that f is continuous or even integrable, then your definition of antiderivative is required.
Not every function with discontinuities is integrable. A function is considered integrable (in the Riemann sense) if the set of its discontinuities has measure zero. Functions with too many or too "wild" discontinuities, such as the Dirichlet function (which is 1 for rational numbers and 0 for irrational numbers), are not Riemann integrable. However, they may still be Lebesgue integrable under certain conditions.
A function may have a finite number of discontinuities and still be integrable according to Riemann (i.e., the Riemann integral exists); it may even have a countable infinite number of discontinuities and still be integrable according to Lebesgue. Any function with a finite amount of discontinuities (that satisfies other requirements, such as being bounded) can serve as an example; an example of a specific function would be the function defined as: f(x) = 1, for x < 10 f(x) = 2, otherwise
yes
Yes. A well-known example is the function defined as: f(x) = * 1, if x is rational * 0, if x is irrational Since this function has infinitely many discontinuities in any interval (it is discontinuous in any point), it doesn't fulfill the conditions for a Riemann-integrable function. Please note that this function IS Lebesgue-integrable. Its Lebesgue-integral over the interval [0, 1], or in fact over any finite interval, is zero.
Yes. The cosine function is continuous. The sine function is also continuous. The tangent function, however, is not continuous.
If the distribution is discrete you need to add together the probabilities of all the values between the two given ones, whereas if the distribution is continuous you will need to integrate the probability distribution function (pdf) between those limits. The above process may require you to use numerical methods if the distribution is not readily integrable. For example, the Gaussian (Normal) distribution is one of the most common continuous pdfs, but it is not analytically integrable. You will need to work with tables that have been computed using numerical methods.
yes it is a continuous function.
Yes, a polynomial function is always continuous
Weistrass function is continuous everywhere but not differentiable everywhere
Yes, that happens with any continuous function. The limit is equal to the function value in this case.Yes, that happens with any continuous function. The limit is equal to the function value in this case.Yes, that happens with any continuous function. The limit is equal to the function value in this case.Yes, that happens with any continuous function. The limit is equal to the function value in this case.