Q: Are continuous functions integrable

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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, a corner is continuous, as long as you don't have to lift your pencil up then it is a continuous function. Continuous functions just have no breaks, gaps, or holes.

They are both continuous, symmetric distribution functions.

The similarities are that they are polynomial functions and therefore continuous and differentiable.A real cubic will has an odd number of roots (and so must have a solution), a quartic has an even number of roots and so may have no solutions.

Identify the degree and leading coefficient of polynomial functions. ... the bird problem, we need to understand a specific type of function. A power ... A power function is a function that can be represented in the form ... Example 3.4.1: Identifying Power Functions ... Comparing Smooth and Continuous Graphs.

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That's true. If a function is continuous, it's (Riemman) integrable, but the converse is not true.

No, all functions are not Riemann 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.

Krzysztof Ciesielski has written: 'I-density continuous functions' -- subject(s): Baire classes, Continuous Functions, Functions, Continuous, Semigroups

Not at all.Y = x2 is a continuous function.

No. Not all functions are continuous. For example, the function f(x) = 1/x is undefined at x = 0.

Yes, all polynomial functions are continuous.

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

No. There are many common functions which are not discrete but the are not continuous everywhere. For example, 1/x is not continuous at x = 0 (it is not even defined there. Then there are curves with step jumps.

Jean Schmets has written: 'Spaces of vector-valued continuous functions' -- subject(s): Continuous Functions, Locally convex spaces, Vector valued functions

Frederick Bagemihl has written: 'Meier points and horocyclic Meier points of continuous functions' -- subject(s): Continuous Functions

Yes, a corner is continuous, as long as you don't have to lift your pencil up then it is a continuous function. Continuous functions just have no breaks, gaps, or holes.