Are all integrable functions continuous?

Answer 1

No, for example you can integrate the function 1/(1+x) into ln(1+x), and that function has a nonremovable discontinuity at x=(-1)

Answer 2

No, continuity is a sufficient, but not necessary, condition for integrability of functions from R to R (over compact intervals).

For example, every monotonic function on an interval [a, b], with finite limits is integrable, even if it's not continuous. So, the function f(x) = x, if 0 <= 0, <1 and f(x) = 3x +1, if 1 <=x <2 is integrable over [0, 2], though it's discontinuous at x =2.

There's a condition for integrabilty. A function f is integrable over a compact interval [a, b] if, and only if, f is bounded on [a, b] and the set of the values at which f is discontinuous has Lebesgue measure 0. The concept of measure is studied in measure theory. Anywa, this implies that, though f doesn't need to be continuous everywhere on [a, b] to be (Riemann) integrable, it can't be "too" discontinuous. But the set of discontinuities don't need to be countable.

Answer 3

Non Continuous Function

Answer 4

No.

Consider the function f on [0, 2] defined by

f(x) = 1, if 0 <= x <= 1
f(x) = 0, if 1 < x <= 2.

This function is clearly not continuous at x = 1.

However, it is integrable on [0, 2]; its integral is 1. (This is "obvious" from the geometric interpretation of the integral as the area under the curve; it is also not difficult to prove it rigorously using the formal definition of the Riemann integral using partitions.)

It is, however, true that all continuous functions are integrable, in the sense that if a function f is continuous on a closed interval [a, b], then that function is integrable on [a, b]. (This is part of what the Fundamental Theorem of Calculus says.)

It is also true that if f is bounded on [a, b] and has at most a countable number of discontinuities on [a, b], then f is integrable on [a, b].

Answer 5

No, continuity is a adequate, yet no longer mandatory, concern for integrability of applications from R to R (over compact periods). to illustrate, each and every monotonic function on an era [a, b], with finite limits is integrable, whether it incredibly is no longer non-end. So, the function f(x) = x, if 0 <= 0,

Answer 6

No. Non-continuous functions can be integrable by parts.

Answer 7

No. Depending on the type of integral you are doing, you can have a "countably infinite" number of discontinuities and still calculate an integral.

Answer 8

No, try x^(-1/2)
which is also 1/(x^2).