Must a Function be Bounded for the Antiderivative to Exist over a Given Interval? In my Calculus cl

desertiev5

desertiev5

Answered question

2022-06-30

Must a Function be Bounded for the Antiderivative to Exist over a Given Interval?
In my Calculus class, we were given the following definition of antiderivative:
Let f be a function defined on an interval.
An antiderivative of f is any function F such that F = f.
The collection of all antiderivatives of f is denoted f ( x ) d x .
My question is, don't we have to say that f should be bounded in the definition?
If not, then f is not integrable by definition, so we can't say anything about f ( x ) d x , right?

Answer & Explanation

pompatzz8

pompatzz8

Beginner2022-07-01Added 11 answers

Step 1
No, the function does not have to be bounded to have an integral.
Consider 0 1 d x x
which is an improper integral because the integrand is not bounded on (0,1).
Step 2
However the anti derivative is 2 x which results in a bounded value for 0 1 d x x = 2.
Gauge Terrell

Gauge Terrell

Beginner2022-07-02Added 5 answers

Step 1
The existence of an antiderivative and being integrable are distinct (although related) concepts.
Take f : [ 0 , 1 ] R x { x 2 sin ( 1 x 2 )  if  x > 0 0  if  x = 0.
Step 2
Then f is differentiable, but f′ is unbounded. But, in particular, f is an antiderivative of f′.

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