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Is the derivative of a continuously differentiable function always integrable?
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I know that $f(x)$ is bounded by the extreme value theorem, because $f(x)$ is defined on a closed bounded interval. Now, how can I prove that its derivative is integrable knowing that it is continuous? Or, is there a counterexample?
integration analysis derivatives continuity
edited 45 mins ago
Brahadeesh
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asked 57 mins ago
Kim
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up vote
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down vote
favorite
I know that $f(x)$ is bounded by the extreme value theorem, because $f(x)$ is defined on a closed bounded interval. Now, how can I prove that its derivative is integrable knowing that it is continuous? Or, is there a counterexample?
integration analysis derivatives continuity
edited 45 mins ago
Brahadeesh
4,78931853
asked 57 mins ago
Kim
61
New contributor
Kim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I know that $f(x)$ is bounded by the extreme value theorem, because $f(x)$ is defined on a closed bounded interval. Now, how can I prove that its derivative is integrable knowing that it is continuous? Or, is there a counterexample?
integration analysis derivatives continuity
edited 45 mins ago
Brahadeesh
4,78931853
asked 57 mins ago
Kim
61
New contributor
Kim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I know that $f(x)$ is bounded by the extreme value theorem, because $f(x)$ is defined on a closed bounded interval. Now, how can I prove that its derivative is integrable knowing that it is continuous? Or, is there a counterexample?
integration analysis derivatives continuity
integration analysis derivatives continuity
edited 45 mins ago
Brahadeesh
4,78931853
asked 57 mins ago
Kim
61
New contributor
Kim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 45 mins ago
Brahadeesh
4,78931853
asked 57 mins ago
Kim
61
New contributor
Kim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 45 mins ago
Brahadeesh
4,78931853
edited 45 mins ago
Brahadeesh
4,78931853
edited 45 mins ago
Brahadeesh
4,78931853
4,78931853
asked 57 mins ago
Kim
61
New contributor
Kim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 57 mins ago
Kim
61
asked 57 mins ago
Kim
61
61
New contributor
Kim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Kim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Kim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
add a comment |Â
2 Answers
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Any continuous function on a closed and bounded interval $[a,b]$ is integrable. Since, $f$ is given to be continuously differentiable, its derivative is continuous on the domain, and hence integrable.
answered 46 mins ago
Brahadeesh
4,78931853
add a comment |Â
up vote
2
down vote
If, according to your question, $f'$ is continuous on a bounded interval (i.e. a compact set), it makes no doubt it is Riemann-integrable (and hence also Lebesgue integrable):
A bounded function on a compact interval $[a, b]$ is Riemann integrable
if and only if it is continuous almost everywhere (the set of its
points of discontinuity has measure zero, in the sense of Lebesgue
measure).
See: Riemann integral
edited 43 mins ago
Brahadeesh
4,78931853
answered 48 mins ago
Picaud Vincent
53615
@Brahadeesh I did not realize I had so much typos in my answer, thanks for the corrections.
– Picaud Vincent
40 mins ago
1
No problem, you're welcome :)
– Brahadeesh
39 mins ago
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Any continuous function on a closed and bounded interval $[a,b]$ is integrable. Since, $f$ is given to be continuously differentiable, its derivative is continuous on the domain, and hence integrable.
answered 46 mins ago
Brahadeesh
4,78931853
add a comment |Â
up vote
2
down vote
Any continuous function on a closed and bounded interval $[a,b]$ is integrable. Since, $f$ is given to be continuously differentiable, its derivative is continuous on the domain, and hence integrable.
answered 46 mins ago
Brahadeesh
4,78931853
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Any continuous function on a closed and bounded interval $[a,b]$ is integrable. Since, $f$ is given to be continuously differentiable, its derivative is continuous on the domain, and hence integrable.
answered 46 mins ago
Brahadeesh
4,78931853
Any continuous function on a closed and bounded interval $[a,b]$ is integrable. Since, $f$ is given to be continuously differentiable, its derivative is continuous on the domain, and hence integrable.
answered 46 mins ago
Brahadeesh
4,78931853
answered 46 mins ago
Brahadeesh
4,78931853
answered 46 mins ago
Brahadeesh
4,78931853
answered 46 mins ago
Brahadeesh
4,78931853
4,78931853
add a comment |Â
add a comment |Â
up vote
2
down vote
If, according to your question, $f'$ is continuous on a bounded interval (i.e. a compact set), it makes no doubt it is Riemann-integrable (and hence also Lebesgue integrable):
A bounded function on a compact interval $[a, b]$ is Riemann integrable
if and only if it is continuous almost everywhere (the set of its
points of discontinuity has measure zero, in the sense of Lebesgue
measure).
See: Riemann integral
edited 43 mins ago
Brahadeesh
4,78931853
answered 48 mins ago
Picaud Vincent
53615
@Brahadeesh I did not realize I had so much typos in my answer, thanks for the corrections.
– Picaud Vincent
40 mins ago
1
No problem, you're welcome :)
– Brahadeesh
39 mins ago
add a comment |Â
up vote
2
down vote
If, according to your question, $f'$ is continuous on a bounded interval (i.e. a compact set), it makes no doubt it is Riemann-integrable (and hence also Lebesgue integrable):
A bounded function on a compact interval $[a, b]$ is Riemann integrable
if and only if it is continuous almost everywhere (the set of its
points of discontinuity has measure zero, in the sense of Lebesgue
measure).
See: Riemann integral
edited 43 mins ago
Brahadeesh
4,78931853
answered 48 mins ago
Picaud Vincent
53615
@Brahadeesh I did not realize I had so much typos in my answer, thanks for the corrections.
– Picaud Vincent
40 mins ago
1
No problem, you're welcome :)
– Brahadeesh
39 mins ago
add a comment |Â
up vote
2
down vote
up vote
2
down vote
If, according to your question, $f'$ is continuous on a bounded interval (i.e. a compact set), it makes no doubt it is Riemann-integrable (and hence also Lebesgue integrable):
A bounded function on a compact interval $[a, b]$ is Riemann integrable
if and only if it is continuous almost everywhere (the set of its
points of discontinuity has measure zero, in the sense of Lebesgue
measure).
See: Riemann integral
edited 43 mins ago
Brahadeesh
4,78931853
answered 48 mins ago
Picaud Vincent
53615
If, according to your question, $f'$ is continuous on a bounded interval (i.e. a compact set), it makes no doubt it is Riemann-integrable (and hence also Lebesgue integrable):
A bounded function on a compact interval $[a, b]$ is Riemann integrable
if and only if it is continuous almost everywhere (the set of its
points of discontinuity has measure zero, in the sense of Lebesgue
measure).
See: Riemann integral
edited 43 mins ago
Brahadeesh
4,78931853
answered 48 mins ago
Picaud Vincent
53615
edited 43 mins ago
Brahadeesh
4,78931853
edited 43 mins ago
Brahadeesh
4,78931853
edited 43 mins ago
Brahadeesh
4,78931853
4,78931853
answered 48 mins ago
Picaud Vincent
53615
answered 48 mins ago
Picaud Vincent
53615
answered 48 mins ago
Picaud Vincent
53615
53615
@Brahadeesh I did not realize I had so much typos in my answer, thanks for the corrections.
– Picaud Vincent
40 mins ago
1
No problem, you're welcome :)
– Brahadeesh
39 mins ago
add a comment |Â
@Brahadeesh I did not realize I had so much typos in my answer, thanks for the corrections.
– Picaud Vincent
40 mins ago
1
No problem, you're welcome :)
– Brahadeesh
39 mins ago
@Brahadeesh I did not realize I had so much typos in my answer, thanks for the corrections.
– Picaud Vincent
40 mins ago
@Brahadeesh I did not realize I had so much typos in my answer, thanks for the corrections.
– Picaud Vincent
40 mins ago
1
1
No problem, you're welcome :)
– Brahadeesh
39 mins ago
No problem, you're welcome :)
– Brahadeesh
39 mins ago
add a comment |Â
Kim is a new contributor. Be nice, and check out our Code of Conduct.
Kim is a new contributor. Be nice, and check out our Code of Conduct.
Kim is a new contributor. Be nice, and check out our Code of Conduct.
Kim is a new contributor. Be nice, and check out our Code of Conduct.
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