If f is continuous on [a,b] and $f(x) geq 0$, for $x in [a,b]$, but $f$ is not the zero function, prove that...
$begingroup$
This question already has an answer here:
Is it true that $int_{a}^b h^2=0implies h=0$?
2 answers
If f is continuous on [a,b] and $f(x) geq 0$, for $x in [a,b]$, but $f$ is not the zero function, prove that $int_{a}^{b} f(x) dx > 0$
Could anyone give me a hint for this proof please?
real-analysis calculus integration riemann-integration
asked Dec 4 '18 at 8:57
hopefullyhopefully
192113
$endgroup$
marked as duplicate by Arthur, egreg calculus
Users with the calculus badge can single-handedly close calculus questions as duplicates and reopen them as needed.
StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
Dec 4 '18 at 9:00
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
Is it true that $int_{a}^b h^2=0implies h=0$?
2 answers
If f is continuous on [a,b] and $f(x) geq 0$, for $x in [a,b]$, but $f$ is not the zero function, prove that $int_{a}^{b} f(x) dx > 0$
Could anyone give me a hint for this proof please?
real-analysis calculus integration riemann-integration
asked Dec 4 '18 at 8:57
hopefullyhopefully
192113
$endgroup$
marked as duplicate by Arthur, egreg calculus
Users with the calculus badge can single-handedly close calculus questions as duplicates and reopen them as needed.
StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
Dec 4 '18 at 9:00
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
Is it true that $int_{a}^b h^2=0implies h=0$?
2 answers
If f is continuous on [a,b] and $f(x) geq 0$, for $x in [a,b]$, but $f$ is not the zero function, prove that $int_{a}^{b} f(x) dx > 0$
Could anyone give me a hint for this proof please?
real-analysis calculus integration riemann-integration
asked Dec 4 '18 at 8:57
hopefullyhopefully
192113
$endgroup$
This question already has an answer here:
Is it true that $int_{a}^b h^2=0implies h=0$?
2 answers
If f is continuous on [a,b] and $f(x) geq 0$, for $x in [a,b]$, but $f$ is not the zero function, prove that $int_{a}^{b} f(x) dx > 0$
Could anyone give me a hint for this proof please?
This question already has an answer here:
Is it true that $int_{a}^b h^2=0implies h=0$?
2 answers
real-analysis calculus integration riemann-integration
real-analysis calculus integration riemann-integration
asked Dec 4 '18 at 8:57
hopefullyhopefully
192113
asked Dec 4 '18 at 8:57
hopefullyhopefully
192113
asked Dec 4 '18 at 8:57
hopefullyhopefully
192113
asked Dec 4 '18 at 8:57
hopefullyhopefully
192113
asked Dec 4 '18 at 8:57
hopefullyhopefully
192113
192113
marked as duplicate by Arthur, egreg calculus
Users with the calculus badge can single-handedly close calculus questions as duplicates and reopen them as needed.
StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
Dec 4 '18 at 9:00
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Arthur, egreg calculus
Users with the calculus badge can single-handedly close calculus questions as duplicates and reopen them as needed.
StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
Dec 4 '18 at 9:00
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
If $f(c) >0$ then there exists $r>0$ such that $f(x) geq frac {f(c)} 2$ for $|x-c| leq r$. Hence $int_a^{b} f geq int_{c-r}^{c+r} f(x) dx geq frac {f(c)} 2 (2r) >0$.
answered Dec 4 '18 at 8:59
Kavi Rama MurthyKavi Rama Murthy
54.4k32055
$endgroup$
$begingroup$
could you explain by words please?
$endgroup$
– hopefully
Dec 4 '18 at 10:25
$begingroup$
I do not understand why $f(x) geq (f(c)/2)$
$endgroup$
– hopefully
Dec 4 '18 at 10:27
$begingroup$
There exists $r$ such that $|f(x)-f(c) |<frac {f(c)} 2$ for $|x-c| <r$. [ This is because $f$ is continuous at $c$]. Now $f(c) = f(x) +(f(c)-f(x)) < f(x)+ frac {f(c)} 2$ which gives $f(x) > f(c) -frac {f(c)} 2 =frac {f(c)} 2$.
$endgroup$
– Kavi Rama Murthy
Dec 4 '18 at 10:30
$begingroup$
but why you take $epsilon $ by this value specifically?
$endgroup$
– hopefully
Dec 4 '18 at 10:31
$begingroup$
@hopefully Any $epsilon$ smaller than $f(c)$ will work fine.
$endgroup$
– Kavi Rama Murthy
Dec 4 '18 at 10:32
|
show 1 more comment
$begingroup$
f is not the zero function, and f ≥ 0, so f >0 somewhere. Since f is continuous, f>0 in a tiny interval, then the integral is positive in this interval.
answered Dec 4 '18 at 9:03
Spade.KSpade.K
111
$endgroup$
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If $f(c) >0$ then there exists $r>0$ such that $f(x) geq frac {f(c)} 2$ for $|x-c| leq r$. Hence $int_a^{b} f geq int_{c-r}^{c+r} f(x) dx geq frac {f(c)} 2 (2r) >0$.
answered Dec 4 '18 at 8:59
Kavi Rama MurthyKavi Rama Murthy
54.4k32055
$endgroup$
$begingroup$
could you explain by words please?
$endgroup$
– hopefully
Dec 4 '18 at 10:25
$begingroup$
I do not understand why $f(x) geq (f(c)/2)$
$endgroup$
– hopefully
Dec 4 '18 at 10:27
$begingroup$
There exists $r$ such that $|f(x)-f(c) |<frac {f(c)} 2$ for $|x-c| <r$. [ This is because $f$ is continuous at $c$]. Now $f(c) = f(x) +(f(c)-f(x)) < f(x)+ frac {f(c)} 2$ which gives $f(x) > f(c) -frac {f(c)} 2 =frac {f(c)} 2$.
$endgroup$
– Kavi Rama Murthy
Dec 4 '18 at 10:30
$begingroup$
but why you take $epsilon $ by this value specifically?
$endgroup$
– hopefully
Dec 4 '18 at 10:31
$begingroup$
@hopefully Any $epsilon$ smaller than $f(c)$ will work fine.
$endgroup$
– Kavi Rama Murthy
Dec 4 '18 at 10:32
|
show 1 more comment
$begingroup$
If $f(c) >0$ then there exists $r>0$ such that $f(x) geq frac {f(c)} 2$ for $|x-c| leq r$. Hence $int_a^{b} f geq int_{c-r}^{c+r} f(x) dx geq frac {f(c)} 2 (2r) >0$.
answered Dec 4 '18 at 8:59
Kavi Rama MurthyKavi Rama Murthy
54.4k32055
$endgroup$
$begingroup$
could you explain by words please?
$endgroup$
– hopefully
Dec 4 '18 at 10:25
$begingroup$
I do not understand why $f(x) geq (f(c)/2)$
$endgroup$
– hopefully
Dec 4 '18 at 10:27
$begingroup$
There exists $r$ such that $|f(x)-f(c) |<frac {f(c)} 2$ for $|x-c| <r$. [ This is because $f$ is continuous at $c$]. Now $f(c) = f(x) +(f(c)-f(x)) < f(x)+ frac {f(c)} 2$ which gives $f(x) > f(c) -frac {f(c)} 2 =frac {f(c)} 2$.
$endgroup$
– Kavi Rama Murthy
Dec 4 '18 at 10:30
$begingroup$
but why you take $epsilon $ by this value specifically?
$endgroup$
– hopefully
Dec 4 '18 at 10:31
$begingroup$
@hopefully Any $epsilon$ smaller than $f(c)$ will work fine.
$endgroup$
– Kavi Rama Murthy
Dec 4 '18 at 10:32
|
show 1 more comment
$begingroup$
If $f(c) >0$ then there exists $r>0$ such that $f(x) geq frac {f(c)} 2$ for $|x-c| leq r$. Hence $int_a^{b} f geq int_{c-r}^{c+r} f(x) dx geq frac {f(c)} 2 (2r) >0$.
answered Dec 4 '18 at 8:59
Kavi Rama MurthyKavi Rama Murthy
54.4k32055
$endgroup$
If $f(c) >0$ then there exists $r>0$ such that $f(x) geq frac {f(c)} 2$ for $|x-c| leq r$. Hence $int_a^{b} f geq int_{c-r}^{c+r} f(x) dx geq frac {f(c)} 2 (2r) >0$.
answered Dec 4 '18 at 8:59
Kavi Rama MurthyKavi Rama Murthy
54.4k32055
answered Dec 4 '18 at 8:59
Kavi Rama MurthyKavi Rama Murthy
54.4k32055
answered Dec 4 '18 at 8:59
Kavi Rama MurthyKavi Rama Murthy
54.4k32055
answered Dec 4 '18 at 8:59
Kavi Rama MurthyKavi Rama Murthy
54.4k32055
54.4k32055
$begingroup$
could you explain by words please?
$endgroup$
– hopefully
Dec 4 '18 at 10:25
$begingroup$
I do not understand why $f(x) geq (f(c)/2)$
$endgroup$
– hopefully
Dec 4 '18 at 10:27
$begingroup$
There exists $r$ such that $|f(x)-f(c) |<frac {f(c)} 2$ for $|x-c| <r$. [ This is because $f$ is continuous at $c$]. Now $f(c) = f(x) +(f(c)-f(x)) < f(x)+ frac {f(c)} 2$ which gives $f(x) > f(c) -frac {f(c)} 2 =frac {f(c)} 2$.
$endgroup$
– Kavi Rama Murthy
Dec 4 '18 at 10:30
$begingroup$
but why you take $epsilon $ by this value specifically?
$endgroup$
– hopefully
Dec 4 '18 at 10:31
$begingroup$
@hopefully Any $epsilon$ smaller than $f(c)$ will work fine.
$endgroup$
– Kavi Rama Murthy
Dec 4 '18 at 10:32
|
show 1 more comment
$begingroup$
could you explain by words please?
$endgroup$
– hopefully
Dec 4 '18 at 10:25
$begingroup$
I do not understand why $f(x) geq (f(c)/2)$
$endgroup$
– hopefully
Dec 4 '18 at 10:27
$begingroup$
There exists $r$ such that $|f(x)-f(c) |<frac {f(c)} 2$ for $|x-c| <r$. [ This is because $f$ is continuous at $c$]. Now $f(c) = f(x) +(f(c)-f(x)) < f(x)+ frac {f(c)} 2$ which gives $f(x) > f(c) -frac {f(c)} 2 =frac {f(c)} 2$.
$endgroup$
– Kavi Rama Murthy
Dec 4 '18 at 10:30
$begingroup$
but why you take $epsilon $ by this value specifically?
$endgroup$
– hopefully
Dec 4 '18 at 10:31
$begingroup$
@hopefully Any $epsilon$ smaller than $f(c)$ will work fine.
$endgroup$
– Kavi Rama Murthy
Dec 4 '18 at 10:32
$begingroup$
could you explain by words please?
$endgroup$
– hopefully
Dec 4 '18 at 10:25
$begingroup$
could you explain by words please?
$endgroup$
– hopefully
Dec 4 '18 at 10:25
$begingroup$
I do not understand why $f(x) geq (f(c)/2)$
$endgroup$
– hopefully
Dec 4 '18 at 10:27
$begingroup$
I do not understand why $f(x) geq (f(c)/2)$
$endgroup$
– hopefully
Dec 4 '18 at 10:27
$begingroup$
There exists $r$ such that $|f(x)-f(c) |<frac {f(c)} 2$ for $|x-c| <r$. [ This is because $f$ is continuous at $c$]. Now $f(c) = f(x) +(f(c)-f(x)) < f(x)+ frac {f(c)} 2$ which gives $f(x) > f(c) -frac {f(c)} 2 =frac {f(c)} 2$.
$endgroup$
– Kavi Rama Murthy
Dec 4 '18 at 10:30
$begingroup$
There exists $r$ such that $|f(x)-f(c) |<frac {f(c)} 2$ for $|x-c| <r$. [ This is because $f$ is continuous at $c$]. Now $f(c) = f(x) +(f(c)-f(x)) < f(x)+ frac {f(c)} 2$ which gives $f(x) > f(c) -frac {f(c)} 2 =frac {f(c)} 2$.
$endgroup$
– Kavi Rama Murthy
Dec 4 '18 at 10:30
$begingroup$
but why you take $epsilon $ by this value specifically?
$endgroup$
– hopefully
Dec 4 '18 at 10:31
$begingroup$
but why you take $epsilon $ by this value specifically?
$endgroup$
– hopefully
Dec 4 '18 at 10:31
$begingroup$
@hopefully Any $epsilon$ smaller than $f(c)$ will work fine.
$endgroup$
– Kavi Rama Murthy
Dec 4 '18 at 10:32
$begingroup$
@hopefully Any $epsilon$ smaller than $f(c)$ will work fine.
$endgroup$
– Kavi Rama Murthy
Dec 4 '18 at 10:32
|
show 1 more comment
$begingroup$
f is not the zero function, and f ≥ 0, so f >0 somewhere. Since f is continuous, f>0 in a tiny interval, then the integral is positive in this interval.
answered Dec 4 '18 at 9:03
Spade.KSpade.K
111
$endgroup$
add a comment |
$begingroup$
f is not the zero function, and f ≥ 0, so f >0 somewhere. Since f is continuous, f>0 in a tiny interval, then the integral is positive in this interval.
answered Dec 4 '18 at 9:03
Spade.KSpade.K
111
$endgroup$
add a comment |
$begingroup$
f is not the zero function, and f ≥ 0, so f >0 somewhere. Since f is continuous, f>0 in a tiny interval, then the integral is positive in this interval.
answered Dec 4 '18 at 9:03
Spade.KSpade.K
111
$endgroup$
f is not the zero function, and f ≥ 0, so f >0 somewhere. Since f is continuous, f>0 in a tiny interval, then the integral is positive in this interval.
answered Dec 4 '18 at 9:03
Spade.KSpade.K
111
answered Dec 4 '18 at 9:03
Spade.KSpade.K
111
answered Dec 4 '18 at 9:03
Spade.KSpade.K
111
answered Dec 4 '18 at 9:03
Spade.KSpade.K
111
111
add a comment |
add a comment |
