let $f:[a,b]rightarrowBbb{R}$ be such that $f(x)=0$ except at finitely many points. Show the integral of...
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This question already has an answer here:
Riemann integration of a function with finite number of non zero points.
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I'm working on the following problem:
Let $f:[a,b]rightarrowBbb{R}$ be such that $f(x)=0$ except at finitely many points. Show that $int_a^b{f(x)}dx=0$.
Am I pointed in the wrong direction for wanting to use contradiction? If I assume $$int_a^b{f(x)}dxneq0$$ then can I assume $f$ must contain a variable x such that $f$ is not a constant? Is it that simple? I'm missing a lot of connections.
real-analysis calculus
asked Dec 15 '18 at 3:50
Jess SavoieJess Savoie
617
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marked as duplicate by copper.hat real-analysis
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Dec 15 '18 at 5:09
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.
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This question already has an answer here:
Riemann integration of a function with finite number of non zero points.
1 answer
I'm working on the following problem:
Let $f:[a,b]rightarrowBbb{R}$ be such that $f(x)=0$ except at finitely many points. Show that $int_a^b{f(x)}dx=0$.
Am I pointed in the wrong direction for wanting to use contradiction? If I assume $$int_a^b{f(x)}dxneq0$$ then can I assume $f$ must contain a variable x such that $f$ is not a constant? Is it that simple? I'm missing a lot of connections.
real-analysis calculus
asked Dec 15 '18 at 3:50
Jess SavoieJess Savoie
617
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marked as duplicate by copper.hat real-analysis
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Dec 15 '18 at 5:09
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.
1
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What does "$f$ must contain a variable $x$ such that $f$ is not a constant" mean?
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– anomaly
Dec 15 '18 at 4:02
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Are you studying Riemann integration or Lebesgue integration ? If it is Lebesgue you have the "almost everywhere" (a.e.) that allows you to modify a function on a set of measure 0, in particular a finite set without modifying the integral.
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– Jean Marie
Dec 15 '18 at 4:07
1
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Using the riemann sum definition, partition the interval and shrink the partitions around the finitely many points containing $0$. Then, as you take the limit, the contribution of these points will go to 0.
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– rubikscube09
Dec 15 '18 at 4:13
add a comment |
$begingroup$
This question already has an answer here:
Riemann integration of a function with finite number of non zero points.
1 answer
I'm working on the following problem:
Let $f:[a,b]rightarrowBbb{R}$ be such that $f(x)=0$ except at finitely many points. Show that $int_a^b{f(x)}dx=0$.
Am I pointed in the wrong direction for wanting to use contradiction? If I assume $$int_a^b{f(x)}dxneq0$$ then can I assume $f$ must contain a variable x such that $f$ is not a constant? Is it that simple? I'm missing a lot of connections.
real-analysis calculus
asked Dec 15 '18 at 3:50
Jess SavoieJess Savoie
617
$endgroup$
This question already has an answer here:
Riemann integration of a function with finite number of non zero points.
1 answer
I'm working on the following problem:
Let $f:[a,b]rightarrowBbb{R}$ be such that $f(x)=0$ except at finitely many points. Show that $int_a^b{f(x)}dx=0$.
Am I pointed in the wrong direction for wanting to use contradiction? If I assume $$int_a^b{f(x)}dxneq0$$ then can I assume $f$ must contain a variable x such that $f$ is not a constant? Is it that simple? I'm missing a lot of connections.
This question already has an answer here:
Riemann integration of a function with finite number of non zero points.
1 answer
real-analysis calculus
real-analysis calculus
asked Dec 15 '18 at 3:50
Jess SavoieJess Savoie
617
asked Dec 15 '18 at 3:50
Jess SavoieJess Savoie
617
asked Dec 15 '18 at 3:50
Jess SavoieJess Savoie
617
asked Dec 15 '18 at 3:50
Jess SavoieJess Savoie
617
asked Dec 15 '18 at 3:50
Jess SavoieJess Savoie
617
617
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Dec 15 '18 at 5:09
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Dec 15 '18 at 5:09
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.
1
$begingroup$
What does "$f$ must contain a variable $x$ such that $f$ is not a constant" mean?
$endgroup$
– anomaly
Dec 15 '18 at 4:02
$begingroup$
Are you studying Riemann integration or Lebesgue integration ? If it is Lebesgue you have the "almost everywhere" (a.e.) that allows you to modify a function on a set of measure 0, in particular a finite set without modifying the integral.
$endgroup$
– Jean Marie
Dec 15 '18 at 4:07
1
$begingroup$
Using the riemann sum definition, partition the interval and shrink the partitions around the finitely many points containing $0$. Then, as you take the limit, the contribution of these points will go to 0.
$endgroup$
– rubikscube09
Dec 15 '18 at 4:13
add a comment |
1
$begingroup$
What does "$f$ must contain a variable $x$ such that $f$ is not a constant" mean?
$endgroup$
– anomaly
Dec 15 '18 at 4:02
$begingroup$
Are you studying Riemann integration or Lebesgue integration ? If it is Lebesgue you have the "almost everywhere" (a.e.) that allows you to modify a function on a set of measure 0, in particular a finite set without modifying the integral.
$endgroup$
– Jean Marie
Dec 15 '18 at 4:07
1
$begingroup$
Using the riemann sum definition, partition the interval and shrink the partitions around the finitely many points containing $0$. Then, as you take the limit, the contribution of these points will go to 0.
$endgroup$
– rubikscube09
Dec 15 '18 at 4:13
1
1
$begingroup$
What does "$f$ must contain a variable $x$ such that $f$ is not a constant" mean?
$endgroup$
– anomaly
Dec 15 '18 at 4:02
$begingroup$
What does "$f$ must contain a variable $x$ such that $f$ is not a constant" mean?
$endgroup$
– anomaly
Dec 15 '18 at 4:02
$begingroup$
Are you studying Riemann integration or Lebesgue integration ? If it is Lebesgue you have the "almost everywhere" (a.e.) that allows you to modify a function on a set of measure 0, in particular a finite set without modifying the integral.
$endgroup$
– Jean Marie
Dec 15 '18 at 4:07
$begingroup$
Are you studying Riemann integration or Lebesgue integration ? If it is Lebesgue you have the "almost everywhere" (a.e.) that allows you to modify a function on a set of measure 0, in particular a finite set without modifying the integral.
$endgroup$
– Jean Marie
Dec 15 '18 at 4:07
1
1
$begingroup$
Using the riemann sum definition, partition the interval and shrink the partitions around the finitely many points containing $0$. Then, as you take the limit, the contribution of these points will go to 0.
$endgroup$
– rubikscube09
Dec 15 '18 at 4:13
$begingroup$
Using the riemann sum definition, partition the interval and shrink the partitions around the finitely many points containing $0$. Then, as you take the limit, the contribution of these points will go to 0.
$endgroup$
– rubikscube09
Dec 15 '18 at 4:13
add a comment |
1 Answer
1
active
oldest
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Take a partition $P$ of $[a,b]$ and calculate the upper and lower Riemann sum with respect to this partition:
$$
begin{align*}
L_P(f)=&sum_{I in P} inf_I(f) |I|\
U_P(f)=&sum_{I in P} sup_I(f) |I|
end{align*}
$$
Then $f$ is Riemann integrable if $forall epsilon>0$ $;exists P_0$ such that $|U_P(f)-L_P(f)|<epsilon$ for all refinements $P$ of $P_0$.
Now let $M = 2max{|f(x)|}$. Then $U_P(f)-L_P(f) leq M left|bigcup_{{I|sup_I(|f|)>0}} right|rightarrow 0$ as the total length of any finite number of intervals in $P$ goes to 0 as you keep refining the partition of $[a,b]$.
answered Dec 15 '18 at 4:53
MatthiasMatthias
2287
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add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Take a partition $P$ of $[a,b]$ and calculate the upper and lower Riemann sum with respect to this partition:
$$
begin{align*}
L_P(f)=&sum_{I in P} inf_I(f) |I|\
U_P(f)=&sum_{I in P} sup_I(f) |I|
end{align*}
$$
Then $f$ is Riemann integrable if $forall epsilon>0$ $;exists P_0$ such that $|U_P(f)-L_P(f)|<epsilon$ for all refinements $P$ of $P_0$.
Now let $M = 2max{|f(x)|}$. Then $U_P(f)-L_P(f) leq M left|bigcup_{{I|sup_I(|f|)>0}} right|rightarrow 0$ as the total length of any finite number of intervals in $P$ goes to 0 as you keep refining the partition of $[a,b]$.
answered Dec 15 '18 at 4:53
MatthiasMatthias
2287
$endgroup$
add a comment |
$begingroup$
Take a partition $P$ of $[a,b]$ and calculate the upper and lower Riemann sum with respect to this partition:
$$
begin{align*}
L_P(f)=&sum_{I in P} inf_I(f) |I|\
U_P(f)=&sum_{I in P} sup_I(f) |I|
end{align*}
$$
Then $f$ is Riemann integrable if $forall epsilon>0$ $;exists P_0$ such that $|U_P(f)-L_P(f)|<epsilon$ for all refinements $P$ of $P_0$.
Now let $M = 2max{|f(x)|}$. Then $U_P(f)-L_P(f) leq M left|bigcup_{{I|sup_I(|f|)>0}} right|rightarrow 0$ as the total length of any finite number of intervals in $P$ goes to 0 as you keep refining the partition of $[a,b]$.
answered Dec 15 '18 at 4:53
MatthiasMatthias
2287
$endgroup$
add a comment |
$begingroup$
Take a partition $P$ of $[a,b]$ and calculate the upper and lower Riemann sum with respect to this partition:
$$
begin{align*}
L_P(f)=&sum_{I in P} inf_I(f) |I|\
U_P(f)=&sum_{I in P} sup_I(f) |I|
end{align*}
$$
Then $f$ is Riemann integrable if $forall epsilon>0$ $;exists P_0$ such that $|U_P(f)-L_P(f)|<epsilon$ for all refinements $P$ of $P_0$.
Now let $M = 2max{|f(x)|}$. Then $U_P(f)-L_P(f) leq M left|bigcup_{{I|sup_I(|f|)>0}} right|rightarrow 0$ as the total length of any finite number of intervals in $P$ goes to 0 as you keep refining the partition of $[a,b]$.
answered Dec 15 '18 at 4:53
MatthiasMatthias
2287
$endgroup$
Take a partition $P$ of $[a,b]$ and calculate the upper and lower Riemann sum with respect to this partition:
$$
begin{align*}
L_P(f)=&sum_{I in P} inf_I(f) |I|\
U_P(f)=&sum_{I in P} sup_I(f) |I|
end{align*}
$$
Then $f$ is Riemann integrable if $forall epsilon>0$ $;exists P_0$ such that $|U_P(f)-L_P(f)|<epsilon$ for all refinements $P$ of $P_0$.
Now let $M = 2max{|f(x)|}$. Then $U_P(f)-L_P(f) leq M left|bigcup_{{I|sup_I(|f|)>0}} right|rightarrow 0$ as the total length of any finite number of intervals in $P$ goes to 0 as you keep refining the partition of $[a,b]$.
answered Dec 15 '18 at 4:53
MatthiasMatthias
2287
answered Dec 15 '18 at 4:53
MatthiasMatthias
2287
answered Dec 15 '18 at 4:53
MatthiasMatthias
2287
answered Dec 15 '18 at 4:53
MatthiasMatthias
2287
2287
add a comment |
add a comment |
1
$begingroup$
What does "$f$ must contain a variable $x$ such that $f$ is not a constant" mean?
$endgroup$
– anomaly
Dec 15 '18 at 4:02
$begingroup$
Are you studying Riemann integration or Lebesgue integration ? If it is Lebesgue you have the "almost everywhere" (a.e.) that allows you to modify a function on a set of measure 0, in particular a finite set without modifying the integral.
$endgroup$
– Jean Marie
Dec 15 '18 at 4:07
1
$begingroup$
Using the riemann sum definition, partition the interval and shrink the partitions around the finitely many points containing $0$. Then, as you take the limit, the contribution of these points will go to 0.
$endgroup$
– rubikscube09
Dec 15 '18 at 4:13