Integral absolute value proof
$begingroup$
Can anyone prove this? I can understand this intuitively, but can't prove it mathematically. Please help me.
$$
left|int_a^b f(x),dx,right| le int_a^b lvert f(x)rvert ,dx
$$
integration absolute-value
edited Dec 8 '18 at 12:03
egreg
181k1485203
asked Dec 8 '18 at 12:00
Sarah Shin Sarah Shin
11
$endgroup$
add a comment |
$begingroup$
Can anyone prove this? I can understand this intuitively, but can't prove it mathematically. Please help me.
$$
left|int_a^b f(x),dx,right| le int_a^b lvert f(x)rvert ,dx
$$
integration absolute-value
edited Dec 8 '18 at 12:03
egreg
181k1485203
asked Dec 8 '18 at 12:00
Sarah Shin Sarah Shin
11
$endgroup$
6
$begingroup$
Hint: $-|f(x)|le f(x)le |f(x)|$. But, please, add your thoughts about the problem. This is a main fact about integrals and you should have it in your textbook.
$endgroup$
– egreg
Dec 8 '18 at 12:04
2
$begingroup$
Is the claim even true? Consider $f(x)=1$, $a=1$, $b=0$.
$endgroup$
– Hagen von Eitzen
Dec 8 '18 at 12:06
$begingroup$
@ egreg Thank you. Actually, my textbook put the same hint but I can't understand how to use it... Whenever I solve questions about this(calculating area) I always regard RHS same or bigger than LHS true as RHS means raising the graph to 1 or 2 quadrants and calculating the area.
$endgroup$
– Sarah Shin
Dec 8 '18 at 12:13
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Hi Sarah, see the application of Egreg's hint ;)
$endgroup$
– Wesley Strik
Dec 8 '18 at 13:33
add a comment |
$begingroup$
Can anyone prove this? I can understand this intuitively, but can't prove it mathematically. Please help me.
$$
left|int_a^b f(x),dx,right| le int_a^b lvert f(x)rvert ,dx
$$
integration absolute-value
edited Dec 8 '18 at 12:03
egreg
181k1485203
asked Dec 8 '18 at 12:00
Sarah Shin Sarah Shin
11
$endgroup$
Can anyone prove this? I can understand this intuitively, but can't prove it mathematically. Please help me.
$$
left|int_a^b f(x),dx,right| le int_a^b lvert f(x)rvert ,dx
$$
integration absolute-value
integration absolute-value
edited Dec 8 '18 at 12:03
egreg
181k1485203
asked Dec 8 '18 at 12:00
Sarah Shin Sarah Shin
11
edited Dec 8 '18 at 12:03
egreg
181k1485203
asked Dec 8 '18 at 12:00
Sarah Shin Sarah Shin
11
edited Dec 8 '18 at 12:03
egreg
181k1485203
edited Dec 8 '18 at 12:03
egreg
181k1485203
edited Dec 8 '18 at 12:03
egreg
181k1485203
181k1485203
asked Dec 8 '18 at 12:00
Sarah Shin Sarah Shin
11
asked Dec 8 '18 at 12:00
Sarah Shin Sarah Shin
11
asked Dec 8 '18 at 12:00
Sarah Shin Sarah Shin
11
11
6
$begingroup$
Hint: $-|f(x)|le f(x)le |f(x)|$. But, please, add your thoughts about the problem. This is a main fact about integrals and you should have it in your textbook.
$endgroup$
– egreg
Dec 8 '18 at 12:04
2
$begingroup$
Is the claim even true? Consider $f(x)=1$, $a=1$, $b=0$.
$endgroup$
– Hagen von Eitzen
Dec 8 '18 at 12:06
$begingroup$
@ egreg Thank you. Actually, my textbook put the same hint but I can't understand how to use it... Whenever I solve questions about this(calculating area) I always regard RHS same or bigger than LHS true as RHS means raising the graph to 1 or 2 quadrants and calculating the area.
$endgroup$
– Sarah Shin
Dec 8 '18 at 12:13
$begingroup$
Hi Sarah, see the application of Egreg's hint ;)
$endgroup$
– Wesley Strik
Dec 8 '18 at 13:33
add a comment |
6
$begingroup$
Hint: $-|f(x)|le f(x)le |f(x)|$. But, please, add your thoughts about the problem. This is a main fact about integrals and you should have it in your textbook.
$endgroup$
– egreg
Dec 8 '18 at 12:04
2
$begingroup$
Is the claim even true? Consider $f(x)=1$, $a=1$, $b=0$.
$endgroup$
– Hagen von Eitzen
Dec 8 '18 at 12:06
$begingroup$
@ egreg Thank you. Actually, my textbook put the same hint but I can't understand how to use it... Whenever I solve questions about this(calculating area) I always regard RHS same or bigger than LHS true as RHS means raising the graph to 1 or 2 quadrants and calculating the area.
$endgroup$
– Sarah Shin
Dec 8 '18 at 12:13
$begingroup$
Hi Sarah, see the application of Egreg's hint ;)
$endgroup$
– Wesley Strik
Dec 8 '18 at 13:33
6
6
$begingroup$
Hint: $-|f(x)|le f(x)le |f(x)|$. But, please, add your thoughts about the problem. This is a main fact about integrals and you should have it in your textbook.
$endgroup$
– egreg
Dec 8 '18 at 12:04
$begingroup$
Hint: $-|f(x)|le f(x)le |f(x)|$. But, please, add your thoughts about the problem. This is a main fact about integrals and you should have it in your textbook.
$endgroup$
– egreg
Dec 8 '18 at 12:04
2
2
$begingroup$
Is the claim even true? Consider $f(x)=1$, $a=1$, $b=0$.
$endgroup$
– Hagen von Eitzen
Dec 8 '18 at 12:06
$begingroup$
Is the claim even true? Consider $f(x)=1$, $a=1$, $b=0$.
$endgroup$
– Hagen von Eitzen
Dec 8 '18 at 12:06
$begingroup$
@ egreg Thank you. Actually, my textbook put the same hint but I can't understand how to use it... Whenever I solve questions about this(calculating area) I always regard RHS same or bigger than LHS true as RHS means raising the graph to 1 or 2 quadrants and calculating the area.
$endgroup$
– Sarah Shin
Dec 8 '18 at 12:13
$begingroup$
@ egreg Thank you. Actually, my textbook put the same hint but I can't understand how to use it... Whenever I solve questions about this(calculating area) I always regard RHS same or bigger than LHS true as RHS means raising the graph to 1 or 2 quadrants and calculating the area.
$endgroup$
– Sarah Shin
Dec 8 '18 at 12:13
$begingroup$
Hi Sarah, see the application of Egreg's hint ;)
$endgroup$
– Wesley Strik
Dec 8 '18 at 13:33
$begingroup$
Hi Sarah, see the application of Egreg's hint ;)
$endgroup$
– Wesley Strik
Dec 8 '18 at 13:33
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Hint: a) If $f$ is a step function, then the inequality follows very easily. b) If $f$ is a regulated function (or a continuous function), use the fact that you can approximate $f$ w.r.t. $| cdot |_infty$ by a sequence ${f_n}$ of step functions. Now use part a).
answered Dec 8 '18 at 12:11
NiklasNiklas
2,203720
$endgroup$
add a comment |
$begingroup$
Using the hint from the comments sections we can provide a nice proof.
If we assume $f$ is a Riemann-integrable function, this inequality is true, we will provide a proof.
Proof
By properties of absolute value, we get
$$-|f(x)|le f(x)le |f(x)|.$$
Since $f$ is continuous, we know that $|f|$ is also continuous, hence $|f|$ is Riemann integrable. We can integrate each side of this inequality and we get
$$-int_a^b |f(x)|dxle int_a^bf(x)dxle int_a^b|f(x)|dx$$
This instantly gives us that:
$$left| int_a^b f(x)dxright|le int_a^b|f(x)|dx. $$ $square$
Alternatively, you are considering a continuous version of the triangle inequality. The very definition of an integral is the limit of discrete sums of (Riemann) intervals. To properly prove this from the definition we must go back to the definition of integration:
For any Riemann sum we get from the usual triangle inequality for the absolute value:
$$left|sum_{k=1}^nf(c_i)(x_i-x_{i-1})right|leqsum_{k=1}^n|f(c_i)|(x_i-x_{i-1}),,,,$$
$${a=x_0<x_1<...<x_n=b},,,,,c_iin[x_{i-1},x_1]$$
Pass now to the limit $,ntoinfty,$ while the maximal length of the subintervals goes to zero (this is what is done to get the Riemann integral from Riemann sums). With courtesy of: The triangle inequality for integrals
answered Dec 8 '18 at 12:07
Wesley StrikWesley Strik
1,761423
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add a comment |
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2 Answers
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active
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2 Answers
2
active
oldest
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oldest
votes
$begingroup$
Hint: a) If $f$ is a step function, then the inequality follows very easily. b) If $f$ is a regulated function (or a continuous function), use the fact that you can approximate $f$ w.r.t. $| cdot |_infty$ by a sequence ${f_n}$ of step functions. Now use part a).
answered Dec 8 '18 at 12:11
NiklasNiklas
2,203720
$endgroup$
add a comment |
$begingroup$
Hint: a) If $f$ is a step function, then the inequality follows very easily. b) If $f$ is a regulated function (or a continuous function), use the fact that you can approximate $f$ w.r.t. $| cdot |_infty$ by a sequence ${f_n}$ of step functions. Now use part a).
answered Dec 8 '18 at 12:11
NiklasNiklas
2,203720
$endgroup$
add a comment |
$begingroup$
Hint: a) If $f$ is a step function, then the inequality follows very easily. b) If $f$ is a regulated function (or a continuous function), use the fact that you can approximate $f$ w.r.t. $| cdot |_infty$ by a sequence ${f_n}$ of step functions. Now use part a).
answered Dec 8 '18 at 12:11
NiklasNiklas
2,203720
$endgroup$
Hint: a) If $f$ is a step function, then the inequality follows very easily. b) If $f$ is a regulated function (or a continuous function), use the fact that you can approximate $f$ w.r.t. $| cdot |_infty$ by a sequence ${f_n}$ of step functions. Now use part a).
answered Dec 8 '18 at 12:11
NiklasNiklas
2,203720
answered Dec 8 '18 at 12:11
NiklasNiklas
2,203720
answered Dec 8 '18 at 12:11
NiklasNiklas
2,203720
answered Dec 8 '18 at 12:11
NiklasNiklas
2,203720
2,203720
add a comment |
add a comment |
$begingroup$
Using the hint from the comments sections we can provide a nice proof.
If we assume $f$ is a Riemann-integrable function, this inequality is true, we will provide a proof.
Proof
By properties of absolute value, we get
$$-|f(x)|le f(x)le |f(x)|.$$
Since $f$ is continuous, we know that $|f|$ is also continuous, hence $|f|$ is Riemann integrable. We can integrate each side of this inequality and we get
$$-int_a^b |f(x)|dxle int_a^bf(x)dxle int_a^b|f(x)|dx$$
This instantly gives us that:
$$left| int_a^b f(x)dxright|le int_a^b|f(x)|dx. $$ $square$
Alternatively, you are considering a continuous version of the triangle inequality. The very definition of an integral is the limit of discrete sums of (Riemann) intervals. To properly prove this from the definition we must go back to the definition of integration:
For any Riemann sum we get from the usual triangle inequality for the absolute value:
$$left|sum_{k=1}^nf(c_i)(x_i-x_{i-1})right|leqsum_{k=1}^n|f(c_i)|(x_i-x_{i-1}),,,,$$
$${a=x_0<x_1<...<x_n=b},,,,,c_iin[x_{i-1},x_1]$$
Pass now to the limit $,ntoinfty,$ while the maximal length of the subintervals goes to zero (this is what is done to get the Riemann integral from Riemann sums). With courtesy of: The triangle inequality for integrals
answered Dec 8 '18 at 12:07
Wesley StrikWesley Strik
1,761423
$endgroup$
add a comment |
$begingroup$
Using the hint from the comments sections we can provide a nice proof.
If we assume $f$ is a Riemann-integrable function, this inequality is true, we will provide a proof.
Proof
By properties of absolute value, we get
$$-|f(x)|le f(x)le |f(x)|.$$
Since $f$ is continuous, we know that $|f|$ is also continuous, hence $|f|$ is Riemann integrable. We can integrate each side of this inequality and we get
$$-int_a^b |f(x)|dxle int_a^bf(x)dxle int_a^b|f(x)|dx$$
This instantly gives us that:
$$left| int_a^b f(x)dxright|le int_a^b|f(x)|dx. $$ $square$
Alternatively, you are considering a continuous version of the triangle inequality. The very definition of an integral is the limit of discrete sums of (Riemann) intervals. To properly prove this from the definition we must go back to the definition of integration:
For any Riemann sum we get from the usual triangle inequality for the absolute value:
$$left|sum_{k=1}^nf(c_i)(x_i-x_{i-1})right|leqsum_{k=1}^n|f(c_i)|(x_i-x_{i-1}),,,,$$
$${a=x_0<x_1<...<x_n=b},,,,,c_iin[x_{i-1},x_1]$$
Pass now to the limit $,ntoinfty,$ while the maximal length of the subintervals goes to zero (this is what is done to get the Riemann integral from Riemann sums). With courtesy of: The triangle inequality for integrals
answered Dec 8 '18 at 12:07
Wesley StrikWesley Strik
1,761423
$endgroup$
add a comment |
$begingroup$
Using the hint from the comments sections we can provide a nice proof.
If we assume $f$ is a Riemann-integrable function, this inequality is true, we will provide a proof.
Proof
By properties of absolute value, we get
$$-|f(x)|le f(x)le |f(x)|.$$
Since $f$ is continuous, we know that $|f|$ is also continuous, hence $|f|$ is Riemann integrable. We can integrate each side of this inequality and we get
$$-int_a^b |f(x)|dxle int_a^bf(x)dxle int_a^b|f(x)|dx$$
This instantly gives us that:
$$left| int_a^b f(x)dxright|le int_a^b|f(x)|dx. $$ $square$
Alternatively, you are considering a continuous version of the triangle inequality. The very definition of an integral is the limit of discrete sums of (Riemann) intervals. To properly prove this from the definition we must go back to the definition of integration:
For any Riemann sum we get from the usual triangle inequality for the absolute value:
$$left|sum_{k=1}^nf(c_i)(x_i-x_{i-1})right|leqsum_{k=1}^n|f(c_i)|(x_i-x_{i-1}),,,,$$
$${a=x_0<x_1<...<x_n=b},,,,,c_iin[x_{i-1},x_1]$$
Pass now to the limit $,ntoinfty,$ while the maximal length of the subintervals goes to zero (this is what is done to get the Riemann integral from Riemann sums). With courtesy of: The triangle inequality for integrals
answered Dec 8 '18 at 12:07
Wesley StrikWesley Strik
1,761423
$endgroup$
Using the hint from the comments sections we can provide a nice proof.
If we assume $f$ is a Riemann-integrable function, this inequality is true, we will provide a proof.
Proof
By properties of absolute value, we get
$$-|f(x)|le f(x)le |f(x)|.$$
Since $f$ is continuous, we know that $|f|$ is also continuous, hence $|f|$ is Riemann integrable. We can integrate each side of this inequality and we get
$$-int_a^b |f(x)|dxle int_a^bf(x)dxle int_a^b|f(x)|dx$$
This instantly gives us that:
$$left| int_a^b f(x)dxright|le int_a^b|f(x)|dx. $$ $square$
Alternatively, you are considering a continuous version of the triangle inequality. The very definition of an integral is the limit of discrete sums of (Riemann) intervals. To properly prove this from the definition we must go back to the definition of integration:
For any Riemann sum we get from the usual triangle inequality for the absolute value:
$$left|sum_{k=1}^nf(c_i)(x_i-x_{i-1})right|leqsum_{k=1}^n|f(c_i)|(x_i-x_{i-1}),,,,$$
$${a=x_0<x_1<...<x_n=b},,,,,c_iin[x_{i-1},x_1]$$
Pass now to the limit $,ntoinfty,$ while the maximal length of the subintervals goes to zero (this is what is done to get the Riemann integral from Riemann sums). With courtesy of: The triangle inequality for integrals
answered Dec 8 '18 at 12:07
Wesley StrikWesley Strik
1,761423
edited Dec 8 '18 at 13:49
answered Dec 8 '18 at 12:07
Wesley StrikWesley Strik
1,761423
answered Dec 8 '18 at 12:07
Wesley StrikWesley Strik
1,761423
answered Dec 8 '18 at 12:07
Wesley StrikWesley Strik
1,761423
1,761423
add a comment |
add a comment |
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$begingroup$
Hint: $-|f(x)|le f(x)le |f(x)|$. But, please, add your thoughts about the problem. This is a main fact about integrals and you should have it in your textbook.
$endgroup$
– egreg
Dec 8 '18 at 12:04
2
$begingroup$
Is the claim even true? Consider $f(x)=1$, $a=1$, $b=0$.
$endgroup$
– Hagen von Eitzen
Dec 8 '18 at 12:06
$begingroup$
@ egreg Thank you. Actually, my textbook put the same hint but I can't understand how to use it... Whenever I solve questions about this(calculating area) I always regard RHS same or bigger than LHS true as RHS means raising the graph to 1 or 2 quadrants and calculating the area.
$endgroup$
– Sarah Shin
Dec 8 '18 at 12:13
$begingroup$
Hi Sarah, see the application of Egreg's hint ;)
$endgroup$
– Wesley Strik
Dec 8 '18 at 13:33