How do i show following equality [closed]
$begingroup$
Let $f(x)$ be monotone on $[a,b]$.Show that there exists $c∈[a,b]$ such that
$$int_a^b f(x)dx = f(a)(c-a) + f(b)(b-c) $$
I want to show it from basics and by using first intermediate value theorem for integrals without directly using second intermediate value theorem for integrals .
real-analysis riemann-integration
asked Dec 2 '18 at 8:23
BhowmickBhowmick
1438
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closed as off-topic by Did, Cesareo, José Carlos Santos, Lord_Farin, user10354138 Dec 3 '18 at 14:48
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, Cesareo, José Carlos Santos, Lord_Farin, user10354138
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Let $f(x)$ be monotone on $[a,b]$.Show that there exists $c∈[a,b]$ such that
$$int_a^b f(x)dx = f(a)(c-a) + f(b)(b-c) $$
I want to show it from basics and by using first intermediate value theorem for integrals without directly using second intermediate value theorem for integrals .
real-analysis riemann-integration
asked Dec 2 '18 at 8:23
BhowmickBhowmick
1438
$endgroup$
closed as off-topic by Did, Cesareo, José Carlos Santos, Lord_Farin, user10354138 Dec 3 '18 at 14:48
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, Cesareo, José Carlos Santos, Lord_Farin, user10354138
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
I feel the $delta (uv) = udelta v + vdelta u$ is somehow involved here.
$endgroup$
– Nick
Dec 2 '18 at 8:36
1
$begingroup$
Is your $f$ continuous as well?
$endgroup$
– Anurag A
Dec 2 '18 at 8:39
$begingroup$
@AnuragA I think we can make two cases one in which $f$ is continuous and other in which it is not .Proving for continuous case might give us hint for non-continuous case .
$endgroup$
– Bhowmick
Dec 2 '18 at 8:46
add a comment |
$begingroup$
Let $f(x)$ be monotone on $[a,b]$.Show that there exists $c∈[a,b]$ such that
$$int_a^b f(x)dx = f(a)(c-a) + f(b)(b-c) $$
I want to show it from basics and by using first intermediate value theorem for integrals without directly using second intermediate value theorem for integrals .
real-analysis riemann-integration
asked Dec 2 '18 at 8:23
BhowmickBhowmick
1438
$endgroup$
Let $f(x)$ be monotone on $[a,b]$.Show that there exists $c∈[a,b]$ such that
$$int_a^b f(x)dx = f(a)(c-a) + f(b)(b-c) $$
I want to show it from basics and by using first intermediate value theorem for integrals without directly using second intermediate value theorem for integrals .
real-analysis riemann-integration
real-analysis riemann-integration
asked Dec 2 '18 at 8:23
BhowmickBhowmick
1438
asked Dec 2 '18 at 8:23
BhowmickBhowmick
1438
asked Dec 2 '18 at 8:23
BhowmickBhowmick
1438
asked Dec 2 '18 at 8:23
BhowmickBhowmick
1438
asked Dec 2 '18 at 8:23
BhowmickBhowmick
1438
1438
closed as off-topic by Did, Cesareo, José Carlos Santos, Lord_Farin, user10354138 Dec 3 '18 at 14:48
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, Cesareo, José Carlos Santos, Lord_Farin, user10354138
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Did, Cesareo, José Carlos Santos, Lord_Farin, user10354138 Dec 3 '18 at 14:48
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, Cesareo, José Carlos Santos, Lord_Farin, user10354138
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
I feel the $delta (uv) = udelta v + vdelta u$ is somehow involved here.
$endgroup$
– Nick
Dec 2 '18 at 8:36
1
$begingroup$
Is your $f$ continuous as well?
$endgroup$
– Anurag A
Dec 2 '18 at 8:39
$begingroup$
@AnuragA I think we can make two cases one in which $f$ is continuous and other in which it is not .Proving for continuous case might give us hint for non-continuous case .
$endgroup$
– Bhowmick
Dec 2 '18 at 8:46
add a comment |
1
$begingroup$
I feel the $delta (uv) = udelta v + vdelta u$ is somehow involved here.
$endgroup$
– Nick
Dec 2 '18 at 8:36
1
$begingroup$
Is your $f$ continuous as well?
$endgroup$
– Anurag A
Dec 2 '18 at 8:39
$begingroup$
@AnuragA I think we can make two cases one in which $f$ is continuous and other in which it is not .Proving for continuous case might give us hint for non-continuous case .
$endgroup$
– Bhowmick
Dec 2 '18 at 8:46
1
1
$begingroup$
I feel the $delta (uv) = udelta v + vdelta u$ is somehow involved here.
$endgroup$
– Nick
Dec 2 '18 at 8:36
$begingroup$
I feel the $delta (uv) = udelta v + vdelta u$ is somehow involved here.
$endgroup$
– Nick
Dec 2 '18 at 8:36
1
1
$begingroup$
Is your $f$ continuous as well?
$endgroup$
– Anurag A
Dec 2 '18 at 8:39
$begingroup$
Is your $f$ continuous as well?
$endgroup$
– Anurag A
Dec 2 '18 at 8:39
$begingroup$
@AnuragA I think we can make two cases one in which $f$ is continuous and other in which it is not .Proving for continuous case might give us hint for non-continuous case .
$endgroup$
– Bhowmick
Dec 2 '18 at 8:46
$begingroup$
@AnuragA I think we can make two cases one in which $f$ is continuous and other in which it is not .Proving for continuous case might give us hint for non-continuous case .
$endgroup$
– Bhowmick
Dec 2 '18 at 8:46
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $g(x)=f(a)(x-a)+f(b)(b-x)$ and $f(x)$ is an increasing function, $aleq xleq b$. If we have
$$
g(a)=f(b)(b-a)>int_a^bf(x)mathrm{d}x>f(a)(b-a)=g(b)
$$
Since $g(x)$ is continous, there exists $cin(a,b)$ s.t.
$$
g(c)=int_a^bf(x)mathrm{d}x.
$$
Else, WLOG, if
$$
int_a^bf(x)mathrm{d}x=f(a)(b-a)
$$
we can still find a point $cin (a,b)$ because there are infinitely many points s.t. $f(x)=f(a)$.
answered Dec 2 '18 at 9:22
yahooyahoo
606412
$endgroup$
$begingroup$
We can solve like this only when $f(x)$ is continuous
$endgroup$
– Bhowmick
Dec 2 '18 at 9:38
1
$begingroup$
In general, you can show that $(b-a) f(a) leq int_a^b f(x) dx leq (b-a) f(b)$ by using the monotonicity of $f$ and the definition of the integral. Then picking $c$ is essentially just a matter of making the weighted average work.
$endgroup$
– platty
Dec 2 '18 at 9:53
$begingroup$
@platty Thanks.you are right .
$endgroup$
– Bhowmick
Dec 2 '18 at 10:03
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $g(x)=f(a)(x-a)+f(b)(b-x)$ and $f(x)$ is an increasing function, $aleq xleq b$. If we have
$$
g(a)=f(b)(b-a)>int_a^bf(x)mathrm{d}x>f(a)(b-a)=g(b)
$$
Since $g(x)$ is continous, there exists $cin(a,b)$ s.t.
$$
g(c)=int_a^bf(x)mathrm{d}x.
$$
Else, WLOG, if
$$
int_a^bf(x)mathrm{d}x=f(a)(b-a)
$$
we can still find a point $cin (a,b)$ because there are infinitely many points s.t. $f(x)=f(a)$.
answered Dec 2 '18 at 9:22
yahooyahoo
606412
$endgroup$
$begingroup$
We can solve like this only when $f(x)$ is continuous
$endgroup$
– Bhowmick
Dec 2 '18 at 9:38
1
$begingroup$
In general, you can show that $(b-a) f(a) leq int_a^b f(x) dx leq (b-a) f(b)$ by using the monotonicity of $f$ and the definition of the integral. Then picking $c$ is essentially just a matter of making the weighted average work.
$endgroup$
– platty
Dec 2 '18 at 9:53
$begingroup$
@platty Thanks.you are right .
$endgroup$
– Bhowmick
Dec 2 '18 at 10:03
add a comment |
$begingroup$
Let $g(x)=f(a)(x-a)+f(b)(b-x)$ and $f(x)$ is an increasing function, $aleq xleq b$. If we have
$$
g(a)=f(b)(b-a)>int_a^bf(x)mathrm{d}x>f(a)(b-a)=g(b)
$$
Since $g(x)$ is continous, there exists $cin(a,b)$ s.t.
$$
g(c)=int_a^bf(x)mathrm{d}x.
$$
Else, WLOG, if
$$
int_a^bf(x)mathrm{d}x=f(a)(b-a)
$$
we can still find a point $cin (a,b)$ because there are infinitely many points s.t. $f(x)=f(a)$.
answered Dec 2 '18 at 9:22
yahooyahoo
606412
$endgroup$
$begingroup$
We can solve like this only when $f(x)$ is continuous
$endgroup$
– Bhowmick
Dec 2 '18 at 9:38
1
$begingroup$
In general, you can show that $(b-a) f(a) leq int_a^b f(x) dx leq (b-a) f(b)$ by using the monotonicity of $f$ and the definition of the integral. Then picking $c$ is essentially just a matter of making the weighted average work.
$endgroup$
– platty
Dec 2 '18 at 9:53
$begingroup$
@platty Thanks.you are right .
$endgroup$
– Bhowmick
Dec 2 '18 at 10:03
add a comment |
$begingroup$
Let $g(x)=f(a)(x-a)+f(b)(b-x)$ and $f(x)$ is an increasing function, $aleq xleq b$. If we have
$$
g(a)=f(b)(b-a)>int_a^bf(x)mathrm{d}x>f(a)(b-a)=g(b)
$$
Since $g(x)$ is continous, there exists $cin(a,b)$ s.t.
$$
g(c)=int_a^bf(x)mathrm{d}x.
$$
Else, WLOG, if
$$
int_a^bf(x)mathrm{d}x=f(a)(b-a)
$$
we can still find a point $cin (a,b)$ because there are infinitely many points s.t. $f(x)=f(a)$.
answered Dec 2 '18 at 9:22
yahooyahoo
606412
$endgroup$
Let $g(x)=f(a)(x-a)+f(b)(b-x)$ and $f(x)$ is an increasing function, $aleq xleq b$. If we have
$$
g(a)=f(b)(b-a)>int_a^bf(x)mathrm{d}x>f(a)(b-a)=g(b)
$$
Since $g(x)$ is continous, there exists $cin(a,b)$ s.t.
$$
g(c)=int_a^bf(x)mathrm{d}x.
$$
Else, WLOG, if
$$
int_a^bf(x)mathrm{d}x=f(a)(b-a)
$$
we can still find a point $cin (a,b)$ because there are infinitely many points s.t. $f(x)=f(a)$.
answered Dec 2 '18 at 9:22
yahooyahoo
606412
edited Dec 2 '18 at 9:59
answered Dec 2 '18 at 9:22
yahooyahoo
606412
answered Dec 2 '18 at 9:22
yahooyahoo
606412
answered Dec 2 '18 at 9:22
yahooyahoo
606412
606412
$begingroup$
We can solve like this only when $f(x)$ is continuous
$endgroup$
– Bhowmick
Dec 2 '18 at 9:38
1
$begingroup$
In general, you can show that $(b-a) f(a) leq int_a^b f(x) dx leq (b-a) f(b)$ by using the monotonicity of $f$ and the definition of the integral. Then picking $c$ is essentially just a matter of making the weighted average work.
$endgroup$
– platty
Dec 2 '18 at 9:53
$begingroup$
@platty Thanks.you are right .
$endgroup$
– Bhowmick
Dec 2 '18 at 10:03
add a comment |
$begingroup$
We can solve like this only when $f(x)$ is continuous
$endgroup$
– Bhowmick
Dec 2 '18 at 9:38
1
$begingroup$
In general, you can show that $(b-a) f(a) leq int_a^b f(x) dx leq (b-a) f(b)$ by using the monotonicity of $f$ and the definition of the integral. Then picking $c$ is essentially just a matter of making the weighted average work.
$endgroup$
– platty
Dec 2 '18 at 9:53
$begingroup$
@platty Thanks.you are right .
$endgroup$
– Bhowmick
Dec 2 '18 at 10:03
$begingroup$
We can solve like this only when $f(x)$ is continuous
$endgroup$
– Bhowmick
Dec 2 '18 at 9:38
$begingroup$
We can solve like this only when $f(x)$ is continuous
$endgroup$
– Bhowmick
Dec 2 '18 at 9:38
1
1
$begingroup$
In general, you can show that $(b-a) f(a) leq int_a^b f(x) dx leq (b-a) f(b)$ by using the monotonicity of $f$ and the definition of the integral. Then picking $c$ is essentially just a matter of making the weighted average work.
$endgroup$
– platty
Dec 2 '18 at 9:53
$begingroup$
In general, you can show that $(b-a) f(a) leq int_a^b f(x) dx leq (b-a) f(b)$ by using the monotonicity of $f$ and the definition of the integral. Then picking $c$ is essentially just a matter of making the weighted average work.
$endgroup$
– platty
Dec 2 '18 at 9:53
$begingroup$
@platty Thanks.you are right .
$endgroup$
– Bhowmick
Dec 2 '18 at 10:03
$begingroup$
@platty Thanks.you are right .
$endgroup$
– Bhowmick
Dec 2 '18 at 10:03
add a comment |
1
$begingroup$
I feel the $delta (uv) = udelta v + vdelta u$ is somehow involved here.
$endgroup$
– Nick
Dec 2 '18 at 8:36
1
$begingroup$
Is your $f$ continuous as well?
$endgroup$
– Anurag A
Dec 2 '18 at 8:39
$begingroup$
@AnuragA I think we can make two cases one in which $f$ is continuous and other in which it is not .Proving for continuous case might give us hint for non-continuous case .
$endgroup$
– Bhowmick
Dec 2 '18 at 8:46