Definite Integration of [arccos(x)] , [.] = GIF [closed]
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How to find integration of [ arccos(x) ] in its domain (-1,1)? Where [.] Is Floor function.
I tried using graph method(area) but end up getting some negative terms there.
integration definite-integrals
asked Dec 19 '18 at 17:09
user627932user627932
73
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closed as off-topic by RRL, Carl Schildkraut, Did, Cesareo, amWhy Dec 20 '18 at 13:01
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Carl Schildkraut, Did, Cesareo, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.
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$begingroup$
How to find integration of [ arccos(x) ] in its domain (-1,1)? Where [.] Is Floor function.
I tried using graph method(area) but end up getting some negative terms there.
integration definite-integrals
asked Dec 19 '18 at 17:09
user627932user627932
73
$endgroup$
closed as off-topic by RRL, Carl Schildkraut, Did, Cesareo, amWhy Dec 20 '18 at 13:01
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Carl Schildkraut, Did, Cesareo, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.
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In what domain?
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– caverac
Dec 19 '18 at 17:12
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Domain of arccos x. from -1 to 1
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– user627932
Dec 19 '18 at 17:13
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This is a step function, so you can just add up the rectangles.
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– eyeballfrog
Dec 19 '18 at 17:19
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Ans is Cos1 + Cos 2 + Cos 3 + 3. So its clear that this has to be solved using graph of arccos x.
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– user627932
Dec 19 '18 at 17:19
1
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@clathratus thanks for pointing it out. I have changed it.
$endgroup$
– user627932
Dec 20 '18 at 18:15
|
show 2 more comments
$begingroup$
How to find integration of [ arccos(x) ] in its domain (-1,1)? Where [.] Is Floor function.
I tried using graph method(area) but end up getting some negative terms there.
integration definite-integrals
asked Dec 19 '18 at 17:09
user627932user627932
73
$endgroup$
How to find integration of [ arccos(x) ] in its domain (-1,1)? Where [.] Is Floor function.
I tried using graph method(area) but end up getting some negative terms there.
integration definite-integrals
integration definite-integrals
asked Dec 19 '18 at 17:09
user627932user627932
73
asked Dec 19 '18 at 17:09
user627932user627932
73
edited Dec 20 '18 at 18:14
user627932
asked Dec 19 '18 at 17:09
user627932user627932
73
asked Dec 19 '18 at 17:09
user627932user627932
73
asked Dec 19 '18 at 17:09
user627932user627932
73
73
closed as off-topic by RRL, Carl Schildkraut, Did, Cesareo, amWhy Dec 20 '18 at 13:01
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Carl Schildkraut, Did, Cesareo, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by RRL, Carl Schildkraut, Did, Cesareo, amWhy Dec 20 '18 at 13:01
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Carl Schildkraut, Did, Cesareo, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.
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In what domain?
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– caverac
Dec 19 '18 at 17:12
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Domain of arccos x. from -1 to 1
$endgroup$
– user627932
Dec 19 '18 at 17:13
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This is a step function, so you can just add up the rectangles.
$endgroup$
– eyeballfrog
Dec 19 '18 at 17:19
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Ans is Cos1 + Cos 2 + Cos 3 + 3. So its clear that this has to be solved using graph of arccos x.
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– user627932
Dec 19 '18 at 17:19
1
$begingroup$
@clathratus thanks for pointing it out. I have changed it.
$endgroup$
– user627932
Dec 20 '18 at 18:15
|
show 2 more comments
$begingroup$
In what domain?
$endgroup$
– caverac
Dec 19 '18 at 17:12
$begingroup$
Domain of arccos x. from -1 to 1
$endgroup$
– user627932
Dec 19 '18 at 17:13
$begingroup$
This is a step function, so you can just add up the rectangles.
$endgroup$
– eyeballfrog
Dec 19 '18 at 17:19
$begingroup$
Ans is Cos1 + Cos 2 + Cos 3 + 3. So its clear that this has to be solved using graph of arccos x.
$endgroup$
– user627932
Dec 19 '18 at 17:19
1
$begingroup$
@clathratus thanks for pointing it out. I have changed it.
$endgroup$
– user627932
Dec 20 '18 at 18:15
$begingroup$
In what domain?
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– caverac
Dec 19 '18 at 17:12
$begingroup$
In what domain?
$endgroup$
– caverac
Dec 19 '18 at 17:12
$begingroup$
Domain of arccos x. from -1 to 1
$endgroup$
– user627932
Dec 19 '18 at 17:13
$begingroup$
Domain of arccos x. from -1 to 1
$endgroup$
– user627932
Dec 19 '18 at 17:13
$begingroup$
This is a step function, so you can just add up the rectangles.
$endgroup$
– eyeballfrog
Dec 19 '18 at 17:19
$begingroup$
This is a step function, so you can just add up the rectangles.
$endgroup$
– eyeballfrog
Dec 19 '18 at 17:19
$begingroup$
Ans is Cos1 + Cos 2 + Cos 3 + 3. So its clear that this has to be solved using graph of arccos x.
$endgroup$
– user627932
Dec 19 '18 at 17:19
$begingroup$
Ans is Cos1 + Cos 2 + Cos 3 + 3. So its clear that this has to be solved using graph of arccos x.
$endgroup$
– user627932
Dec 19 '18 at 17:19
1
1
$begingroup$
@clathratus thanks for pointing it out. I have changed it.
$endgroup$
– user627932
Dec 20 '18 at 18:15
$begingroup$
@clathratus thanks for pointing it out. I have changed it.
$endgroup$
– user627932
Dec 20 '18 at 18:15
|
show 2 more comments
2 Answers
2
active
oldest
votes
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$cos^{-1}x$ is a monotonically decreasing function, such that:
$3lecos^{-1}xlepi$ for $-1le xlecos3$
$2lecos^{-1}xle3$ for $cos3le xlecos2$
$1lecos^{-1}xle2$ for $cos2le xlecos1$
$0lecos^{-1}xle1$ for $cos1le xle 1$
$I=displaystyleint_{-1}^1lfloorcos^{-1}xrfloor dx\=displaystyleint_{-1}^{cos3}3dx+int_{cos3}^{cos2}2dx+int_{cos2}^{cos1}1dx+int_{cos1}^10dx\=cos1+cos2+cos3+3$
answered Dec 19 '18 at 17:25
Shubham JohriShubham Johri
5,192717
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add a comment |
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Maybe a small sketch will help you
It is just a series of steps, so the integral is simple. The trick is to find the transitions, for example, to find the left-most consider $3 = arccos(x)$, whose solution is $x = cos(3)$, so in the region $-1 leq x leq cos(3)$ the function is just three, and the area is
$$
int_{-1}^{cos(3)} lfloor arccos(x) rfloor {rm d}x = 3 [cos(3) + 1]
$$
You can continue with the other steps, the result is
$$
int_{-1}^{1} lfloor arccos(x) rfloor {rm d}x = cos 1 + cos 2 + cos 3 + 3
$$
answered Dec 19 '18 at 17:27
caveraccaverac
14.6k31130
$endgroup$
$begingroup$
Thanks i just found what i was doing wrong
$endgroup$
– user627932
Dec 19 '18 at 17:29
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$cos^{-1}x$ is a monotonically decreasing function, such that:
$3lecos^{-1}xlepi$ for $-1le xlecos3$
$2lecos^{-1}xle3$ for $cos3le xlecos2$
$1lecos^{-1}xle2$ for $cos2le xlecos1$
$0lecos^{-1}xle1$ for $cos1le xle 1$
$I=displaystyleint_{-1}^1lfloorcos^{-1}xrfloor dx\=displaystyleint_{-1}^{cos3}3dx+int_{cos3}^{cos2}2dx+int_{cos2}^{cos1}1dx+int_{cos1}^10dx\=cos1+cos2+cos3+3$
answered Dec 19 '18 at 17:25
Shubham JohriShubham Johri
5,192717
$endgroup$
add a comment |
$begingroup$
$cos^{-1}x$ is a monotonically decreasing function, such that:
$3lecos^{-1}xlepi$ for $-1le xlecos3$
$2lecos^{-1}xle3$ for $cos3le xlecos2$
$1lecos^{-1}xle2$ for $cos2le xlecos1$
$0lecos^{-1}xle1$ for $cos1le xle 1$
$I=displaystyleint_{-1}^1lfloorcos^{-1}xrfloor dx\=displaystyleint_{-1}^{cos3}3dx+int_{cos3}^{cos2}2dx+int_{cos2}^{cos1}1dx+int_{cos1}^10dx\=cos1+cos2+cos3+3$
answered Dec 19 '18 at 17:25
Shubham JohriShubham Johri
5,192717
$endgroup$
add a comment |
$begingroup$
$cos^{-1}x$ is a monotonically decreasing function, such that:
$3lecos^{-1}xlepi$ for $-1le xlecos3$
$2lecos^{-1}xle3$ for $cos3le xlecos2$
$1lecos^{-1}xle2$ for $cos2le xlecos1$
$0lecos^{-1}xle1$ for $cos1le xle 1$
$I=displaystyleint_{-1}^1lfloorcos^{-1}xrfloor dx\=displaystyleint_{-1}^{cos3}3dx+int_{cos3}^{cos2}2dx+int_{cos2}^{cos1}1dx+int_{cos1}^10dx\=cos1+cos2+cos3+3$
answered Dec 19 '18 at 17:25
Shubham JohriShubham Johri
5,192717
$endgroup$
$cos^{-1}x$ is a monotonically decreasing function, such that:
$3lecos^{-1}xlepi$ for $-1le xlecos3$
$2lecos^{-1}xle3$ for $cos3le xlecos2$
$1lecos^{-1}xle2$ for $cos2le xlecos1$
$0lecos^{-1}xle1$ for $cos1le xle 1$
$I=displaystyleint_{-1}^1lfloorcos^{-1}xrfloor dx\=displaystyleint_{-1}^{cos3}3dx+int_{cos3}^{cos2}2dx+int_{cos2}^{cos1}1dx+int_{cos1}^10dx\=cos1+cos2+cos3+3$
answered Dec 19 '18 at 17:25
Shubham JohriShubham Johri
5,192717
answered Dec 19 '18 at 17:25
Shubham JohriShubham Johri
5,192717
answered Dec 19 '18 at 17:25
Shubham JohriShubham Johri
5,192717
answered Dec 19 '18 at 17:25
Shubham JohriShubham Johri
5,192717
5,192717
add a comment |
add a comment |
$begingroup$
Maybe a small sketch will help you
It is just a series of steps, so the integral is simple. The trick is to find the transitions, for example, to find the left-most consider $3 = arccos(x)$, whose solution is $x = cos(3)$, so in the region $-1 leq x leq cos(3)$ the function is just three, and the area is
$$
int_{-1}^{cos(3)} lfloor arccos(x) rfloor {rm d}x = 3 [cos(3) + 1]
$$
You can continue with the other steps, the result is
$$
int_{-1}^{1} lfloor arccos(x) rfloor {rm d}x = cos 1 + cos 2 + cos 3 + 3
$$
answered Dec 19 '18 at 17:27
caveraccaverac
14.6k31130
$endgroup$
$begingroup$
Thanks i just found what i was doing wrong
$endgroup$
– user627932
Dec 19 '18 at 17:29
add a comment |
$begingroup$
Maybe a small sketch will help you
It is just a series of steps, so the integral is simple. The trick is to find the transitions, for example, to find the left-most consider $3 = arccos(x)$, whose solution is $x = cos(3)$, so in the region $-1 leq x leq cos(3)$ the function is just three, and the area is
$$
int_{-1}^{cos(3)} lfloor arccos(x) rfloor {rm d}x = 3 [cos(3) + 1]
$$
You can continue with the other steps, the result is
$$
int_{-1}^{1} lfloor arccos(x) rfloor {rm d}x = cos 1 + cos 2 + cos 3 + 3
$$
answered Dec 19 '18 at 17:27
caveraccaverac
14.6k31130
$endgroup$
$begingroup$
Thanks i just found what i was doing wrong
$endgroup$
– user627932
Dec 19 '18 at 17:29
add a comment |
$begingroup$
Maybe a small sketch will help you
It is just a series of steps, so the integral is simple. The trick is to find the transitions, for example, to find the left-most consider $3 = arccos(x)$, whose solution is $x = cos(3)$, so in the region $-1 leq x leq cos(3)$ the function is just three, and the area is
$$
int_{-1}^{cos(3)} lfloor arccos(x) rfloor {rm d}x = 3 [cos(3) + 1]
$$
You can continue with the other steps, the result is
$$
int_{-1}^{1} lfloor arccos(x) rfloor {rm d}x = cos 1 + cos 2 + cos 3 + 3
$$
answered Dec 19 '18 at 17:27
caveraccaverac
14.6k31130
$endgroup$
Maybe a small sketch will help you
It is just a series of steps, so the integral is simple. The trick is to find the transitions, for example, to find the left-most consider $3 = arccos(x)$, whose solution is $x = cos(3)$, so in the region $-1 leq x leq cos(3)$ the function is just three, and the area is
$$
int_{-1}^{cos(3)} lfloor arccos(x) rfloor {rm d}x = 3 [cos(3) + 1]
$$
You can continue with the other steps, the result is
$$
int_{-1}^{1} lfloor arccos(x) rfloor {rm d}x = cos 1 + cos 2 + cos 3 + 3
$$
answered Dec 19 '18 at 17:27
caveraccaverac
14.6k31130
answered Dec 19 '18 at 17:27
caveraccaverac
14.6k31130
answered Dec 19 '18 at 17:27
caveraccaverac
14.6k31130
answered Dec 19 '18 at 17:27
caveraccaverac
14.6k31130
14.6k31130
$begingroup$
Thanks i just found what i was doing wrong
$endgroup$
– user627932
Dec 19 '18 at 17:29
add a comment |
$begingroup$
Thanks i just found what i was doing wrong
$endgroup$
– user627932
Dec 19 '18 at 17:29
$begingroup$
Thanks i just found what i was doing wrong
$endgroup$
– user627932
Dec 19 '18 at 17:29
$begingroup$
Thanks i just found what i was doing wrong
$endgroup$
– user627932
Dec 19 '18 at 17:29
add a comment |
$begingroup$
In what domain?
$endgroup$
– caverac
Dec 19 '18 at 17:12
$begingroup$
Domain of arccos x. from -1 to 1
$endgroup$
– user627932
Dec 19 '18 at 17:13
$begingroup$
This is a step function, so you can just add up the rectangles.
$endgroup$
– eyeballfrog
Dec 19 '18 at 17:19
$begingroup$
Ans is Cos1 + Cos 2 + Cos 3 + 3. So its clear that this has to be solved using graph of arccos x.
$endgroup$
– user627932
Dec 19 '18 at 17:19
1
$begingroup$
@clathratus thanks for pointing it out. I have changed it.
$endgroup$
– user627932
Dec 20 '18 at 18:15