Mixing
Integration of greatest integer function [closed]
0
$begingroup$
Let $[x]$ denote the greatest integer function, then evaluate
$$int_{-100}^{100}[x^3]dx$$
I have no idea how to go about it, can someone help me out
calculus integration definite-integrals
asked Jan 17 at 14:27
user601297user601297
37119
$endgroup$
closed as off-topic by RRL, Alexander Gruber♦ Jan 17 at 23:21
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, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
0
$begingroup$
Let $[x]$ denote the greatest integer function, then evaluate
$$int_{-100}^{100}[x^3]dx$$
I have no idea how to go about it, can someone help me out
calculus integration definite-integrals
asked Jan 17 at 14:27
user601297user601297
37119
$endgroup$
closed as off-topic by RRL, Alexander Gruber♦ Jan 17 at 23:21
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, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
0
0
0
$begingroup$
Let $[x]$ denote the greatest integer function, then evaluate
$$int_{-100}^{100}[x^3]dx$$
I have no idea how to go about it, can someone help me out
calculus integration definite-integrals
asked Jan 17 at 14:27
user601297user601297
37119
$endgroup$
Let $[x]$ denote the greatest integer function, then evaluate
$$int_{-100}^{100}[x^3]dx$$
I have no idea how to go about it, can someone help me out
calculus integration definite-integrals
calculus integration definite-integrals
asked Jan 17 at 14:27
user601297user601297
37119
asked Jan 17 at 14:27
user601297user601297
37119
asked Jan 17 at 14:27
user601297user601297
37119
asked Jan 17 at 14:27
user601297user601297
37119
asked Jan 17 at 14:27
user601297user601297
37119
37119
closed as off-topic by RRL, Alexander Gruber♦ Jan 17 at 23:21
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, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by RRL, Alexander Gruber♦ Jan 17 at 23:21
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, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
4
$begingroup$
Hint: If $ninmathbb Z$, then$$int_{sqrt[3]n}^{sqrt[3]{n+1}}lfloor x^3rfloor,mathrm dx=int_{sqrt[3]n}^{sqrt[3]{n+1}}n,mathrm dx=ntimesleft(sqrt[3]{n+1}-sqrt[3]nright).$$
answered Jan 17 at 14:32
José Carlos SantosJosé Carlos Santos
163k22131234
$endgroup$
$begingroup$
Are you saying that between -2 to -1 the function will just be -2 and so on???
$endgroup$
– user601297
Jan 17 at 17:47
$begingroup$
Well… yes, I am! (Well, not quite. It will be $-8$ there.)
$endgroup$
– José Carlos Santos
Jan 17 at 17:57
$begingroup$
I’m sorry I don’t get how it will be -8, can you elaborate please
$endgroup$
– user601297
Jan 17 at 17:58
$begingroup$
There was an error in my answer and I've edited it. What do you think now?
$endgroup$
– José Carlos Santos
Jan 17 at 18:04
$begingroup$
Yup definitely makes sense now, your previous answer i think would be correct if the power was 1.
$endgroup$
– user601297
Jan 17 at 18:05
|
show 3 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
4
$begingroup$
Hint: If $ninmathbb Z$, then$$int_{sqrt[3]n}^{sqrt[3]{n+1}}lfloor x^3rfloor,mathrm dx=int_{sqrt[3]n}^{sqrt[3]{n+1}}n,mathrm dx=ntimesleft(sqrt[3]{n+1}-sqrt[3]nright).$$
answered Jan 17 at 14:32
José Carlos SantosJosé Carlos Santos
163k22131234
$endgroup$
$begingroup$
Are you saying that between -2 to -1 the function will just be -2 and so on???
$endgroup$
– user601297
Jan 17 at 17:47
$begingroup$
Well… yes, I am! (Well, not quite. It will be $-8$ there.)
$endgroup$
– José Carlos Santos
Jan 17 at 17:57
$begingroup$
I’m sorry I don’t get how it will be -8, can you elaborate please
$endgroup$
– user601297
Jan 17 at 17:58
$begingroup$
There was an error in my answer and I've edited it. What do you think now?
$endgroup$
– José Carlos Santos
Jan 17 at 18:04
$begingroup$
Yup definitely makes sense now, your previous answer i think would be correct if the power was 1.
$endgroup$
– user601297
Jan 17 at 18:05
|
show 3 more comments
4
$begingroup$
Hint: If $ninmathbb Z$, then$$int_{sqrt[3]n}^{sqrt[3]{n+1}}lfloor x^3rfloor,mathrm dx=int_{sqrt[3]n}^{sqrt[3]{n+1}}n,mathrm dx=ntimesleft(sqrt[3]{n+1}-sqrt[3]nright).$$
answered Jan 17 at 14:32
José Carlos SantosJosé Carlos Santos
163k22131234
$endgroup$
$begingroup$
Are you saying that between -2 to -1 the function will just be -2 and so on???
$endgroup$
– user601297
Jan 17 at 17:47
$begingroup$
Well… yes, I am! (Well, not quite. It will be $-8$ there.)
$endgroup$
– José Carlos Santos
Jan 17 at 17:57
$begingroup$
I’m sorry I don’t get how it will be -8, can you elaborate please
$endgroup$
– user601297
Jan 17 at 17:58
$begingroup$
There was an error in my answer and I've edited it. What do you think now?
$endgroup$
– José Carlos Santos
Jan 17 at 18:04
$begingroup$
Yup definitely makes sense now, your previous answer i think would be correct if the power was 1.
$endgroup$
– user601297
Jan 17 at 18:05
|
show 3 more comments
4
4
4
$begingroup$
Hint: If $ninmathbb Z$, then$$int_{sqrt[3]n}^{sqrt[3]{n+1}}lfloor x^3rfloor,mathrm dx=int_{sqrt[3]n}^{sqrt[3]{n+1}}n,mathrm dx=ntimesleft(sqrt[3]{n+1}-sqrt[3]nright).$$
answered Jan 17 at 14:32
José Carlos SantosJosé Carlos Santos
163k22131234
$endgroup$
Hint: If $ninmathbb Z$, then$$int_{sqrt[3]n}^{sqrt[3]{n+1}}lfloor x^3rfloor,mathrm dx=int_{sqrt[3]n}^{sqrt[3]{n+1}}n,mathrm dx=ntimesleft(sqrt[3]{n+1}-sqrt[3]nright).$$
answered Jan 17 at 14:32
José Carlos SantosJosé Carlos Santos
163k22131234
edited Jan 17 at 18:03
answered Jan 17 at 14:32
José Carlos SantosJosé Carlos Santos
163k22131234
answered Jan 17 at 14:32
José Carlos SantosJosé Carlos Santos
163k22131234
answered Jan 17 at 14:32
José Carlos SantosJosé Carlos Santos
163k22131234
163k22131234
$begingroup$
Are you saying that between -2 to -1 the function will just be -2 and so on???
$endgroup$
– user601297
Jan 17 at 17:47
$begingroup$
Well… yes, I am! (Well, not quite. It will be $-8$ there.)
$endgroup$
– José Carlos Santos
Jan 17 at 17:57
$begingroup$
I’m sorry I don’t get how it will be -8, can you elaborate please
$endgroup$
– user601297
Jan 17 at 17:58
$begingroup$
There was an error in my answer and I've edited it. What do you think now?
$endgroup$
– José Carlos Santos
Jan 17 at 18:04
$begingroup$
Yup definitely makes sense now, your previous answer i think would be correct if the power was 1.
$endgroup$
– user601297
Jan 17 at 18:05
|
show 3 more comments
$begingroup$
Are you saying that between -2 to -1 the function will just be -2 and so on???
$endgroup$
– user601297
Jan 17 at 17:47
$begingroup$
Well… yes, I am! (Well, not quite. It will be $-8$ there.)
$endgroup$
– José Carlos Santos
Jan 17 at 17:57
$begingroup$
I’m sorry I don’t get how it will be -8, can you elaborate please
$endgroup$
– user601297
Jan 17 at 17:58
$begingroup$
There was an error in my answer and I've edited it. What do you think now?
$endgroup$
– José Carlos Santos
Jan 17 at 18:04
$begingroup$
Yup definitely makes sense now, your previous answer i think would be correct if the power was 1.
$endgroup$
– user601297
Jan 17 at 18:05
$begingroup$
Are you saying that between -2 to -1 the function will just be -2 and so on???
$endgroup$
– user601297
Jan 17 at 17:47
$begingroup$
Are you saying that between -2 to -1 the function will just be -2 and so on???
$endgroup$
– user601297
Jan 17 at 17:47
$begingroup$
Well… yes, I am! (Well, not quite. It will be $-8$ there.)
$endgroup$
– José Carlos Santos
Jan 17 at 17:57
$begingroup$
Well… yes, I am! (Well, not quite. It will be $-8$ there.)
$endgroup$
– José Carlos Santos
Jan 17 at 17:57
$begingroup$
I’m sorry I don’t get how it will be -8, can you elaborate please
$endgroup$
– user601297
Jan 17 at 17:58
$begingroup$
I’m sorry I don’t get how it will be -8, can you elaborate please
$endgroup$
– user601297
Jan 17 at 17:58
$begingroup$
There was an error in my answer and I've edited it. What do you think now?
$endgroup$
– José Carlos Santos
Jan 17 at 18:04
$begingroup$
There was an error in my answer and I've edited it. What do you think now?
$endgroup$
– José Carlos Santos
Jan 17 at 18:04
$begingroup$
Yup definitely makes sense now, your previous answer i think would be correct if the power was 1.
$endgroup$
– user601297
Jan 17 at 18:05
$begingroup$
Yup definitely makes sense now, your previous answer i think would be correct if the power was 1.
$endgroup$
– user601297
Jan 17 at 18:05
|
show 3 more comments
