If one region is above the $x$-axis, and the other below it, should “the ratio of their areas” use...
$begingroup$
Good morning mathematicians,
I apologize for asking such a basic question. When you are to find the ratio of "Area 1" and "Area 2", such that "Area 1" is the area of a region above the x-axis and "Area 2" is the area of a region below the x-axis, are you looking for an absolute value of the ratio, or do you leave the value negative?
integration ratio
edited Jan 13 at 8:21
Blue
49.1k870156
asked Jan 13 at 8:15
Adam PáltikAdam Páltik
1149
$endgroup$
add a comment |
$begingroup$
Good morning mathematicians,
I apologize for asking such a basic question. When you are to find the ratio of "Area 1" and "Area 2", such that "Area 1" is the area of a region above the x-axis and "Area 2" is the area of a region below the x-axis, are you looking for an absolute value of the ratio, or do you leave the value negative?
integration ratio
edited Jan 13 at 8:21
Blue
49.1k870156
asked Jan 13 at 8:15
Adam PáltikAdam Páltik
1149
$endgroup$
2
$begingroup$
It depends on context. After all, treating below-the-$x$-axis regions as having negative area is a key element in how integration works, so "maybe" the ratio here should be considered negative if this is an integration exercise. On the other hand, if you simply happen to be using integration as a means of computing the areas of a couple of geometric figures, one of which just happens to be below the $x$-axis, then it's likely that "area" should be taken in absolute value. Authors sometimes avoid confusion by writing, say, "the ratio of the signed areas".
$endgroup$
– Blue
Jan 13 at 8:28
add a comment |
$begingroup$
Good morning mathematicians,
I apologize for asking such a basic question. When you are to find the ratio of "Area 1" and "Area 2", such that "Area 1" is the area of a region above the x-axis and "Area 2" is the area of a region below the x-axis, are you looking for an absolute value of the ratio, or do you leave the value negative?
integration ratio
edited Jan 13 at 8:21
Blue
49.1k870156
asked Jan 13 at 8:15
Adam PáltikAdam Páltik
1149
$endgroup$
Good morning mathematicians,
I apologize for asking such a basic question. When you are to find the ratio of "Area 1" and "Area 2", such that "Area 1" is the area of a region above the x-axis and "Area 2" is the area of a region below the x-axis, are you looking for an absolute value of the ratio, or do you leave the value negative?
integration ratio
integration ratio
edited Jan 13 at 8:21
Blue
49.1k870156
asked Jan 13 at 8:15
Adam PáltikAdam Páltik
1149
edited Jan 13 at 8:21
Blue
49.1k870156
asked Jan 13 at 8:15
Adam PáltikAdam Páltik
1149
edited Jan 13 at 8:21
Blue
49.1k870156
edited Jan 13 at 8:21
Blue
49.1k870156
edited Jan 13 at 8:21
Blue
49.1k870156
49.1k870156
asked Jan 13 at 8:15
Adam PáltikAdam Páltik
1149
asked Jan 13 at 8:15
Adam PáltikAdam Páltik
1149
asked Jan 13 at 8:15
Adam PáltikAdam Páltik
1149
1149
2
$begingroup$
It depends on context. After all, treating below-the-$x$-axis regions as having negative area is a key element in how integration works, so "maybe" the ratio here should be considered negative if this is an integration exercise. On the other hand, if you simply happen to be using integration as a means of computing the areas of a couple of geometric figures, one of which just happens to be below the $x$-axis, then it's likely that "area" should be taken in absolute value. Authors sometimes avoid confusion by writing, say, "the ratio of the signed areas".
$endgroup$
– Blue
Jan 13 at 8:28
add a comment |
2
$begingroup$
It depends on context. After all, treating below-the-$x$-axis regions as having negative area is a key element in how integration works, so "maybe" the ratio here should be considered negative if this is an integration exercise. On the other hand, if you simply happen to be using integration as a means of computing the areas of a couple of geometric figures, one of which just happens to be below the $x$-axis, then it's likely that "area" should be taken in absolute value. Authors sometimes avoid confusion by writing, say, "the ratio of the signed areas".
$endgroup$
– Blue
Jan 13 at 8:28
2
2
$begingroup$
It depends on context. After all, treating below-the-$x$-axis regions as having negative area is a key element in how integration works, so "maybe" the ratio here should be considered negative if this is an integration exercise. On the other hand, if you simply happen to be using integration as a means of computing the areas of a couple of geometric figures, one of which just happens to be below the $x$-axis, then it's likely that "area" should be taken in absolute value. Authors sometimes avoid confusion by writing, say, "the ratio of the signed areas".
$endgroup$
– Blue
Jan 13 at 8:28
$begingroup$
It depends on context. After all, treating below-the-$x$-axis regions as having negative area is a key element in how integration works, so "maybe" the ratio here should be considered negative if this is an integration exercise. On the other hand, if you simply happen to be using integration as a means of computing the areas of a couple of geometric figures, one of which just happens to be below the $x$-axis, then it's likely that "area" should be taken in absolute value. Authors sometimes avoid confusion by writing, say, "the ratio of the signed areas".
$endgroup$
– Blue
Jan 13 at 8:28
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Typically, areas are always non-negative numbers, so a ratio of areas should also be non-negative (i.e. if the result is negative, the absolute value is taken).
answered Jan 13 at 8:20
Parcly TaxelParcly Taxel
44.6k1376109
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3071797%2fif-one-region-is-above-the-x-axis-and-the-other-below-it-should-the-ratio-o%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Typically, areas are always non-negative numbers, so a ratio of areas should also be non-negative (i.e. if the result is negative, the absolute value is taken).
answered Jan 13 at 8:20
Parcly TaxelParcly Taxel
44.6k1376109
$endgroup$
add a comment |
$begingroup$
Typically, areas are always non-negative numbers, so a ratio of areas should also be non-negative (i.e. if the result is negative, the absolute value is taken).
answered Jan 13 at 8:20
Parcly TaxelParcly Taxel
44.6k1376109
$endgroup$
add a comment |
$begingroup$
Typically, areas are always non-negative numbers, so a ratio of areas should also be non-negative (i.e. if the result is negative, the absolute value is taken).
answered Jan 13 at 8:20
Parcly TaxelParcly Taxel
44.6k1376109
$endgroup$
Typically, areas are always non-negative numbers, so a ratio of areas should also be non-negative (i.e. if the result is negative, the absolute value is taken).
answered Jan 13 at 8:20
Parcly TaxelParcly Taxel
44.6k1376109
answered Jan 13 at 8:20
Parcly TaxelParcly Taxel
44.6k1376109
answered Jan 13 at 8:20
Parcly TaxelParcly Taxel
44.6k1376109
answered Jan 13 at 8:20
Parcly TaxelParcly Taxel
44.6k1376109
44.6k1376109
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3071797%2fif-one-region-is-above-the-x-axis-and-the-other-below-it-should-the-ratio-o%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
2
$begingroup$
It depends on context. After all, treating below-the-$x$-axis regions as having negative area is a key element in how integration works, so "maybe" the ratio here should be considered negative if this is an integration exercise. On the other hand, if you simply happen to be using integration as a means of computing the areas of a couple of geometric figures, one of which just happens to be below the $x$-axis, then it's likely that "area" should be taken in absolute value. Authors sometimes avoid confusion by writing, say, "the ratio of the signed areas".
$endgroup$
– Blue
Jan 13 at 8:28