Finding the area between two curves with Integrate The 2019 Stack Overflow Developer Survey Results Are InHow to evaluate this indefinite integral $csc(4x)sin(x)$Finding the centroid of the area between two curvesRevolving the area between two functions about an axisArea enclosed by two functionsComputing the area between two curvesIntegrate to calculate enclosed areaInteresting discrepencies between integrate functionsFinding the volume enclosed by two surfaces of revolutionFinding an area enclosed by 4 curvesApproximate the relationship between 6 nonlinear functions involving elliptic integrals
Is an up-to-date browser secure on an out-of-date OS?
What is the meaning of Triage in Cybersec world?
How can I have a shield and a way of attacking with a ranged weapon at the same time?
Output the Arecibo Message
Button changing its text & action. Good or terrible?
What is this business jet?
Can I have a signal generator on while it's not connected?
What is the motivation for a law requiring 2 parties to consent for recording a conversation
Is it ok to offer lower paid work as a trial period before negotiating for a full-time job?
Short story: child made less intelligent and less attractive
What is the most efficient way to store a numeric range?
Does adding complexity mean a more secure cipher?
Match Roman Numerals
For what reasons would an animal species NOT cross a *horizontal* land bridge?
Geography at the pixel level
Correct punctuation for showing a character's confusion
Why can't devices on different VLANs, but on the same subnet, communicate?
How do I free up internal storage if I don't have any apps downloaded?
How can I define good in a religion that claims no moral authority?
Will it cause any balance problems to have PCs level up and gain the benefits of a long rest mid-fight?
What is this sharp, curved notch on my knife for?
Getting crown tickets for Statue of Liberty
Is bread bad for ducks?
Why doesn't UInt have a toDouble()?
Finding the area between two curves with Integrate
The 2019 Stack Overflow Developer Survey Results Are InHow to evaluate this indefinite integral $csc(4x)sin(x)$Finding the centroid of the area between two curvesRevolving the area between two functions about an axisArea enclosed by two functionsComputing the area between two curvesIntegrate to calculate enclosed areaInteresting discrepencies between integrate functionsFinding the volume enclosed by two surfaces of revolutionFinding an area enclosed by 4 curvesApproximate the relationship between 6 nonlinear functions involving elliptic integrals
$begingroup$
I'm trying to solve and approximate the area between the two graphs. Right now, my functions are stored as
f[x_] := 3 Sin[x]
g[x_] := x - 1
and then I tried to integrate by evaluating
Integrate[Abs[f[x] - g[x]], x]
Instead of getting an answer, I just get the exact same thing I inputted
Integrate[Abs[f[x] - g[x]], x]
How do I fix this?
calculus-and-analysis
New contributor
$endgroup$
add a comment |
$begingroup$
I'm trying to solve and approximate the area between the two graphs. Right now, my functions are stored as
f[x_] := 3 Sin[x]
g[x_] := x - 1
and then I tried to integrate by evaluating
Integrate[Abs[f[x] - g[x]], x]
Instead of getting an answer, I just get the exact same thing I inputted
Integrate[Abs[f[x] - g[x]], x]
How do I fix this?
calculus-and-analysis
New contributor
$endgroup$
$begingroup$
You can format inline code and code blocks by selecting the code and clicking thebutton above the edit window. The edit window help button
?
is useful for learning how to format your questions and answers. You may also find this meta Q&A helpful
$endgroup$
– Michael E2
2 hours ago
add a comment |
$begingroup$
I'm trying to solve and approximate the area between the two graphs. Right now, my functions are stored as
f[x_] := 3 Sin[x]
g[x_] := x - 1
and then I tried to integrate by evaluating
Integrate[Abs[f[x] - g[x]], x]
Instead of getting an answer, I just get the exact same thing I inputted
Integrate[Abs[f[x] - g[x]], x]
How do I fix this?
calculus-and-analysis
New contributor
$endgroup$
I'm trying to solve and approximate the area between the two graphs. Right now, my functions are stored as
f[x_] := 3 Sin[x]
g[x_] := x - 1
and then I tried to integrate by evaluating
Integrate[Abs[f[x] - g[x]], x]
Instead of getting an answer, I just get the exact same thing I inputted
Integrate[Abs[f[x] - g[x]], x]
How do I fix this?
calculus-and-analysis
calculus-and-analysis
New contributor
New contributor
edited 2 hours ago
m_goldberg
88.6k873200
88.6k873200
New contributor
asked 2 hours ago
RyanRyan
111
111
New contributor
New contributor
$begingroup$
You can format inline code and code blocks by selecting the code and clicking thebutton above the edit window. The edit window help button
?
is useful for learning how to format your questions and answers. You may also find this meta Q&A helpful
$endgroup$
– Michael E2
2 hours ago
add a comment |
$begingroup$
You can format inline code and code blocks by selecting the code and clicking thebutton above the edit window. The edit window help button
?
is useful for learning how to format your questions and answers. You may also find this meta Q&A helpful
$endgroup$
– Michael E2
2 hours ago
$begingroup$
You can format inline code and code blocks by selecting the code and clicking the
button above the edit window. The edit window help button ?
is useful for learning how to format your questions and answers. You may also find this meta Q&A helpful$endgroup$
– Michael E2
2 hours ago
$begingroup$
You can format inline code and code blocks by selecting the code and clicking the
button above the edit window. The edit window help button ?
is useful for learning how to format your questions and answers. You may also find this meta Q&A helpful$endgroup$
– Michael E2
2 hours ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Use Assumptions
:
Integrate[Abs[f[x] - g[x]], x, Assumptions -> x [Element] Reals]
Or try RealAbs
instead of Abs
:
Integrate[RealAbs[f[x] - g[x]], x]
(They are equivalent antiderivatives.)
To get the area between the graphs, you need also to solve for the points of intersection.
area = Integrate[
Abs[f[x] - g[x]], x, Sequence @@ MinMax[x /. Solve[f[x] == g[x], x, Reals]]]
The area is approximately:
N[area]
(* 5.57475 *)
$endgroup$
$begingroup$
RealAbs
is awesome to know about! :O
$endgroup$
– Kagaratsch
2 hours ago
add a comment |
$begingroup$
You need to add assumptions, like this
Integrate[Abs[f[x] - g[x]], x, Assumptions :> Element[x, Reals]]
$endgroup$
add a comment |
$begingroup$
Assuming your functions
f[x_] := 3 Sin[x]
g[x_] := x - 1
are real valued, you can use square root of square to parametrize the absolute value. This then gives:
Integrate[Sqrt[(f[x] - g[x])^2], x]
(((-2 + x) x + 6 Cos[x]) Sqrt[(-1 + x - 3 Sin[x])^2])/(2 (-1 + x -
3 Sin[x]))
$endgroup$
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: "387"
;
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: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
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
,
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Ryan is a new contributor. Be nice, and check out our Code of Conduct.
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%2fmathematica.stackexchange.com%2fquestions%2f195049%2ffinding-the-area-between-two-curves-with-integrate%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Use Assumptions
:
Integrate[Abs[f[x] - g[x]], x, Assumptions -> x [Element] Reals]
Or try RealAbs
instead of Abs
:
Integrate[RealAbs[f[x] - g[x]], x]
(They are equivalent antiderivatives.)
To get the area between the graphs, you need also to solve for the points of intersection.
area = Integrate[
Abs[f[x] - g[x]], x, Sequence @@ MinMax[x /. Solve[f[x] == g[x], x, Reals]]]
The area is approximately:
N[area]
(* 5.57475 *)
$endgroup$
$begingroup$
RealAbs
is awesome to know about! :O
$endgroup$
– Kagaratsch
2 hours ago
add a comment |
$begingroup$
Use Assumptions
:
Integrate[Abs[f[x] - g[x]], x, Assumptions -> x [Element] Reals]
Or try RealAbs
instead of Abs
:
Integrate[RealAbs[f[x] - g[x]], x]
(They are equivalent antiderivatives.)
To get the area between the graphs, you need also to solve for the points of intersection.
area = Integrate[
Abs[f[x] - g[x]], x, Sequence @@ MinMax[x /. Solve[f[x] == g[x], x, Reals]]]
The area is approximately:
N[area]
(* 5.57475 *)
$endgroup$
$begingroup$
RealAbs
is awesome to know about! :O
$endgroup$
– Kagaratsch
2 hours ago
add a comment |
$begingroup$
Use Assumptions
:
Integrate[Abs[f[x] - g[x]], x, Assumptions -> x [Element] Reals]
Or try RealAbs
instead of Abs
:
Integrate[RealAbs[f[x] - g[x]], x]
(They are equivalent antiderivatives.)
To get the area between the graphs, you need also to solve for the points of intersection.
area = Integrate[
Abs[f[x] - g[x]], x, Sequence @@ MinMax[x /. Solve[f[x] == g[x], x, Reals]]]
The area is approximately:
N[area]
(* 5.57475 *)
$endgroup$
Use Assumptions
:
Integrate[Abs[f[x] - g[x]], x, Assumptions -> x [Element] Reals]
Or try RealAbs
instead of Abs
:
Integrate[RealAbs[f[x] - g[x]], x]
(They are equivalent antiderivatives.)
To get the area between the graphs, you need also to solve for the points of intersection.
area = Integrate[
Abs[f[x] - g[x]], x, Sequence @@ MinMax[x /. Solve[f[x] == g[x], x, Reals]]]
The area is approximately:
N[area]
(* 5.57475 *)
edited 2 hours ago
answered 2 hours ago
Michael E2Michael E2
150k12203482
150k12203482
$begingroup$
RealAbs
is awesome to know about! :O
$endgroup$
– Kagaratsch
2 hours ago
add a comment |
$begingroup$
RealAbs
is awesome to know about! :O
$endgroup$
– Kagaratsch
2 hours ago
$begingroup$
RealAbs
is awesome to know about! :O$endgroup$
– Kagaratsch
2 hours ago
$begingroup$
RealAbs
is awesome to know about! :O$endgroup$
– Kagaratsch
2 hours ago
add a comment |
$begingroup$
You need to add assumptions, like this
Integrate[Abs[f[x] - g[x]], x, Assumptions :> Element[x, Reals]]
$endgroup$
add a comment |
$begingroup$
You need to add assumptions, like this
Integrate[Abs[f[x] - g[x]], x, Assumptions :> Element[x, Reals]]
$endgroup$
add a comment |
$begingroup$
You need to add assumptions, like this
Integrate[Abs[f[x] - g[x]], x, Assumptions :> Element[x, Reals]]
$endgroup$
You need to add assumptions, like this
Integrate[Abs[f[x] - g[x]], x, Assumptions :> Element[x, Reals]]
answered 2 hours ago
NasserNasser
58.7k490206
58.7k490206
add a comment |
add a comment |
$begingroup$
Assuming your functions
f[x_] := 3 Sin[x]
g[x_] := x - 1
are real valued, you can use square root of square to parametrize the absolute value. This then gives:
Integrate[Sqrt[(f[x] - g[x])^2], x]
(((-2 + x) x + 6 Cos[x]) Sqrt[(-1 + x - 3 Sin[x])^2])/(2 (-1 + x -
3 Sin[x]))
$endgroup$
add a comment |
$begingroup$
Assuming your functions
f[x_] := 3 Sin[x]
g[x_] := x - 1
are real valued, you can use square root of square to parametrize the absolute value. This then gives:
Integrate[Sqrt[(f[x] - g[x])^2], x]
(((-2 + x) x + 6 Cos[x]) Sqrt[(-1 + x - 3 Sin[x])^2])/(2 (-1 + x -
3 Sin[x]))
$endgroup$
add a comment |
$begingroup$
Assuming your functions
f[x_] := 3 Sin[x]
g[x_] := x - 1
are real valued, you can use square root of square to parametrize the absolute value. This then gives:
Integrate[Sqrt[(f[x] - g[x])^2], x]
(((-2 + x) x + 6 Cos[x]) Sqrt[(-1 + x - 3 Sin[x])^2])/(2 (-1 + x -
3 Sin[x]))
$endgroup$
Assuming your functions
f[x_] := 3 Sin[x]
g[x_] := x - 1
are real valued, you can use square root of square to parametrize the absolute value. This then gives:
Integrate[Sqrt[(f[x] - g[x])^2], x]
(((-2 + x) x + 6 Cos[x]) Sqrt[(-1 + x - 3 Sin[x])^2])/(2 (-1 + x -
3 Sin[x]))
answered 2 hours ago
KagaratschKagaratsch
4,83831348
4,83831348
add a comment |
add a comment |
Ryan is a new contributor. Be nice, and check out our Code of Conduct.
Ryan is a new contributor. Be nice, and check out our Code of Conduct.
Ryan is a new contributor. Be nice, and check out our Code of Conduct.
Ryan is a new contributor. Be nice, and check out our Code of Conduct.
Thanks for contributing an answer to Mathematica 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%2fmathematica.stackexchange.com%2fquestions%2f195049%2ffinding-the-area-between-two-curves-with-integrate%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
$begingroup$
You can format inline code and code blocks by selecting the code and clicking the
button above the edit window. The edit window help button
?
is useful for learning how to format your questions and answers. You may also find this meta Q&A helpful$endgroup$
– Michael E2
2 hours ago