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
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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
edited 53 mins ago
m_goldberg
88.6k873200
asked 1 hour ago
RyanRyan
111
New contributor
Ryan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
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
edited 53 mins ago
m_goldberg
88.6k873200
asked 1 hour ago
RyanRyan
111
New contributor
Ryan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
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
1 hour 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
edited 53 mins ago
m_goldberg
88.6k873200
asked 1 hour ago
RyanRyan
111
New contributor
Ryan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
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
edited 53 mins ago
m_goldberg
88.6k873200
asked 1 hour ago
RyanRyan
111
New contributor
Ryan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 53 mins ago
m_goldberg
88.6k873200
asked 1 hour ago
RyanRyan
111
New contributor
Ryan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 53 mins ago
m_goldberg
88.6k873200
edited 53 mins ago
m_goldberg
88.6k873200
edited 53 mins ago
m_goldberg
88.6k873200
88.6k873200
asked 1 hour ago
RyanRyan
111
New contributor
Ryan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
RyanRyan
111
asked 1 hour ago
RyanRyan
111
111
New contributor
Ryan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Ryan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Ryan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
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
1 hour 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
1 hour ago
$begingroup$
You can format inline code and code blocks by selecting the code and clicking the
? is useful for learning how to format your questions and answers. You may also find this meta Q&A helpful$endgroup$
– Michael E2
1 hour ago
$begingroup$
You can format inline code and code blocks by selecting the code and clicking the
? is useful for learning how to format your questions and answers. You may also find this meta Q&A helpful$endgroup$
– Michael E2
1 hour 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 *)
answered 1 hour ago
Michael E2Michael E2
150k12203482
$endgroup$
$begingroup$
RealAbsis awesome to know about! :O
$endgroup$
– Kagaratsch
1 hour ago
add a comment |
$begingroup$
You need to add assumptions, like this
Integrate[Abs[f[x] - g[x]], x, Assumptions :> Element[x, Reals]]
answered 1 hour ago
NasserNasser
58.7k490206
$endgroup$
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]))
answered 1 hour ago
KagaratschKagaratsch
4,83831348
$endgroup$
add a comment |
Your Answer
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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 *)
answered 1 hour ago
Michael E2Michael E2
150k12203482
$endgroup$
$begingroup$
RealAbsis awesome to know about! :O
$endgroup$
– Kagaratsch
1 hour 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 *)
answered 1 hour ago
Michael E2Michael E2
150k12203482
$endgroup$
$begingroup$
RealAbsis awesome to know about! :O
$endgroup$
– Kagaratsch
1 hour 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 *)
answered 1 hour ago
Michael E2Michael E2
150k12203482
$endgroup$
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 *)
answered 1 hour ago
Michael E2Michael E2
150k12203482
edited 1 hour ago
answered 1 hour ago
Michael E2Michael E2
150k12203482
answered 1 hour ago
Michael E2Michael E2
150k12203482
answered 1 hour ago
Michael E2Michael E2
150k12203482
150k12203482
$begingroup$
RealAbsis awesome to know about! :O
$endgroup$
– Kagaratsch
1 hour ago
add a comment |
$begingroup$
RealAbsis awesome to know about! :O
$endgroup$
– Kagaratsch
1 hour ago
$begingroup$
RealAbs is awesome to know about! :O$endgroup$
– Kagaratsch
1 hour ago
$begingroup$
RealAbs is awesome to know about! :O$endgroup$
– Kagaratsch
1 hour ago
add a comment |
$begingroup$
You need to add assumptions, like this
Integrate[Abs[f[x] - g[x]], x, Assumptions :> Element[x, Reals]]
answered 1 hour ago
NasserNasser
58.7k490206
$endgroup$
add a comment |
$begingroup$
You need to add assumptions, like this
Integrate[Abs[f[x] - g[x]], x, Assumptions :> Element[x, Reals]]
answered 1 hour ago
NasserNasser
58.7k490206
$endgroup$
add a comment |
$begingroup$
You need to add assumptions, like this
Integrate[Abs[f[x] - g[x]], x, Assumptions :> Element[x, Reals]]
answered 1 hour ago
NasserNasser
58.7k490206
$endgroup$
You need to add assumptions, like this
Integrate[Abs[f[x] - g[x]], x, Assumptions :> Element[x, Reals]]
answered 1 hour ago
NasserNasser
58.7k490206
answered 1 hour ago
NasserNasser
58.7k490206
answered 1 hour ago
NasserNasser
58.7k490206
answered 1 hour 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]))
answered 1 hour ago
KagaratschKagaratsch
4,83831348
$endgroup$
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]))
answered 1 hour ago
KagaratschKagaratsch
4,83831348
$endgroup$
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]))
answered 1 hour ago
KagaratschKagaratsch
4,83831348
$endgroup$
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 1 hour ago
KagaratschKagaratsch
4,83831348
answered 1 hour ago
KagaratschKagaratsch
4,83831348
answered 1 hour ago
KagaratschKagaratsch
4,83831348
answered 1 hour 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.
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$begingroup$
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
1 hour ago