Finding the area between two curves with Integrate The 2019 Stack Overflow Developer Survey...
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Finding the area between two curves with Integrate
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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
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 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
46 mins 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 28 mins ago
m_goldberg
88.6k873200
88.6k873200
New contributor
asked 1 hour ago
RyanRyan
111
111
New contributor
New contributor
$begingroup$
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
46 mins ago
add a comment |
$begingroup$
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
46 mins 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
46 mins 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
46 mins 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
42 mins 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
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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 *)
$endgroup$
$begingroup$
RealAbs
is awesome to know about! :O
$endgroup$
– Kagaratsch
42 mins 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
42 mins 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 39 mins ago
answered 42 mins ago
Michael E2Michael E2
150k12203482
150k12203482
$begingroup$
RealAbs
is awesome to know about! :O
$endgroup$
– Kagaratsch
42 mins ago
add a comment |
$begingroup$
RealAbs
is awesome to know about! :O
$endgroup$
– Kagaratsch
42 mins ago
$begingroup$
RealAbs
is awesome to know about! :O$endgroup$
– Kagaratsch
42 mins ago
$begingroup$
RealAbs
is awesome to know about! :O$endgroup$
– Kagaratsch
42 mins 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 43 mins 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 44 mins 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
46 mins ago