Finding the area between two curves with Integrate
$begingroup$
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 52 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.
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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 52 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.
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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
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– 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 52 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 52 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 52 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 52 mins ago
m_goldberg
88.6k873200
edited 52 mins ago
m_goldberg
88.6k873200
edited 52 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 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
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
1 hour 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
1 hour 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
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 |
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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