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Finding the area between two curves with Integrate

2

$begingroup$

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

share|improve this question

edited 44 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 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 | 

2

$begingroup$

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

share|improve this question

edited 44 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 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 | 

$begingroup$

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

share|improve this question

edited 44 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.

$endgroup$

f[x_] := 3 Sin[x]
g[x_] := x - 1

Integrate[Abs[f[x] - g[x]], x]

Integrate[Abs[f[x] - g[x]], x]

share|improve this question

edited 44 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.

share|improve this question

edited 44 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 44 mins ago

m_goldberg

88.6k873200

edited 44 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.

asked 1 hour ago

RyanRyan

111

3 Answers
3

active

oldest

votes

2

$begingroup$

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 *)

share|improve this answer

edited 54 mins ago

answered 58 mins ago

Michael E2Michael E2

150k12203482

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  RealAbs is awesome to know about! :O
  $endgroup$
  – Kagaratsch
  57 mins ago
  

  
  
  
  

add a comment | 

1

$begingroup$

You need to add assumptions, like this

 Integrate[Abs[f[x] - g[x]], x, Assumptions :> Element[x, Reals]]

share|improve this answer

answered 58 mins ago

NasserNasser

58.7k490206

$endgroup$

add a comment | 

0

$begingroup$

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]))

share|improve this answer

answered 59 mins ago

KagaratschKagaratsch

4,83831348

$endgroup$

add a comment | 

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2

$begingroup$

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 *)

share|improve this answer

edited 54 mins ago

answered 58 mins ago

Michael E2Michael E2

150k12203482

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  RealAbs is awesome to know about! :O
  $endgroup$
  – Kagaratsch
  57 mins ago
  

  
  
  
  

add a comment | 

2

$begingroup$

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 *)

share|improve this answer

edited 54 mins ago

answered 58 mins ago

Michael E2Michael E2

150k12203482

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  RealAbs is awesome to know about! :O
  $endgroup$
  – Kagaratsch
  57 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 *)

share|improve this answer

edited 54 mins ago

answered 58 mins ago

Michael E2Michael E2

150k12203482

$endgroup$

Integrate[Abs[f[x] - g[x]], x, Assumptions -> x [Element] Reals]

Integrate[RealAbs[f[x] - g[x]], x]

area = Integrate[
 Abs[f[x] - g[x]], x, Sequence @@ MinMax[x /. Solve[f[x] == g[x], x, Reals]]]

N[area]
(* 5.57475 *)

share|improve this answer

edited 54 mins ago

answered 58 mins ago

Michael E2Michael E2

150k12203482

answered 58 mins ago

Michael E2Michael E2

150k12203482

answered 58 mins ago

Michael E2Michael E2

150k12203482

1

$begingroup$

You need to add assumptions, like this

 Integrate[Abs[f[x] - g[x]], x, Assumptions :> Element[x, Reals]]

share|improve this answer

answered 58 mins ago

NasserNasser

58.7k490206

$endgroup$

add a comment | 

1

$begingroup$

You need to add assumptions, like this

 Integrate[Abs[f[x] - g[x]], x, Assumptions :> Element[x, Reals]]

share|improve this answer

answered 58 mins 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]]

share|improve this answer

answered 58 mins ago

NasserNasser

58.7k490206

$endgroup$

 Integrate[Abs[f[x] - g[x]], x, Assumptions :> Element[x, Reals]]

share|improve this answer

answered 58 mins ago

NasserNasser

58.7k490206

answered 58 mins ago

NasserNasser

58.7k490206

answered 58 mins ago

NasserNasser

58.7k490206

0

$begingroup$

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]))

share|improve this answer

answered 59 mins ago

KagaratschKagaratsch

4,83831348

$endgroup$

add a comment | 

0

$begingroup$

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]))

share|improve this answer

answered 59 mins 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]))

share|improve this answer

answered 59 mins ago

KagaratschKagaratsch

4,83831348

$endgroup$

f[x_] := 3 Sin[x] 
g[x_] := x - 1

Integrate[Sqrt[(f[x] - g[x])^2], x]

share|improve this answer

answered 59 mins ago

KagaratschKagaratsch

4,83831348

answered 59 mins ago

KagaratschKagaratsch

4,83831348

answered 59 mins ago

KagaratschKagaratsch

4,83831348