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Finding the area between two curves with Integrate

3

$begingroup$

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 Apr 12 at 2:53

m_goldberg

88.7k873200

asked Apr 12 at 2:17

RyanRyan

161

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
  Apr 12 at 2:35
  

  
  
  
  

add a comment | 

3

$begingroup$

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 Apr 12 at 2:53

m_goldberg

88.7k873200

asked Apr 12 at 2:17

RyanRyan

161

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
  Apr 12 at 2:35
  

  
  
  
  

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 Apr 12 at 2:53

m_goldberg

88.7k873200

asked Apr 12 at 2:17

RyanRyan

161

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 Apr 12 at 2:53

m_goldberg

88.7k873200

asked Apr 12 at 2:17

RyanRyan

161

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 Apr 12 at 2:53

m_goldberg

88.7k873200

asked Apr 12 at 2:17

RyanRyan

161

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 Apr 12 at 2:53

m_goldberg

88.7k873200

edited Apr 12 at 2:53

m_goldberg

88.7k873200

asked Apr 12 at 2:17

RyanRyan

161

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 Apr 12 at 2:17

RyanRyan

161

3 Answers
3

active

oldest

votes

4

$begingroup$

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 Apr 12 at 2:43

answered Apr 12 at 2:39

Michael E2Michael E2

151k12203482

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  RealAbs is awesome to know about! :O
  $endgroup$
  – Kagaratsch
  Apr 12 at 2:40
  

  
  
  
  

add a comment | 

2

$begingroup$

You need to add assumptions, like this

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

share|improve this answer

answered Apr 12 at 2:39

NasserNasser

58.8k490206

$endgroup$

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 Apr 12 at 2:38

KagaratschKagaratsch

4,83831348

$endgroup$

add a comment | 

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4

$begingroup$

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 Apr 12 at 2:43

answered Apr 12 at 2:39

Michael E2Michael E2

151k12203482

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  RealAbs is awesome to know about! :O
  $endgroup$
  – Kagaratsch
  Apr 12 at 2:40
  

  
  
  
  

add a comment | 

4

$begingroup$

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 Apr 12 at 2:43

answered Apr 12 at 2:39

Michael E2Michael E2

151k12203482

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  RealAbs is awesome to know about! :O
  $endgroup$
  – Kagaratsch
  Apr 12 at 2:40
  

  
  
  
  

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 Apr 12 at 2:43

answered Apr 12 at 2:39

Michael E2Michael E2

151k12203482

$endgroup$

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 Apr 12 at 2:43

answered Apr 12 at 2:39

Michael E2Michael E2

151k12203482

answered Apr 12 at 2:39

Michael E2Michael E2

151k12203482

answered Apr 12 at 2:39

Michael E2Michael E2

151k12203482

2

$begingroup$

You need to add assumptions, like this

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

share|improve this answer

answered Apr 12 at 2:39

NasserNasser

58.8k490206

$endgroup$

add a comment | 

2

$begingroup$

You need to add assumptions, like this

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

share|improve this answer

answered Apr 12 at 2:39

NasserNasser

58.8k490206

$endgroup$

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 Apr 12 at 2:39

NasserNasser

58.8k490206

$endgroup$

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

share|improve this answer

answered Apr 12 at 2:39

NasserNasser

58.8k490206

answered Apr 12 at 2:39

NasserNasser

58.8k490206

answered Apr 12 at 2:39

NasserNasser

58.8k490206

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 Apr 12 at 2:38

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 Apr 12 at 2:38

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 Apr 12 at 2:38

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 Apr 12 at 2:38

KagaratschKagaratsch

4,83831348

answered Apr 12 at 2:38

KagaratschKagaratsch

4,83831348

answered Apr 12 at 2:38

KagaratschKagaratsch

4,83831348