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









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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
















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









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














2












2








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









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






share|improve this question









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









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




share|improve this question








edited 52 mins ago









m_goldberg

88.6k873200




88.6k873200






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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.
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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


















  • $begingroup$
    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










3 Answers
3






active

oldest

votes


















2












$begingroup$

Use Assumptions:



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


Mathematica graphics



Or try RealAbs instead of Abs:



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


Mathematica graphics



(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]]}]


Mathematica graphics



The area is approximately:



N[area]

(*  5.57475  *)





share|improve this answer











$endgroup$













  • $begingroup$
    RealAbs is awesome to know about! :O
    $endgroup$
    – Kagaratsch
    1 hour ago



















1












$begingroup$

You need to add assumptions, like this



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


Mathematica graphics






share|improve this answer









$endgroup$





















    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









    $endgroup$














      Your Answer





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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      Use Assumptions:



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


      Mathematica graphics



      Or try RealAbs instead of Abs:



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


      Mathematica graphics



      (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]]}]


      Mathematica graphics



      The area is approximately:



      N[area]
      
(*  5.57475  *)





      share|improve this answer











      $endgroup$













      • $begingroup$
        RealAbs is awesome to know about! :O
        $endgroup$
        – Kagaratsch
        1 hour ago
















      2












      $begingroup$

      Use Assumptions:



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


      Mathematica graphics



      Or try RealAbs instead of Abs:



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


      Mathematica graphics



      (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]]}]


      Mathematica graphics



      The area is approximately:



      N[area]
      
(*  5.57475  *)





      share|improve this answer











      $endgroup$













      • $begingroup$
        RealAbs is awesome to know about! :O
        $endgroup$
        – Kagaratsch
        1 hour ago














      2












      2








      2





      $begingroup$

      Use Assumptions:



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


      Mathematica graphics



      Or try RealAbs instead of Abs:



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


      Mathematica graphics



      (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]]}]


      Mathematica graphics



      The area is approximately:



      N[area]
      
(*  5.57475  *)





      share|improve this answer











      $endgroup$



      Use Assumptions:



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


      Mathematica graphics



      Or try RealAbs instead of Abs:



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


      Mathematica graphics



      (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]]}]


      Mathematica graphics



      The area is approximately:



      N[area]
      
(*  5.57475  *)






      share|improve this answer














      share|improve this answer



      share|improve this answer








      edited 1 hour ago

























      answered 1 hour ago









      Michael E2Michael E2

      150k12203482




      150k12203482












      • $begingroup$
        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
















      $begingroup$
      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











      1












      $begingroup$

      You need to add assumptions, like this



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


      Mathematica graphics






      share|improve this answer









      $endgroup$


















        1












        $begingroup$

        You need to add assumptions, like this



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


        Mathematica graphics






        share|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          You need to add assumptions, like this



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


          Mathematica graphics






          share|improve this answer









          $endgroup$



          You need to add assumptions, like this



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


          Mathematica graphics







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered 1 hour ago









          NasserNasser

          58.7k490206




          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









              $endgroup$


















                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









                $endgroup$
















                  0












                  0








                  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









                  $endgroup$



                  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]))








                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered 1 hour ago









                  KagaratschKagaratsch

                  4,83831348




                  4,83831348






















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