Integrating an absolute function using Mathematica





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I was unsure as to how to integrate an absolute function, such as $int_0^3 left|sqrt{9-x^2}right|dx$, so I tried:



Integrate[Abs[(9-x^2)^(1/2)],x]


But it returned:




$int sqrt{Abs[9-x^2]}mathbb{d}x$




Could someone please show me how to do so correctly? And also, is there a way of evaluating the integral without having to make it a function and sub in the values of the terminals?










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    Welcome to the Mathematica StackExchange. We would typically prefer that you post your code rather than pictures of your code, or that you use LaTeX to format mathematical expressions. I transcribed your pictures to both ends, but please feel free to edit your post again if you feel I've changed the intent somehow.
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    – eyorble
    May 26 at 3:23










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    Thanks for letting me know. I'll try to keep that in mind next time I ask a question, @eyerble
    $endgroup$
    – wendy
    May 26 at 3:24


















2












$begingroup$


I was unsure as to how to integrate an absolute function, such as $int_0^3 left|sqrt{9-x^2}right|dx$, so I tried:



Integrate[Abs[(9-x^2)^(1/2)],x]


But it returned:




$int sqrt{Abs[9-x^2]}mathbb{d}x$




Could someone please show me how to do so correctly? And also, is there a way of evaluating the integral without having to make it a function and sub in the values of the terminals?










share|improve this question











$endgroup$














  • $begingroup$
    Welcome to the Mathematica StackExchange. We would typically prefer that you post your code rather than pictures of your code, or that you use LaTeX to format mathematical expressions. I transcribed your pictures to both ends, but please feel free to edit your post again if you feel I've changed the intent somehow.
    $endgroup$
    – eyorble
    May 26 at 3:23










  • $begingroup$
    Thanks for letting me know. I'll try to keep that in mind next time I ask a question, @eyerble
    $endgroup$
    – wendy
    May 26 at 3:24














2












2








2





$begingroup$


I was unsure as to how to integrate an absolute function, such as $int_0^3 left|sqrt{9-x^2}right|dx$, so I tried:



Integrate[Abs[(9-x^2)^(1/2)],x]


But it returned:




$int sqrt{Abs[9-x^2]}mathbb{d}x$




Could someone please show me how to do so correctly? And also, is there a way of evaluating the integral without having to make it a function and sub in the values of the terminals?










share|improve this question











$endgroup$




I was unsure as to how to integrate an absolute function, such as $int_0^3 left|sqrt{9-x^2}right|dx$, so I tried:



Integrate[Abs[(9-x^2)^(1/2)],x]


But it returned:




$int sqrt{Abs[9-x^2]}mathbb{d}x$




Could someone please show me how to do so correctly? And also, is there a way of evaluating the integral without having to make it a function and sub in the values of the terminals?







calculus-and-analysis






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edited May 26 at 19:55









Michael E2

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asked May 26 at 3:04









wendywendy

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  • $begingroup$
    Welcome to the Mathematica StackExchange. We would typically prefer that you post your code rather than pictures of your code, or that you use LaTeX to format mathematical expressions. I transcribed your pictures to both ends, but please feel free to edit your post again if you feel I've changed the intent somehow.
    $endgroup$
    – eyorble
    May 26 at 3:23










  • $begingroup$
    Thanks for letting me know. I'll try to keep that in mind next time I ask a question, @eyerble
    $endgroup$
    – wendy
    May 26 at 3:24


















  • $begingroup$
    Welcome to the Mathematica StackExchange. We would typically prefer that you post your code rather than pictures of your code, or that you use LaTeX to format mathematical expressions. I transcribed your pictures to both ends, but please feel free to edit your post again if you feel I've changed the intent somehow.
    $endgroup$
    – eyorble
    May 26 at 3:23










  • $begingroup$
    Thanks for letting me know. I'll try to keep that in mind next time I ask a question, @eyerble
    $endgroup$
    – wendy
    May 26 at 3:24
















$begingroup$
Welcome to the Mathematica StackExchange. We would typically prefer that you post your code rather than pictures of your code, or that you use LaTeX to format mathematical expressions. I transcribed your pictures to both ends, but please feel free to edit your post again if you feel I've changed the intent somehow.
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– eyorble
May 26 at 3:23




$begingroup$
Welcome to the Mathematica StackExchange. We would typically prefer that you post your code rather than pictures of your code, or that you use LaTeX to format mathematical expressions. I transcribed your pictures to both ends, but please feel free to edit your post again if you feel I've changed the intent somehow.
$endgroup$
– eyorble
May 26 at 3:23












$begingroup$
Thanks for letting me know. I'll try to keep that in mind next time I ask a question, @eyerble
$endgroup$
– wendy
May 26 at 3:24




$begingroup$
Thanks for letting me know. I'll try to keep that in mind next time I ask a question, @eyerble
$endgroup$
– wendy
May 26 at 3:24










4 Answers
4






active

oldest

votes


















6












$begingroup$

Use RealAbs



Integrate[RealAbs[(9 - x^2)^(1/2)], x]


enter image description here






share|improve this answer









$endgroup$























    6












    $begingroup$

    Use Assumptions:



    Integrate[Abs[(9 - x^2)^(1/2)], x, Assumptions -> x [Element] Reals]
    Differences[% /. {{x -> 0}, {x -> 3}}]


    Mathematica graphics



    Explanation of the math: Mathematica defaults to complex analysis, but many users I suspect tend to think in terms of real-variable calculus, often just in terms of single-variable calculus on problems like the OP's. Unlike the real absolute value, the complex absolute value does not have an antiderivative. More particularly, integrals of (continuous) functions of the real absolute value are "path independent"1): that is, there is a "potential function" F[x] such that the integral from a to b is given by the values of at the end points F[b] - F[a], and the values of F[x] along the path from a to b do not matter. For integrals on the 1D real line, the potential function is the same as the antiderivative. However, integrals over paths in the complex plane of functions of the complex absolute value Abs are generally not path independent. Hence there cannot be a potential function that can be used to evaluate the integral by subtracting values at the end points. Here's a numerical example, in which the end points of the paths are the same but the values of the integrals are not (Integrate and NIntegrate use straight-lined paths through the complex numbers specified):



    NIntegrate[Abs[(9 - x^2)^(1/2)], {x, 0, 3}]
    (* 7.06858 *)

    NIntegrate[Abs[(9 - x^2)^(1/2)], {x, 0, 2 + 2 I, 3}]
    (* 8.55748 + 1.50584 I *)


    Consequently, the answer to the OP's integral,



    Integrate[Abs[(9 - x^2)^(1/2)], x]


    does not exist.



    1) Note: I'm borrowing the terms "path independent" and "potential function" from vector calculus. They are usually only introduced in dimension 2 and higher, because dimension 1 may be understood in simpler terms in first-year calculus. In a comparison of complex analysis with single-variable real calculus, it is natural to consider the real integral $int_a^b f(x) , dx$ as a path integral in the complex plane whose path happens to lie on the real axis. It is also valid to consider it a path integral of a 1D vector field on the 1D real line.






    share|improve this answer











    $endgroup$























      4












      $begingroup$

      Integrate's variable declaration can be either a variable by itself, as you currently have it (x), or as a variable with its endpoints. The latter form will evaluate in full automatically:



      Integrate[Abs[(9 - x^2)^(1/2)], {x, 0, 3}]



      $frac{9pi}{4}$




      This is the second usage shown in Integrate's documentation, so please check there if you have any further questions on it.



      In general, the indefinite integrals of functions containing an absolute value are messier than strictly necessary. However, if you specify appropriate assumptions (particularly that x is real), Mathematica can divide it into a piecewise function:



      Integrate[Abs[(9 - x^2)^(1/2)], x, Assumptions -> {x [Element] Reals}]


      The result is a mess, so I'd recommend looking at it on your own machine. If complex values are allowed, then ComplexExpand would ordinarily be helpful.



      Integrate[ComplexExpand[Abs[(9 - x^2)^(1/2)], x], x]


      In this case, however, it just returns a different, more complicated looking integral.






      share|improve this answer









      $endgroup$















      • $begingroup$
        thank you so much for answering! sorry for replying late, just came out of class.
        $endgroup$
        – wendy
        May 26 at 5:46



















      3












      $begingroup$

      For indefinite integrals replace Abs by the square root of the square:



      Integrate[Sqrt[((9 - x^2)^(1/2))^2], x]
      (* 1/2 x Sqrt[9 - x^2] + 9/2 ArcSin[x/3] *)





      share|improve this answer









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        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        6












        $begingroup$

        Use RealAbs



        Integrate[RealAbs[(9 - x^2)^(1/2)], x]


        enter image description here






        share|improve this answer









        $endgroup$




















          6












          $begingroup$

          Use RealAbs



          Integrate[RealAbs[(9 - x^2)^(1/2)], x]


          enter image description here






          share|improve this answer









          $endgroup$


















            6












            6








            6





            $begingroup$

            Use RealAbs



            Integrate[RealAbs[(9 - x^2)^(1/2)], x]


            enter image description here






            share|improve this answer









            $endgroup$



            Use RealAbs



            Integrate[RealAbs[(9 - x^2)^(1/2)], x]


            enter image description here







            share|improve this answer












            share|improve this answer



            share|improve this answer










            answered May 26 at 9:42









            rmwrmw

            6581 silver badge7 bronze badges




            6581 silver badge7 bronze badges




























                6












                $begingroup$

                Use Assumptions:



                Integrate[Abs[(9 - x^2)^(1/2)], x, Assumptions -> x [Element] Reals]
                Differences[% /. {{x -> 0}, {x -> 3}}]


                Mathematica graphics



                Explanation of the math: Mathematica defaults to complex analysis, but many users I suspect tend to think in terms of real-variable calculus, often just in terms of single-variable calculus on problems like the OP's. Unlike the real absolute value, the complex absolute value does not have an antiderivative. More particularly, integrals of (continuous) functions of the real absolute value are "path independent"1): that is, there is a "potential function" F[x] such that the integral from a to b is given by the values of at the end points F[b] - F[a], and the values of F[x] along the path from a to b do not matter. For integrals on the 1D real line, the potential function is the same as the antiderivative. However, integrals over paths in the complex plane of functions of the complex absolute value Abs are generally not path independent. Hence there cannot be a potential function that can be used to evaluate the integral by subtracting values at the end points. Here's a numerical example, in which the end points of the paths are the same but the values of the integrals are not (Integrate and NIntegrate use straight-lined paths through the complex numbers specified):



                NIntegrate[Abs[(9 - x^2)^(1/2)], {x, 0, 3}]
                (* 7.06858 *)

                NIntegrate[Abs[(9 - x^2)^(1/2)], {x, 0, 2 + 2 I, 3}]
                (* 8.55748 + 1.50584 I *)


                Consequently, the answer to the OP's integral,



                Integrate[Abs[(9 - x^2)^(1/2)], x]


                does not exist.



                1) Note: I'm borrowing the terms "path independent" and "potential function" from vector calculus. They are usually only introduced in dimension 2 and higher, because dimension 1 may be understood in simpler terms in first-year calculus. In a comparison of complex analysis with single-variable real calculus, it is natural to consider the real integral $int_a^b f(x) , dx$ as a path integral in the complex plane whose path happens to lie on the real axis. It is also valid to consider it a path integral of a 1D vector field on the 1D real line.






                share|improve this answer











                $endgroup$




















                  6












                  $begingroup$

                  Use Assumptions:



                  Integrate[Abs[(9 - x^2)^(1/2)], x, Assumptions -> x [Element] Reals]
                  Differences[% /. {{x -> 0}, {x -> 3}}]


                  Mathematica graphics



                  Explanation of the math: Mathematica defaults to complex analysis, but many users I suspect tend to think in terms of real-variable calculus, often just in terms of single-variable calculus on problems like the OP's. Unlike the real absolute value, the complex absolute value does not have an antiderivative. More particularly, integrals of (continuous) functions of the real absolute value are "path independent"1): that is, there is a "potential function" F[x] such that the integral from a to b is given by the values of at the end points F[b] - F[a], and the values of F[x] along the path from a to b do not matter. For integrals on the 1D real line, the potential function is the same as the antiderivative. However, integrals over paths in the complex plane of functions of the complex absolute value Abs are generally not path independent. Hence there cannot be a potential function that can be used to evaluate the integral by subtracting values at the end points. Here's a numerical example, in which the end points of the paths are the same but the values of the integrals are not (Integrate and NIntegrate use straight-lined paths through the complex numbers specified):



                  NIntegrate[Abs[(9 - x^2)^(1/2)], {x, 0, 3}]
                  (* 7.06858 *)

                  NIntegrate[Abs[(9 - x^2)^(1/2)], {x, 0, 2 + 2 I, 3}]
                  (* 8.55748 + 1.50584 I *)


                  Consequently, the answer to the OP's integral,



                  Integrate[Abs[(9 - x^2)^(1/2)], x]


                  does not exist.



                  1) Note: I'm borrowing the terms "path independent" and "potential function" from vector calculus. They are usually only introduced in dimension 2 and higher, because dimension 1 may be understood in simpler terms in first-year calculus. In a comparison of complex analysis with single-variable real calculus, it is natural to consider the real integral $int_a^b f(x) , dx$ as a path integral in the complex plane whose path happens to lie on the real axis. It is also valid to consider it a path integral of a 1D vector field on the 1D real line.






                  share|improve this answer











                  $endgroup$


















                    6












                    6








                    6





                    $begingroup$

                    Use Assumptions:



                    Integrate[Abs[(9 - x^2)^(1/2)], x, Assumptions -> x [Element] Reals]
                    Differences[% /. {{x -> 0}, {x -> 3}}]


                    Mathematica graphics



                    Explanation of the math: Mathematica defaults to complex analysis, but many users I suspect tend to think in terms of real-variable calculus, often just in terms of single-variable calculus on problems like the OP's. Unlike the real absolute value, the complex absolute value does not have an antiderivative. More particularly, integrals of (continuous) functions of the real absolute value are "path independent"1): that is, there is a "potential function" F[x] such that the integral from a to b is given by the values of at the end points F[b] - F[a], and the values of F[x] along the path from a to b do not matter. For integrals on the 1D real line, the potential function is the same as the antiderivative. However, integrals over paths in the complex plane of functions of the complex absolute value Abs are generally not path independent. Hence there cannot be a potential function that can be used to evaluate the integral by subtracting values at the end points. Here's a numerical example, in which the end points of the paths are the same but the values of the integrals are not (Integrate and NIntegrate use straight-lined paths through the complex numbers specified):



                    NIntegrate[Abs[(9 - x^2)^(1/2)], {x, 0, 3}]
                    (* 7.06858 *)

                    NIntegrate[Abs[(9 - x^2)^(1/2)], {x, 0, 2 + 2 I, 3}]
                    (* 8.55748 + 1.50584 I *)


                    Consequently, the answer to the OP's integral,



                    Integrate[Abs[(9 - x^2)^(1/2)], x]


                    does not exist.



                    1) Note: I'm borrowing the terms "path independent" and "potential function" from vector calculus. They are usually only introduced in dimension 2 and higher, because dimension 1 may be understood in simpler terms in first-year calculus. In a comparison of complex analysis with single-variable real calculus, it is natural to consider the real integral $int_a^b f(x) , dx$ as a path integral in the complex plane whose path happens to lie on the real axis. It is also valid to consider it a path integral of a 1D vector field on the 1D real line.






                    share|improve this answer











                    $endgroup$



                    Use Assumptions:



                    Integrate[Abs[(9 - x^2)^(1/2)], x, Assumptions -> x [Element] Reals]
                    Differences[% /. {{x -> 0}, {x -> 3}}]


                    Mathematica graphics



                    Explanation of the math: Mathematica defaults to complex analysis, but many users I suspect tend to think in terms of real-variable calculus, often just in terms of single-variable calculus on problems like the OP's. Unlike the real absolute value, the complex absolute value does not have an antiderivative. More particularly, integrals of (continuous) functions of the real absolute value are "path independent"1): that is, there is a "potential function" F[x] such that the integral from a to b is given by the values of at the end points F[b] - F[a], and the values of F[x] along the path from a to b do not matter. For integrals on the 1D real line, the potential function is the same as the antiderivative. However, integrals over paths in the complex plane of functions of the complex absolute value Abs are generally not path independent. Hence there cannot be a potential function that can be used to evaluate the integral by subtracting values at the end points. Here's a numerical example, in which the end points of the paths are the same but the values of the integrals are not (Integrate and NIntegrate use straight-lined paths through the complex numbers specified):



                    NIntegrate[Abs[(9 - x^2)^(1/2)], {x, 0, 3}]
                    (* 7.06858 *)

                    NIntegrate[Abs[(9 - x^2)^(1/2)], {x, 0, 2 + 2 I, 3}]
                    (* 8.55748 + 1.50584 I *)


                    Consequently, the answer to the OP's integral,



                    Integrate[Abs[(9 - x^2)^(1/2)], x]


                    does not exist.



                    1) Note: I'm borrowing the terms "path independent" and "potential function" from vector calculus. They are usually only introduced in dimension 2 and higher, because dimension 1 may be understood in simpler terms in first-year calculus. In a comparison of complex analysis with single-variable real calculus, it is natural to consider the real integral $int_a^b f(x) , dx$ as a path integral in the complex plane whose path happens to lie on the real axis. It is also valid to consider it a path integral of a 1D vector field on the 1D real line.







                    share|improve this answer














                    share|improve this answer



                    share|improve this answer








                    edited May 26 at 13:26

























                    answered May 26 at 12:47









                    Michael E2Michael E2

                    158k13 gold badges216 silver badges513 bronze badges




                    158k13 gold badges216 silver badges513 bronze badges


























                        4












                        $begingroup$

                        Integrate's variable declaration can be either a variable by itself, as you currently have it (x), or as a variable with its endpoints. The latter form will evaluate in full automatically:



                        Integrate[Abs[(9 - x^2)^(1/2)], {x, 0, 3}]



                        $frac{9pi}{4}$




                        This is the second usage shown in Integrate's documentation, so please check there if you have any further questions on it.



                        In general, the indefinite integrals of functions containing an absolute value are messier than strictly necessary. However, if you specify appropriate assumptions (particularly that x is real), Mathematica can divide it into a piecewise function:



                        Integrate[Abs[(9 - x^2)^(1/2)], x, Assumptions -> {x [Element] Reals}]


                        The result is a mess, so I'd recommend looking at it on your own machine. If complex values are allowed, then ComplexExpand would ordinarily be helpful.



                        Integrate[ComplexExpand[Abs[(9 - x^2)^(1/2)], x], x]


                        In this case, however, it just returns a different, more complicated looking integral.






                        share|improve this answer









                        $endgroup$















                        • $begingroup$
                          thank you so much for answering! sorry for replying late, just came out of class.
                          $endgroup$
                          – wendy
                          May 26 at 5:46
















                        4












                        $begingroup$

                        Integrate's variable declaration can be either a variable by itself, as you currently have it (x), or as a variable with its endpoints. The latter form will evaluate in full automatically:



                        Integrate[Abs[(9 - x^2)^(1/2)], {x, 0, 3}]



                        $frac{9pi}{4}$




                        This is the second usage shown in Integrate's documentation, so please check there if you have any further questions on it.



                        In general, the indefinite integrals of functions containing an absolute value are messier than strictly necessary. However, if you specify appropriate assumptions (particularly that x is real), Mathematica can divide it into a piecewise function:



                        Integrate[Abs[(9 - x^2)^(1/2)], x, Assumptions -> {x [Element] Reals}]


                        The result is a mess, so I'd recommend looking at it on your own machine. If complex values are allowed, then ComplexExpand would ordinarily be helpful.



                        Integrate[ComplexExpand[Abs[(9 - x^2)^(1/2)], x], x]


                        In this case, however, it just returns a different, more complicated looking integral.






                        share|improve this answer









                        $endgroup$















                        • $begingroup$
                          thank you so much for answering! sorry for replying late, just came out of class.
                          $endgroup$
                          – wendy
                          May 26 at 5:46














                        4












                        4








                        4





                        $begingroup$

                        Integrate's variable declaration can be either a variable by itself, as you currently have it (x), or as a variable with its endpoints. The latter form will evaluate in full automatically:



                        Integrate[Abs[(9 - x^2)^(1/2)], {x, 0, 3}]



                        $frac{9pi}{4}$




                        This is the second usage shown in Integrate's documentation, so please check there if you have any further questions on it.



                        In general, the indefinite integrals of functions containing an absolute value are messier than strictly necessary. However, if you specify appropriate assumptions (particularly that x is real), Mathematica can divide it into a piecewise function:



                        Integrate[Abs[(9 - x^2)^(1/2)], x, Assumptions -> {x [Element] Reals}]


                        The result is a mess, so I'd recommend looking at it on your own machine. If complex values are allowed, then ComplexExpand would ordinarily be helpful.



                        Integrate[ComplexExpand[Abs[(9 - x^2)^(1/2)], x], x]


                        In this case, however, it just returns a different, more complicated looking integral.






                        share|improve this answer









                        $endgroup$



                        Integrate's variable declaration can be either a variable by itself, as you currently have it (x), or as a variable with its endpoints. The latter form will evaluate in full automatically:



                        Integrate[Abs[(9 - x^2)^(1/2)], {x, 0, 3}]



                        $frac{9pi}{4}$




                        This is the second usage shown in Integrate's documentation, so please check there if you have any further questions on it.



                        In general, the indefinite integrals of functions containing an absolute value are messier than strictly necessary. However, if you specify appropriate assumptions (particularly that x is real), Mathematica can divide it into a piecewise function:



                        Integrate[Abs[(9 - x^2)^(1/2)], x, Assumptions -> {x [Element] Reals}]


                        The result is a mess, so I'd recommend looking at it on your own machine. If complex values are allowed, then ComplexExpand would ordinarily be helpful.



                        Integrate[ComplexExpand[Abs[(9 - x^2)^(1/2)], x], x]


                        In this case, however, it just returns a different, more complicated looking integral.







                        share|improve this answer












                        share|improve this answer



                        share|improve this answer










                        answered May 26 at 3:34









                        eyorbleeyorble

                        6,5231 gold badge11 silver badges30 bronze badges




                        6,5231 gold badge11 silver badges30 bronze badges















                        • $begingroup$
                          thank you so much for answering! sorry for replying late, just came out of class.
                          $endgroup$
                          – wendy
                          May 26 at 5:46


















                        • $begingroup$
                          thank you so much for answering! sorry for replying late, just came out of class.
                          $endgroup$
                          – wendy
                          May 26 at 5:46
















                        $begingroup$
                        thank you so much for answering! sorry for replying late, just came out of class.
                        $endgroup$
                        – wendy
                        May 26 at 5:46




                        $begingroup$
                        thank you so much for answering! sorry for replying late, just came out of class.
                        $endgroup$
                        – wendy
                        May 26 at 5:46











                        3












                        $begingroup$

                        For indefinite integrals replace Abs by the square root of the square:



                        Integrate[Sqrt[((9 - x^2)^(1/2))^2], x]
                        (* 1/2 x Sqrt[9 - x^2] + 9/2 ArcSin[x/3] *)





                        share|improve this answer









                        $endgroup$




















                          3












                          $begingroup$

                          For indefinite integrals replace Abs by the square root of the square:



                          Integrate[Sqrt[((9 - x^2)^(1/2))^2], x]
                          (* 1/2 x Sqrt[9 - x^2] + 9/2 ArcSin[x/3] *)





                          share|improve this answer









                          $endgroup$


















                            3












                            3








                            3





                            $begingroup$

                            For indefinite integrals replace Abs by the square root of the square:



                            Integrate[Sqrt[((9 - x^2)^(1/2))^2], x]
                            (* 1/2 x Sqrt[9 - x^2] + 9/2 ArcSin[x/3] *)





                            share|improve this answer









                            $endgroup$



                            For indefinite integrals replace Abs by the square root of the square:



                            Integrate[Sqrt[((9 - x^2)^(1/2))^2], x]
                            (* 1/2 x Sqrt[9 - x^2] + 9/2 ArcSin[x/3] *)






                            share|improve this answer












                            share|improve this answer



                            share|improve this answer










                            answered May 26 at 6:38









                            CoolwaterCoolwater

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