Integrate $int_{-1}^{1}x^2sin^{2019}(x)e^{-x^4}dx$












1












$begingroup$


I'm following a course on analysis and I am supposed to compute the following integral:



$int_{-1}^{1}x^2sin^{2019}(x)e^{-x^4}dx$.



I've been trying to use the integration by part but I've been stuck for a while. Any help would be highly appreciated.
Thanks










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  • $begingroup$
    Use math.stackexchange.com/questions/439851/… and the given function is odd
    $endgroup$
    – lab bhattacharjee
    Dec 31 '18 at 14:18
















1












$begingroup$


I'm following a course on analysis and I am supposed to compute the following integral:



$int_{-1}^{1}x^2sin^{2019}(x)e^{-x^4}dx$.



I've been trying to use the integration by part but I've been stuck for a while. Any help would be highly appreciated.
Thanks










share|cite|improve this question











$endgroup$












  • $begingroup$
    Use math.stackexchange.com/questions/439851/… and the given function is odd
    $endgroup$
    – lab bhattacharjee
    Dec 31 '18 at 14:18














1












1








1





$begingroup$


I'm following a course on analysis and I am supposed to compute the following integral:



$int_{-1}^{1}x^2sin^{2019}(x)e^{-x^4}dx$.



I've been trying to use the integration by part but I've been stuck for a while. Any help would be highly appreciated.
Thanks










share|cite|improve this question











$endgroup$




I'm following a course on analysis and I am supposed to compute the following integral:



$int_{-1}^{1}x^2sin^{2019}(x)e^{-x^4}dx$.



I've been trying to use the integration by part but I've been stuck for a while. Any help would be highly appreciated.
Thanks







real-analysis calculus integration






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share|cite|improve this question













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share|cite|improve this question








edited Dec 31 '18 at 14:21









Bernard

119k740113




119k740113










asked Dec 31 '18 at 14:15









jffijffi

828




828












  • $begingroup$
    Use math.stackexchange.com/questions/439851/… and the given function is odd
    $endgroup$
    – lab bhattacharjee
    Dec 31 '18 at 14:18


















  • $begingroup$
    Use math.stackexchange.com/questions/439851/… and the given function is odd
    $endgroup$
    – lab bhattacharjee
    Dec 31 '18 at 14:18
















$begingroup$
Use math.stackexchange.com/questions/439851/… and the given function is odd
$endgroup$
– lab bhattacharjee
Dec 31 '18 at 14:18




$begingroup$
Use math.stackexchange.com/questions/439851/… and the given function is odd
$endgroup$
– lab bhattacharjee
Dec 31 '18 at 14:18










2 Answers
2






active

oldest

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9












$begingroup$

Just substitute $x=-t$ and you will get that: $$I=int_{-1}^{1}x^2sin^{2019}(x)e^{-x^4}dx=-int_{-1}^{1}t^2sin^{2019}(t)e^{-t^4}dt=-I$$
$$I=-IRightarrow I=0$$






share|cite|improve this answer









$endgroup$









  • 2




    $begingroup$
    Oh... right! I can't believe I was actually trying to compute the primitive. Thanks
    $endgroup$
    – jffi
    Dec 31 '18 at 14:26








  • 1




    $begingroup$
    I believe it's a good thing that you already tried to compute the primitive, that is how in the future you will be able to see faster when you can find a primitive for an integral, by trying and failing in the past and earning experience. So no worries at all, keep it up!
    $endgroup$
    – Zacky
    Dec 31 '18 at 14:32





















6












$begingroup$

The integrand is an odd function.



For any odd (integrable) function $f$, we have



$$int_{-a}^a f = 0$$



Therefore, the integral is equal to $0$.






share|cite|improve this answer









$endgroup$













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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    9












    $begingroup$

    Just substitute $x=-t$ and you will get that: $$I=int_{-1}^{1}x^2sin^{2019}(x)e^{-x^4}dx=-int_{-1}^{1}t^2sin^{2019}(t)e^{-t^4}dt=-I$$
    $$I=-IRightarrow I=0$$






    share|cite|improve this answer









    $endgroup$









    • 2




      $begingroup$
      Oh... right! I can't believe I was actually trying to compute the primitive. Thanks
      $endgroup$
      – jffi
      Dec 31 '18 at 14:26








    • 1




      $begingroup$
      I believe it's a good thing that you already tried to compute the primitive, that is how in the future you will be able to see faster when you can find a primitive for an integral, by trying and failing in the past and earning experience. So no worries at all, keep it up!
      $endgroup$
      – Zacky
      Dec 31 '18 at 14:32


















    9












    $begingroup$

    Just substitute $x=-t$ and you will get that: $$I=int_{-1}^{1}x^2sin^{2019}(x)e^{-x^4}dx=-int_{-1}^{1}t^2sin^{2019}(t)e^{-t^4}dt=-I$$
    $$I=-IRightarrow I=0$$






    share|cite|improve this answer









    $endgroup$









    • 2




      $begingroup$
      Oh... right! I can't believe I was actually trying to compute the primitive. Thanks
      $endgroup$
      – jffi
      Dec 31 '18 at 14:26








    • 1




      $begingroup$
      I believe it's a good thing that you already tried to compute the primitive, that is how in the future you will be able to see faster when you can find a primitive for an integral, by trying and failing in the past and earning experience. So no worries at all, keep it up!
      $endgroup$
      – Zacky
      Dec 31 '18 at 14:32
















    9












    9








    9





    $begingroup$

    Just substitute $x=-t$ and you will get that: $$I=int_{-1}^{1}x^2sin^{2019}(x)e^{-x^4}dx=-int_{-1}^{1}t^2sin^{2019}(t)e^{-t^4}dt=-I$$
    $$I=-IRightarrow I=0$$






    share|cite|improve this answer









    $endgroup$



    Just substitute $x=-t$ and you will get that: $$I=int_{-1}^{1}x^2sin^{2019}(x)e^{-x^4}dx=-int_{-1}^{1}t^2sin^{2019}(t)e^{-t^4}dt=-I$$
    $$I=-IRightarrow I=0$$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 31 '18 at 14:20









    ZackyZacky

    5,4661856




    5,4661856








    • 2




      $begingroup$
      Oh... right! I can't believe I was actually trying to compute the primitive. Thanks
      $endgroup$
      – jffi
      Dec 31 '18 at 14:26








    • 1




      $begingroup$
      I believe it's a good thing that you already tried to compute the primitive, that is how in the future you will be able to see faster when you can find a primitive for an integral, by trying and failing in the past and earning experience. So no worries at all, keep it up!
      $endgroup$
      – Zacky
      Dec 31 '18 at 14:32
















    • 2




      $begingroup$
      Oh... right! I can't believe I was actually trying to compute the primitive. Thanks
      $endgroup$
      – jffi
      Dec 31 '18 at 14:26








    • 1




      $begingroup$
      I believe it's a good thing that you already tried to compute the primitive, that is how in the future you will be able to see faster when you can find a primitive for an integral, by trying and failing in the past and earning experience. So no worries at all, keep it up!
      $endgroup$
      – Zacky
      Dec 31 '18 at 14:32










    2




    2




    $begingroup$
    Oh... right! I can't believe I was actually trying to compute the primitive. Thanks
    $endgroup$
    – jffi
    Dec 31 '18 at 14:26






    $begingroup$
    Oh... right! I can't believe I was actually trying to compute the primitive. Thanks
    $endgroup$
    – jffi
    Dec 31 '18 at 14:26






    1




    1




    $begingroup$
    I believe it's a good thing that you already tried to compute the primitive, that is how in the future you will be able to see faster when you can find a primitive for an integral, by trying and failing in the past and earning experience. So no worries at all, keep it up!
    $endgroup$
    – Zacky
    Dec 31 '18 at 14:32






    $begingroup$
    I believe it's a good thing that you already tried to compute the primitive, that is how in the future you will be able to see faster when you can find a primitive for an integral, by trying and failing in the past and earning experience. So no worries at all, keep it up!
    $endgroup$
    – Zacky
    Dec 31 '18 at 14:32













    6












    $begingroup$

    The integrand is an odd function.



    For any odd (integrable) function $f$, we have



    $$int_{-a}^a f = 0$$



    Therefore, the integral is equal to $0$.






    share|cite|improve this answer









    $endgroup$


















      6












      $begingroup$

      The integrand is an odd function.



      For any odd (integrable) function $f$, we have



      $$int_{-a}^a f = 0$$



      Therefore, the integral is equal to $0$.






      share|cite|improve this answer









      $endgroup$
















        6












        6








        6





        $begingroup$

        The integrand is an odd function.



        For any odd (integrable) function $f$, we have



        $$int_{-a}^a f = 0$$



        Therefore, the integral is equal to $0$.






        share|cite|improve this answer









        $endgroup$



        The integrand is an odd function.



        For any odd (integrable) function $f$, we have



        $$int_{-a}^a f = 0$$



        Therefore, the integral is equal to $0$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 31 '18 at 14:20









        Math_QEDMath_QED

        7,39531450




        7,39531450






























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