Fundamental Theorem of Calculus I help!












4












$begingroup$


If $F(x)=int_2^{x^3}sqrt{t^2+t^4}dt$



a.) The integral of $F(x)$ is $3x^2sqrt{x^2+x^4}$.



b.) The derivative of $F(x)$ does not exist.



c.) $F'(x)=3x^2sqrt{x^6+x^{12}}$.



I can't seem to find the answer. I found that
$$F(x)=int_2^{x^3}sqrt{t^2+t^4}dt= frac{(x^6+1)^{3/2}}{3}-frac{5
sqrt{5}}{3}$$



and $$frac{d}{dx}F(x)=int_2^{x^3}sqrt{t^2+t^4}dt=3x^5sqrt{x^6+1}$$



so, to me the answer is not provided. Any help would be much appreciated.






real-analysis calculus definite-integrals







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Ryan Pennell is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







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    4












    $begingroup$


    If $F(x)=int_2^{x^3}sqrt{t^2+t^4}dt$



    a.) The integral of $F(x)$ is $3x^2sqrt{x^2+x^4}$.



    b.) The derivative of $F(x)$ does not exist.



    c.) $F'(x)=3x^2sqrt{x^6+x^{12}}$.



    I can't seem to find the answer. I found that
    $$F(x)=int_2^{x^3}sqrt{t^2+t^4}dt= frac{(x^6+1)^{3/2}}{3}-frac{5
    sqrt{5}}{3}$$



    and $$frac{d}{dx}F(x)=int_2^{x^3}sqrt{t^2+t^4}dt=3x^5sqrt{x^6+1}$$



    so, to me the answer is not provided. Any help would be much appreciated.






    real-analysis calculus definite-integrals







    share|cite|improve this question







    New contributor




    Ryan Pennell is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.







    $endgroup$















      4












      4








      4


      1



      $begingroup$


      If $F(x)=int_2^{x^3}sqrt{t^2+t^4}dt$



      a.) The integral of $F(x)$ is $3x^2sqrt{x^2+x^4}$.



      b.) The derivative of $F(x)$ does not exist.



      c.) $F'(x)=3x^2sqrt{x^6+x^{12}}$.



      I can't seem to find the answer. I found that
      $$F(x)=int_2^{x^3}sqrt{t^2+t^4}dt= frac{(x^6+1)^{3/2}}{3}-frac{5
      sqrt{5}}{3}$$



      and $$frac{d}{dx}F(x)=int_2^{x^3}sqrt{t^2+t^4}dt=3x^5sqrt{x^6+1}$$



      so, to me the answer is not provided. Any help would be much appreciated.






      real-analysis calculus definite-integrals







      share|cite|improve this question







      New contributor




      Ryan Pennell is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.







      $endgroup$




      If $F(x)=int_2^{x^3}sqrt{t^2+t^4}dt$



      a.) The integral of $F(x)$ is $3x^2sqrt{x^2+x^4}$.



      b.) The derivative of $F(x)$ does not exist.



      c.) $F'(x)=3x^2sqrt{x^6+x^{12}}$.



      I can't seem to find the answer. I found that
      $$F(x)=int_2^{x^3}sqrt{t^2+t^4}dt= frac{(x^6+1)^{3/2}}{3}-frac{5
      sqrt{5}}{3}$$



      and $$frac{d}{dx}F(x)=int_2^{x^3}sqrt{t^2+t^4}dt=3x^5sqrt{x^6+1}$$



      so, to me the answer is not provided. Any help would be much appreciated.






      real-analysis calculus definite-integrals




      real-analysis calculus definite-integrals






      share|cite|improve this question







      New contributor




      Ryan Pennell is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|cite|improve this question







      New contributor




      Ryan Pennell is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      share|cite|improve this question




      share|cite|improve this question






      New contributor




      Ryan Pennell is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      asked 4 hours ago









      Ryan PennellRyan Pennell

      212




      212




      New contributor




      Ryan Pennell is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.





      New contributor





      Ryan Pennell is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






      Ryan Pennell is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






















          2 Answers
          2






          active

          oldest

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          5












          $begingroup$

          Let's say $G(x)$ is an antiderivative of $sqrt{x^2+x^4}$. (we know about its existence from the fundamental theorem of calculus). So now from Newton-Leibniz formula we have $F(x)=G(x^3)-G(2)$. Now if we differentiate by $x$ using the chain rule we get:



          $F'(x)=3x^2G'(x^3)=3x^2sqrt{x^6+x^{12}}$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you Mark!
            $endgroup$
            – Ryan Pennell
            4 hours ago



















          2












          $begingroup$

          Hint:
          If $F(x)= int_a^{g(x)} f(t) dt$ , then
          $$F’(x) = f(g(x))g’(x)$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you Jose!
            $endgroup$
            – Ryan Pennell
            4 hours ago










          • $begingroup$
            Your welcome, notice that with this method is not necessary to calculate an undefined integral, which is useful because is not always possible (see $int e^{-x^2} dx$) @RyanPennell
            $endgroup$
            – JoseSquare
            3 hours ago













          Your Answer





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          2 Answers
          2






          active

          oldest

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          2 Answers
          2






          active

          oldest

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          active

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          active

          oldest

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          5












          $begingroup$

          Let's say $G(x)$ is an antiderivative of $sqrt{x^2+x^4}$. (we know about its existence from the fundamental theorem of calculus). So now from Newton-Leibniz formula we have $F(x)=G(x^3)-G(2)$. Now if we differentiate by $x$ using the chain rule we get:



          $F'(x)=3x^2G'(x^3)=3x^2sqrt{x^6+x^{12}}$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you Mark!
            $endgroup$
            – Ryan Pennell
            4 hours ago
















          5












          $begingroup$

          Let's say $G(x)$ is an antiderivative of $sqrt{x^2+x^4}$. (we know about its existence from the fundamental theorem of calculus). So now from Newton-Leibniz formula we have $F(x)=G(x^3)-G(2)$. Now if we differentiate by $x$ using the chain rule we get:



          $F'(x)=3x^2G'(x^3)=3x^2sqrt{x^6+x^{12}}$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you Mark!
            $endgroup$
            – Ryan Pennell
            4 hours ago














          5












          5








          5





          $begingroup$

          Let's say $G(x)$ is an antiderivative of $sqrt{x^2+x^4}$. (we know about its existence from the fundamental theorem of calculus). So now from Newton-Leibniz formula we have $F(x)=G(x^3)-G(2)$. Now if we differentiate by $x$ using the chain rule we get:



          $F'(x)=3x^2G'(x^3)=3x^2sqrt{x^6+x^{12}}$






          share|cite|improve this answer









          $endgroup$



          Let's say $G(x)$ is an antiderivative of $sqrt{x^2+x^4}$. (we know about its existence from the fundamental theorem of calculus). So now from Newton-Leibniz formula we have $F(x)=G(x^3)-G(2)$. Now if we differentiate by $x$ using the chain rule we get:



          $F'(x)=3x^2G'(x^3)=3x^2sqrt{x^6+x^{12}}$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 4 hours ago









          MarkMark

          8,907521




          8,907521












          • $begingroup$
            Thank you Mark!
            $endgroup$
            – Ryan Pennell
            4 hours ago


















          • $begingroup$
            Thank you Mark!
            $endgroup$
            – Ryan Pennell
            4 hours ago
















          $begingroup$
          Thank you Mark!
          $endgroup$
          – Ryan Pennell
          4 hours ago




          $begingroup$
          Thank you Mark!
          $endgroup$
          – Ryan Pennell
          4 hours ago











          2












          $begingroup$

          Hint:
          If $F(x)= int_a^{g(x)} f(t) dt$ , then
          $$F’(x) = f(g(x))g’(x)$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you Jose!
            $endgroup$
            – Ryan Pennell
            4 hours ago










          • $begingroup$
            Your welcome, notice that with this method is not necessary to calculate an undefined integral, which is useful because is not always possible (see $int e^{-x^2} dx$) @RyanPennell
            $endgroup$
            – JoseSquare
            3 hours ago


















          2












          $begingroup$

          Hint:
          If $F(x)= int_a^{g(x)} f(t) dt$ , then
          $$F’(x) = f(g(x))g’(x)$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you Jose!
            $endgroup$
            – Ryan Pennell
            4 hours ago










          • $begingroup$
            Your welcome, notice that with this method is not necessary to calculate an undefined integral, which is useful because is not always possible (see $int e^{-x^2} dx$) @RyanPennell
            $endgroup$
            – JoseSquare
            3 hours ago
















          2












          2








          2





          $begingroup$

          Hint:
          If $F(x)= int_a^{g(x)} f(t) dt$ , then
          $$F’(x) = f(g(x))g’(x)$$






          share|cite|improve this answer









          $endgroup$



          Hint:
          If $F(x)= int_a^{g(x)} f(t) dt$ , then
          $$F’(x) = f(g(x))g’(x)$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 4 hours ago









          JoseSquareJoseSquare

          66012




          66012












          • $begingroup$
            Thank you Jose!
            $endgroup$
            – Ryan Pennell
            4 hours ago










          • $begingroup$
            Your welcome, notice that with this method is not necessary to calculate an undefined integral, which is useful because is not always possible (see $int e^{-x^2} dx$) @RyanPennell
            $endgroup$
            – JoseSquare
            3 hours ago




















          • $begingroup$
            Thank you Jose!
            $endgroup$
            – Ryan Pennell
            4 hours ago










          • $begingroup$
            Your welcome, notice that with this method is not necessary to calculate an undefined integral, which is useful because is not always possible (see $int e^{-x^2} dx$) @RyanPennell
            $endgroup$
            – JoseSquare
            3 hours ago


















          $begingroup$
          Thank you Jose!
          $endgroup$
          – Ryan Pennell
          4 hours ago




          $begingroup$
          Thank you Jose!
          $endgroup$
          – Ryan Pennell
          4 hours ago












          $begingroup$
          Your welcome, notice that with this method is not necessary to calculate an undefined integral, which is useful because is not always possible (see $int e^{-x^2} dx$) @RyanPennell
          $endgroup$
          – JoseSquare
          3 hours ago






          $begingroup$
          Your welcome, notice that with this method is not necessary to calculate an undefined integral, which is useful because is not always possible (see $int e^{-x^2} dx$) @RyanPennell
          $endgroup$
          – JoseSquare
          3 hours ago












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