Fundamental Theorem of Calculus with Different Variables












2












$begingroup$


How can one explain that $$frac{d}{dx}left(int_0^x{cos(t^2+t)dt}right) = cos(x^2+x)$$
Without solving the integral?



I know it's related to the fundamental theorem of calculus, but here we have a derivative with respect to $x$, while the antiderivative is with respect to $t$.



Thank you.






calculus integration derivatives definite-integrals







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    2












    $begingroup$


    How can one explain that $$frac{d}{dx}left(int_0^x{cos(t^2+t)dt}right) = cos(x^2+x)$$
    Without solving the integral?



    I know it's related to the fundamental theorem of calculus, but here we have a derivative with respect to $x$, while the antiderivative is with respect to $t$.



    Thank you.






    calculus integration derivatives definite-integrals







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      How can one explain that $$frac{d}{dx}left(int_0^x{cos(t^2+t)dt}right) = cos(x^2+x)$$
      Without solving the integral?



      I know it's related to the fundamental theorem of calculus, but here we have a derivative with respect to $x$, while the antiderivative is with respect to $t$.



      Thank you.






      calculus integration derivatives definite-integrals







      share|cite|improve this question











      $endgroup$




      How can one explain that $$frac{d}{dx}left(int_0^x{cos(t^2+t)dt}right) = cos(x^2+x)$$
      Without solving the integral?



      I know it's related to the fundamental theorem of calculus, but here we have a derivative with respect to $x$, while the antiderivative is with respect to $t$.



      Thank you.






      calculus integration derivatives definite-integrals




      calculus integration derivatives definite-integrals






      share|cite|improve this question















      share|cite|improve this question













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      edited 21 mins ago









      KM101

      5,9161423




      5,9161423










      asked 31 mins ago









      NetanelNetanel

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          $begingroup$

          Say $f(t)=cos(t^2+t)$ and an antiderivative is $F(t)$. The integral in question is, by the fundamental theorem of calculus,
          $$F(x)-F(0)$$
          $F(0)$ is a constant and disappears upon differentiating with respect to $x$, whereas $F(x)$ becomes $f(x)$ once again. Thus, after differentiation we must have the RHS as $cos(x^2+x)$.






          share|cite|improve this answer









          $endgroup$





















            0












            $begingroup$

            By the fundamental theorem of calculus, $int_a^b{f(t)dt}=F(b)-F(a)$, where $F'(x)=f(x)$. In this case:
            $$int_0^x{cos(t^2+t)dt}=F(x)-F(0)$$
            $$frac{d}{dx}(F(x)-F(0))=F'(x)=f(x)=cos(x^2+x)$$
            $F(0)$ disappears after differentiation because it is a constant






            share|cite|improve this answer









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              2 Answers
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              2 Answers
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              $begingroup$

              Say $f(t)=cos(t^2+t)$ and an antiderivative is $F(t)$. The integral in question is, by the fundamental theorem of calculus,
              $$F(x)-F(0)$$
              $F(0)$ is a constant and disappears upon differentiating with respect to $x$, whereas $F(x)$ becomes $f(x)$ once again. Thus, after differentiation we must have the RHS as $cos(x^2+x)$.






              share|cite|improve this answer









              $endgroup$


















                3












                $begingroup$

                Say $f(t)=cos(t^2+t)$ and an antiderivative is $F(t)$. The integral in question is, by the fundamental theorem of calculus,
                $$F(x)-F(0)$$
                $F(0)$ is a constant and disappears upon differentiating with respect to $x$, whereas $F(x)$ becomes $f(x)$ once again. Thus, after differentiation we must have the RHS as $cos(x^2+x)$.






                share|cite|improve this answer









                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  Say $f(t)=cos(t^2+t)$ and an antiderivative is $F(t)$. The integral in question is, by the fundamental theorem of calculus,
                  $$F(x)-F(0)$$
                  $F(0)$ is a constant and disappears upon differentiating with respect to $x$, whereas $F(x)$ becomes $f(x)$ once again. Thus, after differentiation we must have the RHS as $cos(x^2+x)$.






                  share|cite|improve this answer









                  $endgroup$



                  Say $f(t)=cos(t^2+t)$ and an antiderivative is $F(t)$. The integral in question is, by the fundamental theorem of calculus,
                  $$F(x)-F(0)$$
                  $F(0)$ is a constant and disappears upon differentiating with respect to $x$, whereas $F(x)$ becomes $f(x)$ once again. Thus, after differentiation we must have the RHS as $cos(x^2+x)$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 23 mins ago









                  Parcly TaxelParcly Taxel

                  41.5k137299




                  41.5k137299























                      0












                      $begingroup$

                      By the fundamental theorem of calculus, $int_a^b{f(t)dt}=F(b)-F(a)$, where $F'(x)=f(x)$. In this case:
                      $$int_0^x{cos(t^2+t)dt}=F(x)-F(0)$$
                      $$frac{d}{dx}(F(x)-F(0))=F'(x)=f(x)=cos(x^2+x)$$
                      $F(0)$ disappears after differentiation because it is a constant






                      share|cite|improve this answer









                      $endgroup$


















                        0












                        $begingroup$

                        By the fundamental theorem of calculus, $int_a^b{f(t)dt}=F(b)-F(a)$, where $F'(x)=f(x)$. In this case:
                        $$int_0^x{cos(t^2+t)dt}=F(x)-F(0)$$
                        $$frac{d}{dx}(F(x)-F(0))=F'(x)=f(x)=cos(x^2+x)$$
                        $F(0)$ disappears after differentiation because it is a constant






                        share|cite|improve this answer









                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          By the fundamental theorem of calculus, $int_a^b{f(t)dt}=F(b)-F(a)$, where $F'(x)=f(x)$. In this case:
                          $$int_0^x{cos(t^2+t)dt}=F(x)-F(0)$$
                          $$frac{d}{dx}(F(x)-F(0))=F'(x)=f(x)=cos(x^2+x)$$
                          $F(0)$ disappears after differentiation because it is a constant






                          share|cite|improve this answer









                          $endgroup$



                          By the fundamental theorem of calculus, $int_a^b{f(t)dt}=F(b)-F(a)$, where $F'(x)=f(x)$. In this case:
                          $$int_0^x{cos(t^2+t)dt}=F(x)-F(0)$$
                          $$frac{d}{dx}(F(x)-F(0))=F'(x)=f(x)=cos(x^2+x)$$
                          $F(0)$ disappears after differentiation because it is a constant







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered 16 mins ago









                          Lorenzo B.Lorenzo B.

                          1,8202520




                          1,8202520






























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