Limit of integral as length of interval approaches 0












1














Let $u$ be a continuous real function.



Prove that, for any $t in mathbb{R}$ :



begin{equation} ^{text{lim}}_{h rightarrow{} 0} frac{1}{h} int^{t + h}_{t} u(tau) d tau = u(t)
end{equation}



I would just like to be pointed in the right direction for how to investigate this. Could someone please offer a hint for me to begin? Thank you very much.






real-analysis integration limits functions







share|cite|improve this question




















  • 1




    The denominator probably has $h$ instead of $t$. Try to use L'Hopital's rule.
    – Shubham Johri
    2 hours ago










  • That makes much more sense. The question must have been written incorrectly. Thank you very much
    – David Hughes
    1 hour ago
















1














Let $u$ be a continuous real function.



Prove that, for any $t in mathbb{R}$ :



begin{equation} ^{text{lim}}_{h rightarrow{} 0} frac{1}{h} int^{t + h}_{t} u(tau) d tau = u(t)
end{equation}



I would just like to be pointed in the right direction for how to investigate this. Could someone please offer a hint for me to begin? Thank you very much.






real-analysis integration limits functions







share|cite|improve this question




















  • 1




    The denominator probably has $h$ instead of $t$. Try to use L'Hopital's rule.
    – Shubham Johri
    2 hours ago










  • That makes much more sense. The question must have been written incorrectly. Thank you very much
    – David Hughes
    1 hour ago














1












1








1







Let $u$ be a continuous real function.



Prove that, for any $t in mathbb{R}$ :



begin{equation} ^{text{lim}}_{h rightarrow{} 0} frac{1}{h} int^{t + h}_{t} u(tau) d tau = u(t)
end{equation}



I would just like to be pointed in the right direction for how to investigate this. Could someone please offer a hint for me to begin? Thank you very much.






real-analysis integration limits functions







share|cite|improve this question















Let $u$ be a continuous real function.



Prove that, for any $t in mathbb{R}$ :



begin{equation} ^{text{lim}}_{h rightarrow{} 0} frac{1}{h} int^{t + h}_{t} u(tau) d tau = u(t)
end{equation}



I would just like to be pointed in the right direction for how to investigate this. Could someone please offer a hint for me to begin? Thank you very much.






real-analysis integration limits functions




real-analysis integration limits functions






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share|cite|improve this question













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edited 2 hours ago









clathratus

3,173331




3,173331










asked 2 hours ago









David Hughes

1136




1136








  • 1




    The denominator probably has $h$ instead of $t$. Try to use L'Hopital's rule.
    – Shubham Johri
    2 hours ago










  • That makes much more sense. The question must have been written incorrectly. Thank you very much
    – David Hughes
    1 hour ago














  • 1




    The denominator probably has $h$ instead of $t$. Try to use L'Hopital's rule.
    – Shubham Johri
    2 hours ago










  • That makes much more sense. The question must have been written incorrectly. Thank you very much
    – David Hughes
    1 hour ago








1




1




The denominator probably has $h$ instead of $t$. Try to use L'Hopital's rule.
– Shubham Johri
2 hours ago




The denominator probably has $h$ instead of $t$. Try to use L'Hopital's rule.
– Shubham Johri
2 hours ago












That makes much more sense. The question must have been written incorrectly. Thank you very much
– David Hughes
1 hour ago




That makes much more sense. The question must have been written incorrectly. Thank you very much
– David Hughes
1 hour ago










4 Answers
4






active

oldest

votes


















4














Let $;F;$ be the primitive function of $;u;$ , then



$$frac1hint_t^{t+h} u(tau),dtau=frac{F(t+h)-F(t)}h$$



Now just do the limit...






share|cite|improve this answer





























    2














    The limit has the $0/0$ form. Use the L'Hopital's rule,
    $$lim_{hto 0}frac{displaystyleint^{t + h}_tu(tau)dtau}h=lim_{hto 0}u(t+h)=u(t),$$ since $u(x)$ is continuous over $Bbb R$.






    share|cite|improve this answer





























      1














      This is what is usually called The Fundamental Theorem of Calculus part 1:




      FTOC Part 1: Let the function $f:[a, b] to mathbb {R} $ be Riemann integrable on $[a, b] $. Then the function $F:[a, b] tomathbb {R} $ defined by $$F(x) =int_{a} ^{x} f(t) , dt$$ is continuous on $[a, b] $ and if $f$ is continuous at some point $cin[a, b] $ then $F$ is differentiable at $c$ with derivative $F'(c) =f(c) $.




      For your question let $$U(t) =int _{0}^{t}u(tau),dtau$$ then the limit in your question is (by definition of derivative) $U'(t) $ and by FTOC part 1 this equals $u(t) $.






      share|cite|improve this answer





























        0














        MVT:



        $displaystyle{int_{t}^{t+h}}u(x)dx=hu(xi)$, where $xi in [t,t+h].$



        $lim_{h rightarrow } xi (h)= t.$



        Since $u$ is continuous:



        $lim_{h rightarrow 0}(1/h)displaystyle{int_{t}^{t+h}}u(x)dx =$



        $lim_{h rightarrow 0}u(xi(h))=$



        $u(lim_{h rightarrow 0}xi(h))=u(t).$






        share|cite





















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          4 Answers
          4






          active

          oldest

          votes








          4 Answers
          4






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          4














          Let $;F;$ be the primitive function of $;u;$ , then



          $$frac1hint_t^{t+h} u(tau),dtau=frac{F(t+h)-F(t)}h$$



          Now just do the limit...






          share|cite|improve this answer


























            4














            Let $;F;$ be the primitive function of $;u;$ , then



            $$frac1hint_t^{t+h} u(tau),dtau=frac{F(t+h)-F(t)}h$$



            Now just do the limit...






            share|cite|improve this answer
























              4












              4








              4






              Let $;F;$ be the primitive function of $;u;$ , then



              $$frac1hint_t^{t+h} u(tau),dtau=frac{F(t+h)-F(t)}h$$



              Now just do the limit...






              share|cite|improve this answer












              Let $;F;$ be the primitive function of $;u;$ , then



              $$frac1hint_t^{t+h} u(tau),dtau=frac{F(t+h)-F(t)}h$$



              Now just do the limit...







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered 2 hours ago









              DonAntonio

              177k1491225




              177k1491225























                  2














                  The limit has the $0/0$ form. Use the L'Hopital's rule,
                  $$lim_{hto 0}frac{displaystyleint^{t + h}_tu(tau)dtau}h=lim_{hto 0}u(t+h)=u(t),$$ since $u(x)$ is continuous over $Bbb R$.






                  share|cite|improve this answer


























                    2














                    The limit has the $0/0$ form. Use the L'Hopital's rule,
                    $$lim_{hto 0}frac{displaystyleint^{t + h}_tu(tau)dtau}h=lim_{hto 0}u(t+h)=u(t),$$ since $u(x)$ is continuous over $Bbb R$.






                    share|cite|improve this answer
























                      2












                      2








                      2






                      The limit has the $0/0$ form. Use the L'Hopital's rule,
                      $$lim_{hto 0}frac{displaystyleint^{t + h}_tu(tau)dtau}h=lim_{hto 0}u(t+h)=u(t),$$ since $u(x)$ is continuous over $Bbb R$.






                      share|cite|improve this answer












                      The limit has the $0/0$ form. Use the L'Hopital's rule,
                      $$lim_{hto 0}frac{displaystyleint^{t + h}_tu(tau)dtau}h=lim_{hto 0}u(t+h)=u(t),$$ since $u(x)$ is continuous over $Bbb R$.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered 2 hours ago









                      Shubham Johri

                      3,918716




                      3,918716























                          1














                          This is what is usually called The Fundamental Theorem of Calculus part 1:




                          FTOC Part 1: Let the function $f:[a, b] to mathbb {R} $ be Riemann integrable on $[a, b] $. Then the function $F:[a, b] tomathbb {R} $ defined by $$F(x) =int_{a} ^{x} f(t) , dt$$ is continuous on $[a, b] $ and if $f$ is continuous at some point $cin[a, b] $ then $F$ is differentiable at $c$ with derivative $F'(c) =f(c) $.




                          For your question let $$U(t) =int _{0}^{t}u(tau),dtau$$ then the limit in your question is (by definition of derivative) $U'(t) $ and by FTOC part 1 this equals $u(t) $.






                          share|cite|improve this answer


























                            1














                            This is what is usually called The Fundamental Theorem of Calculus part 1:




                            FTOC Part 1: Let the function $f:[a, b] to mathbb {R} $ be Riemann integrable on $[a, b] $. Then the function $F:[a, b] tomathbb {R} $ defined by $$F(x) =int_{a} ^{x} f(t) , dt$$ is continuous on $[a, b] $ and if $f$ is continuous at some point $cin[a, b] $ then $F$ is differentiable at $c$ with derivative $F'(c) =f(c) $.




                            For your question let $$U(t) =int _{0}^{t}u(tau),dtau$$ then the limit in your question is (by definition of derivative) $U'(t) $ and by FTOC part 1 this equals $u(t) $.






                            share|cite|improve this answer
























                              1












                              1








                              1






                              This is what is usually called The Fundamental Theorem of Calculus part 1:




                              FTOC Part 1: Let the function $f:[a, b] to mathbb {R} $ be Riemann integrable on $[a, b] $. Then the function $F:[a, b] tomathbb {R} $ defined by $$F(x) =int_{a} ^{x} f(t) , dt$$ is continuous on $[a, b] $ and if $f$ is continuous at some point $cin[a, b] $ then $F$ is differentiable at $c$ with derivative $F'(c) =f(c) $.




                              For your question let $$U(t) =int _{0}^{t}u(tau),dtau$$ then the limit in your question is (by definition of derivative) $U'(t) $ and by FTOC part 1 this equals $u(t) $.






                              share|cite|improve this answer












                              This is what is usually called The Fundamental Theorem of Calculus part 1:




                              FTOC Part 1: Let the function $f:[a, b] to mathbb {R} $ be Riemann integrable on $[a, b] $. Then the function $F:[a, b] tomathbb {R} $ defined by $$F(x) =int_{a} ^{x} f(t) , dt$$ is continuous on $[a, b] $ and if $f$ is continuous at some point $cin[a, b] $ then $F$ is differentiable at $c$ with derivative $F'(c) =f(c) $.




                              For your question let $$U(t) =int _{0}^{t}u(tau),dtau$$ then the limit in your question is (by definition of derivative) $U'(t) $ and by FTOC part 1 this equals $u(t) $.







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered 44 mins ago









                              Paramanand Singh

                              48.9k555159




                              48.9k555159























                                  0














                                  MVT:



                                  $displaystyle{int_{t}^{t+h}}u(x)dx=hu(xi)$, where $xi in [t,t+h].$



                                  $lim_{h rightarrow } xi (h)= t.$



                                  Since $u$ is continuous:



                                  $lim_{h rightarrow 0}(1/h)displaystyle{int_{t}^{t+h}}u(x)dx =$



                                  $lim_{h rightarrow 0}u(xi(h))=$



                                  $u(lim_{h rightarrow 0}xi(h))=u(t).$






                                  share|cite


























                                    0














                                    MVT:



                                    $displaystyle{int_{t}^{t+h}}u(x)dx=hu(xi)$, where $xi in [t,t+h].$



                                    $lim_{h rightarrow } xi (h)= t.$



                                    Since $u$ is continuous:



                                    $lim_{h rightarrow 0}(1/h)displaystyle{int_{t}^{t+h}}u(x)dx =$



                                    $lim_{h rightarrow 0}u(xi(h))=$



                                    $u(lim_{h rightarrow 0}xi(h))=u(t).$






                                    share|cite
























                                      0












                                      0








                                      0






                                      MVT:



                                      $displaystyle{int_{t}^{t+h}}u(x)dx=hu(xi)$, where $xi in [t,t+h].$



                                      $lim_{h rightarrow } xi (h)= t.$



                                      Since $u$ is continuous:



                                      $lim_{h rightarrow 0}(1/h)displaystyle{int_{t}^{t+h}}u(x)dx =$



                                      $lim_{h rightarrow 0}u(xi(h))=$



                                      $u(lim_{h rightarrow 0}xi(h))=u(t).$






                                      share|cite












                                      MVT:



                                      $displaystyle{int_{t}^{t+h}}u(x)dx=hu(xi)$, where $xi in [t,t+h].$



                                      $lim_{h rightarrow } xi (h)= t.$



                                      Since $u$ is continuous:



                                      $lim_{h rightarrow 0}(1/h)displaystyle{int_{t}^{t+h}}u(x)dx =$



                                      $lim_{h rightarrow 0}u(xi(h))=$



                                      $u(lim_{h rightarrow 0}xi(h))=u(t).$







                                      share|cite












                                      share|cite



                                      share|cite










                                      answered 3 mins ago









                                      Peter Szilas

                                      10.7k2720




                                      10.7k2720






























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