Suppose $f:Bbb{R}rightarrowBbb{R}$ is continuous. Define $G(x)=int_0^x(x-t)f(t)dt$ for all $xinBbb{R}$. Prove...












4












$begingroup$


This is a tricky example of the Fundamental Theorem of Calculus 1. I am looking for confirmation of my proof or answer to see if I am doing it correctly because this is different than the basic applications.



$underline{Proof}$



$G(x)=int_0^x(x-t)f(t)dtrightarrow G(x)=int_0^xxf(t)dt-int_0^xtf(t)dt$.



So, let $Q(x)=int_0^xf(t)dt$ and let $P(x)=int_0^xtf(t)dt$ then we have,



$G(x)=xQ(x)-P(x)$



$G'(x)=xQ'(x)+Q(x)-P'(x)=xf(x)+Q(x)-xf(x)=Q(x)$



hence,



$G''(x)=Q'(x)=f(x)$



∎






real-analysis calculus







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$endgroup$












  • $begingroup$
    Looks fine. You can also use the Leibniz integral rule.
    $endgroup$
    – Mattos
    Dec 6 '18 at 6:15












  • $begingroup$
    @Mattos, thank you I will also try this problem with Leibniz for practise
    $endgroup$
    – Albert Diaz
    Dec 6 '18 at 6:22
















4












$begingroup$


This is a tricky example of the Fundamental Theorem of Calculus 1. I am looking for confirmation of my proof or answer to see if I am doing it correctly because this is different than the basic applications.



$underline{Proof}$



$G(x)=int_0^x(x-t)f(t)dtrightarrow G(x)=int_0^xxf(t)dt-int_0^xtf(t)dt$.



So, let $Q(x)=int_0^xf(t)dt$ and let $P(x)=int_0^xtf(t)dt$ then we have,



$G(x)=xQ(x)-P(x)$



$G'(x)=xQ'(x)+Q(x)-P'(x)=xf(x)+Q(x)-xf(x)=Q(x)$



hence,



$G''(x)=Q'(x)=f(x)$



∎






real-analysis calculus







share|cite|improve this question









$endgroup$












  • $begingroup$
    Looks fine. You can also use the Leibniz integral rule.
    $endgroup$
    – Mattos
    Dec 6 '18 at 6:15












  • $begingroup$
    @Mattos, thank you I will also try this problem with Leibniz for practise
    $endgroup$
    – Albert Diaz
    Dec 6 '18 at 6:22














4












4








4





$begingroup$


This is a tricky example of the Fundamental Theorem of Calculus 1. I am looking for confirmation of my proof or answer to see if I am doing it correctly because this is different than the basic applications.



$underline{Proof}$



$G(x)=int_0^x(x-t)f(t)dtrightarrow G(x)=int_0^xxf(t)dt-int_0^xtf(t)dt$.



So, let $Q(x)=int_0^xf(t)dt$ and let $P(x)=int_0^xtf(t)dt$ then we have,



$G(x)=xQ(x)-P(x)$



$G'(x)=xQ'(x)+Q(x)-P'(x)=xf(x)+Q(x)-xf(x)=Q(x)$



hence,



$G''(x)=Q'(x)=f(x)$



∎






real-analysis calculus







share|cite|improve this question









$endgroup$




This is a tricky example of the Fundamental Theorem of Calculus 1. I am looking for confirmation of my proof or answer to see if I am doing it correctly because this is different than the basic applications.



$underline{Proof}$



$G(x)=int_0^x(x-t)f(t)dtrightarrow G(x)=int_0^xxf(t)dt-int_0^xtf(t)dt$.



So, let $Q(x)=int_0^xf(t)dt$ and let $P(x)=int_0^xtf(t)dt$ then we have,



$G(x)=xQ(x)-P(x)$



$G'(x)=xQ'(x)+Q(x)-P'(x)=xf(x)+Q(x)-xf(x)=Q(x)$



hence,



$G''(x)=Q'(x)=f(x)$



∎






real-analysis calculus




real-analysis calculus






share|cite|improve this question













share|cite|improve this question











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asked Dec 6 '18 at 6:12









Albert DiazAlbert Diaz

925




925












  • $begingroup$
    Looks fine. You can also use the Leibniz integral rule.
    $endgroup$
    – Mattos
    Dec 6 '18 at 6:15












  • $begingroup$
    @Mattos, thank you I will also try this problem with Leibniz for practise
    $endgroup$
    – Albert Diaz
    Dec 6 '18 at 6:22


















  • $begingroup$
    Looks fine. You can also use the Leibniz integral rule.
    $endgroup$
    – Mattos
    Dec 6 '18 at 6:15












  • $begingroup$
    @Mattos, thank you I will also try this problem with Leibniz for practise
    $endgroup$
    – Albert Diaz
    Dec 6 '18 at 6:22
















$begingroup$
Looks fine. You can also use the Leibniz integral rule.
$endgroup$
– Mattos
Dec 6 '18 at 6:15






$begingroup$
Looks fine. You can also use the Leibniz integral rule.
$endgroup$
– Mattos
Dec 6 '18 at 6:15














$begingroup$
@Mattos, thank you I will also try this problem with Leibniz for practise
$endgroup$
– Albert Diaz
Dec 6 '18 at 6:22




$begingroup$
@Mattos, thank you I will also try this problem with Leibniz for practise
$endgroup$
– Albert Diaz
Dec 6 '18 at 6:22










1 Answer
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$begingroup$

Your proof is correct. An alternative is to use integration by parts. With
$$
f_1(x) = int_0^x f(t) , dt , , quad f_2(x) = int_0^x f_1(t) , dt
$$
we have
$$
G(x) = bigllbrack (x-t)f_1(t) bigrrbrack_{t=0}^{t=x} +
int_0^x f_1(t) , dt = f_2(x)
$$
and therefore $G'' = f_2'' = f_1' = f$.



This generalizes to higher anti-derivatives:
$$
int_0^x frac{(x-t)^{n-1}}{(n-1)!} f(t) , dt
= int_0^x int_0^{t_1} cdots int_0^{t_{n-1}} f(t_n) , dt_n cdots dt_1
$$
for $n ge 1$.






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    $begingroup$

    Your proof is correct. An alternative is to use integration by parts. With
    $$
    f_1(x) = int_0^x f(t) , dt , , quad f_2(x) = int_0^x f_1(t) , dt
    $$
    we have
    $$
    G(x) = bigllbrack (x-t)f_1(t) bigrrbrack_{t=0}^{t=x} +
    int_0^x f_1(t) , dt = f_2(x)
    $$
    and therefore $G'' = f_2'' = f_1' = f$.



    This generalizes to higher anti-derivatives:
    $$
    int_0^x frac{(x-t)^{n-1}}{(n-1)!} f(t) , dt
    = int_0^x int_0^{t_1} cdots int_0^{t_{n-1}} f(t_n) , dt_n cdots dt_1
    $$
    for $n ge 1$.






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      Your proof is correct. An alternative is to use integration by parts. With
      $$
      f_1(x) = int_0^x f(t) , dt , , quad f_2(x) = int_0^x f_1(t) , dt
      $$
      we have
      $$
      G(x) = bigllbrack (x-t)f_1(t) bigrrbrack_{t=0}^{t=x} +
      int_0^x f_1(t) , dt = f_2(x)
      $$
      and therefore $G'' = f_2'' = f_1' = f$.



      This generalizes to higher anti-derivatives:
      $$
      int_0^x frac{(x-t)^{n-1}}{(n-1)!} f(t) , dt
      = int_0^x int_0^{t_1} cdots int_0^{t_{n-1}} f(t_n) , dt_n cdots dt_1
      $$
      for $n ge 1$.






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        Your proof is correct. An alternative is to use integration by parts. With
        $$
        f_1(x) = int_0^x f(t) , dt , , quad f_2(x) = int_0^x f_1(t) , dt
        $$
        we have
        $$
        G(x) = bigllbrack (x-t)f_1(t) bigrrbrack_{t=0}^{t=x} +
        int_0^x f_1(t) , dt = f_2(x)
        $$
        and therefore $G'' = f_2'' = f_1' = f$.



        This generalizes to higher anti-derivatives:
        $$
        int_0^x frac{(x-t)^{n-1}}{(n-1)!} f(t) , dt
        = int_0^x int_0^{t_1} cdots int_0^{t_{n-1}} f(t_n) , dt_n cdots dt_1
        $$
        for $n ge 1$.






        share|cite|improve this answer











        $endgroup$



        Your proof is correct. An alternative is to use integration by parts. With
        $$
        f_1(x) = int_0^x f(t) , dt , , quad f_2(x) = int_0^x f_1(t) , dt
        $$
        we have
        $$
        G(x) = bigllbrack (x-t)f_1(t) bigrrbrack_{t=0}^{t=x} +
        int_0^x f_1(t) , dt = f_2(x)
        $$
        and therefore $G'' = f_2'' = f_1' = f$.



        This generalizes to higher anti-derivatives:
        $$
        int_0^x frac{(x-t)^{n-1}}{(n-1)!} f(t) , dt
        = int_0^x int_0^{t_1} cdots int_0^{t_{n-1}} f(t_n) , dt_n cdots dt_1
        $$
        for $n ge 1$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 6 '18 at 7:37

























        answered Dec 6 '18 at 6:53









        Martin RMartin R

        27.9k33255




        27.9k33255






























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