Fundamental Theorem of Calculus: Derivative of integral with respect to parameter in integration limits












1












$begingroup$


I have the following function:



$$F(t) = int_{a-t}^{b+t} f(x) dx$$



In my notes when this is differentiated with respect to $t$ we get:
$$frac{d F}{d t} = f(b + t) + f(a - t)$$



How can we obtain this result?



My approach



The approach that I was thinking about was about splitting the integral into two based on the integration limits, for instance, something like:



$$F_1(t) = int_{a-t}^c f(x) dx quad quad F_2(t) = int_{c}^{b+t} f(x)dx$$



So that $F(t) = F_1(t) + F_2(t)$, however I don't know how else to proceed.






calculus integration ordinary-differential-equations derivatives







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    I have the following function:



    $$F(t) = int_{a-t}^{b+t} f(x) dx$$



    In my notes when this is differentiated with respect to $t$ we get:
    $$frac{d F}{d t} = f(b + t) + f(a - t)$$



    How can we obtain this result?



    My approach



    The approach that I was thinking about was about splitting the integral into two based on the integration limits, for instance, something like:



    $$F_1(t) = int_{a-t}^c f(x) dx quad quad F_2(t) = int_{c}^{b+t} f(x)dx$$



    So that $F(t) = F_1(t) + F_2(t)$, however I don't know how else to proceed.






    calculus integration ordinary-differential-equations derivatives







    share|cite|improve this question









    $endgroup$















      1












      1








      1


      1



      $begingroup$


      I have the following function:



      $$F(t) = int_{a-t}^{b+t} f(x) dx$$



      In my notes when this is differentiated with respect to $t$ we get:
      $$frac{d F}{d t} = f(b + t) + f(a - t)$$



      How can we obtain this result?



      My approach



      The approach that I was thinking about was about splitting the integral into two based on the integration limits, for instance, something like:



      $$F_1(t) = int_{a-t}^c f(x) dx quad quad F_2(t) = int_{c}^{b+t} f(x)dx$$



      So that $F(t) = F_1(t) + F_2(t)$, however I don't know how else to proceed.






      calculus integration ordinary-differential-equations derivatives







      share|cite|improve this question









      $endgroup$




      I have the following function:



      $$F(t) = int_{a-t}^{b+t} f(x) dx$$



      In my notes when this is differentiated with respect to $t$ we get:
      $$frac{d F}{d t} = f(b + t) + f(a - t)$$



      How can we obtain this result?



      My approach



      The approach that I was thinking about was about splitting the integral into two based on the integration limits, for instance, something like:



      $$F_1(t) = int_{a-t}^c f(x) dx quad quad F_2(t) = int_{c}^{b+t} f(x)dx$$



      So that $F(t) = F_1(t) + F_2(t)$, however I don't know how else to proceed.






      calculus integration ordinary-differential-equations derivatives




      calculus integration ordinary-differential-equations derivatives






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 1 '18 at 19:48









      Euler_SalterEuler_Salter

      2,0571335




      2,0571335






















          3 Answers
          3






          active

          oldest

          votes


















          1












          $begingroup$

          You're on the right track. You can proceed by using the chain rule.



          Observe that (with your notation)
          $$ F_1(t) = G(H(t)), $$
          where
          $$ H(t) = a-t, quad text{and} quad G(t) = int_t^c f(x) , mathrm dx = -int_c^t f(x) , mathrm dx. $$
          Now
          $$ H'(t) = -1, quad text{and} quad G'(t) = -f(t), $$
          where we invoked FTC when differentiating $G$.
          Hence, the chain rule gives
          $$ F_1'(t) = G'(H(t))H'(t) = -G'(a-t)f(t) = f(a-t). $$
          Can you deal with $F_2$ on your own?



          P.S. You could also directly refer to the more general Leibniz integral rule.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            Thanks for directing me to Leibniz integral rule, seems super handy!
            $endgroup$
            – Euler_Salter
            Dec 1 '18 at 20:32










          • $begingroup$
            @Euler_Salter Indeed!
            $endgroup$
            – MisterRiemann
            Dec 1 '18 at 20:32



















          1












          $begingroup$

          If we have a function defined by $$Q'(x)=f(x)$$
          Then we have that $$F(t)=Q(b+t)-Q(a-t)$$
          Taking $frac{mathrm d}{mathrm dt}(cdot)$ on both sides:
          $$F'(t)=frac{mathrm d}{mathrm dt}Q(b+t)-frac{mathrm d}{mathrm dt}Q(a-t)$$
          $$F'(t)=Q'(b+t)frac{mathrm d}{mathrm dt}(b+t)-Q'(a-t)frac{mathrm d}{mathrm dt}(a-t)$$
          $$F'(t)=Q'(b+t)+Q'(a-t)$$
          $$F'(t)=f(b+t)+f(a-t)$$
          QED






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            This is amazingly simple!
            $endgroup$
            – Euler_Salter
            Dec 1 '18 at 20:29






          • 2




            $begingroup$
            @Euler_Salter Thank you very much. I have often found that many seemingly complex problems in calculus have amazingly simple results.
            $endgroup$
            – clathratus
            Dec 1 '18 at 20:32






          • 1




            $begingroup$
            True! I think in this case I was just subconsciously deceived by the usual notation F on the LHS of the Fundamental Theorem of Calculus
            $endgroup$
            – Euler_Salter
            Dec 1 '18 at 20:34



















          0












          $begingroup$

          $$frac{dF}{dt}=frac{d(b+t)}{dt}frac{d}{d(b+t)}int_c^{b+t}f(x)dx-frac{d(a-t)}{dt}frac{d}{d(a-t)}int_c^{a-t}f(x)dx\=1cdot f(b+t)-(-1)cdot f(a-t)=f(b+t)+f(a-t).$$We use the chain rule at the first $=$, then the FTC at the second $=$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            can you please state which results you're using?
            $endgroup$
            – Euler_Salter
            Dec 1 '18 at 20:23






          • 1




            $begingroup$
            @Euler_Salter See edit.
            $endgroup$
            – J.G.
            Dec 1 '18 at 20:29










          • $begingroup$
            Thank you for explaining it more!
            $endgroup$
            – Euler_Salter
            Dec 1 '18 at 20:30











          Your Answer





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          3 Answers
          3






          active

          oldest

          votes








          3 Answers
          3






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          You're on the right track. You can proceed by using the chain rule.



          Observe that (with your notation)
          $$ F_1(t) = G(H(t)), $$
          where
          $$ H(t) = a-t, quad text{and} quad G(t) = int_t^c f(x) , mathrm dx = -int_c^t f(x) , mathrm dx. $$
          Now
          $$ H'(t) = -1, quad text{and} quad G'(t) = -f(t), $$
          where we invoked FTC when differentiating $G$.
          Hence, the chain rule gives
          $$ F_1'(t) = G'(H(t))H'(t) = -G'(a-t)f(t) = f(a-t). $$
          Can you deal with $F_2$ on your own?



          P.S. You could also directly refer to the more general Leibniz integral rule.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            Thanks for directing me to Leibniz integral rule, seems super handy!
            $endgroup$
            – Euler_Salter
            Dec 1 '18 at 20:32










          • $begingroup$
            @Euler_Salter Indeed!
            $endgroup$
            – MisterRiemann
            Dec 1 '18 at 20:32
















          1












          $begingroup$

          You're on the right track. You can proceed by using the chain rule.



          Observe that (with your notation)
          $$ F_1(t) = G(H(t)), $$
          where
          $$ H(t) = a-t, quad text{and} quad G(t) = int_t^c f(x) , mathrm dx = -int_c^t f(x) , mathrm dx. $$
          Now
          $$ H'(t) = -1, quad text{and} quad G'(t) = -f(t), $$
          where we invoked FTC when differentiating $G$.
          Hence, the chain rule gives
          $$ F_1'(t) = G'(H(t))H'(t) = -G'(a-t)f(t) = f(a-t). $$
          Can you deal with $F_2$ on your own?



          P.S. You could also directly refer to the more general Leibniz integral rule.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            Thanks for directing me to Leibniz integral rule, seems super handy!
            $endgroup$
            – Euler_Salter
            Dec 1 '18 at 20:32










          • $begingroup$
            @Euler_Salter Indeed!
            $endgroup$
            – MisterRiemann
            Dec 1 '18 at 20:32














          1












          1








          1





          $begingroup$

          You're on the right track. You can proceed by using the chain rule.



          Observe that (with your notation)
          $$ F_1(t) = G(H(t)), $$
          where
          $$ H(t) = a-t, quad text{and} quad G(t) = int_t^c f(x) , mathrm dx = -int_c^t f(x) , mathrm dx. $$
          Now
          $$ H'(t) = -1, quad text{and} quad G'(t) = -f(t), $$
          where we invoked FTC when differentiating $G$.
          Hence, the chain rule gives
          $$ F_1'(t) = G'(H(t))H'(t) = -G'(a-t)f(t) = f(a-t). $$
          Can you deal with $F_2$ on your own?



          P.S. You could also directly refer to the more general Leibniz integral rule.






          share|cite|improve this answer











          $endgroup$



          You're on the right track. You can proceed by using the chain rule.



          Observe that (with your notation)
          $$ F_1(t) = G(H(t)), $$
          where
          $$ H(t) = a-t, quad text{and} quad G(t) = int_t^c f(x) , mathrm dx = -int_c^t f(x) , mathrm dx. $$
          Now
          $$ H'(t) = -1, quad text{and} quad G'(t) = -f(t), $$
          where we invoked FTC when differentiating $G$.
          Hence, the chain rule gives
          $$ F_1'(t) = G'(H(t))H'(t) = -G'(a-t)f(t) = f(a-t). $$
          Can you deal with $F_2$ on your own?



          P.S. You could also directly refer to the more general Leibniz integral rule.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 1 '18 at 20:03

























          answered Dec 1 '18 at 19:52









          MisterRiemannMisterRiemann

          5,8841624




          5,8841624








          • 1




            $begingroup$
            Thanks for directing me to Leibniz integral rule, seems super handy!
            $endgroup$
            – Euler_Salter
            Dec 1 '18 at 20:32










          • $begingroup$
            @Euler_Salter Indeed!
            $endgroup$
            – MisterRiemann
            Dec 1 '18 at 20:32














          • 1




            $begingroup$
            Thanks for directing me to Leibniz integral rule, seems super handy!
            $endgroup$
            – Euler_Salter
            Dec 1 '18 at 20:32










          • $begingroup$
            @Euler_Salter Indeed!
            $endgroup$
            – MisterRiemann
            Dec 1 '18 at 20:32








          1




          1




          $begingroup$
          Thanks for directing me to Leibniz integral rule, seems super handy!
          $endgroup$
          – Euler_Salter
          Dec 1 '18 at 20:32




          $begingroup$
          Thanks for directing me to Leibniz integral rule, seems super handy!
          $endgroup$
          – Euler_Salter
          Dec 1 '18 at 20:32












          $begingroup$
          @Euler_Salter Indeed!
          $endgroup$
          – MisterRiemann
          Dec 1 '18 at 20:32




          $begingroup$
          @Euler_Salter Indeed!
          $endgroup$
          – MisterRiemann
          Dec 1 '18 at 20:32











          1












          $begingroup$

          If we have a function defined by $$Q'(x)=f(x)$$
          Then we have that $$F(t)=Q(b+t)-Q(a-t)$$
          Taking $frac{mathrm d}{mathrm dt}(cdot)$ on both sides:
          $$F'(t)=frac{mathrm d}{mathrm dt}Q(b+t)-frac{mathrm d}{mathrm dt}Q(a-t)$$
          $$F'(t)=Q'(b+t)frac{mathrm d}{mathrm dt}(b+t)-Q'(a-t)frac{mathrm d}{mathrm dt}(a-t)$$
          $$F'(t)=Q'(b+t)+Q'(a-t)$$
          $$F'(t)=f(b+t)+f(a-t)$$
          QED






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            This is amazingly simple!
            $endgroup$
            – Euler_Salter
            Dec 1 '18 at 20:29






          • 2




            $begingroup$
            @Euler_Salter Thank you very much. I have often found that many seemingly complex problems in calculus have amazingly simple results.
            $endgroup$
            – clathratus
            Dec 1 '18 at 20:32






          • 1




            $begingroup$
            True! I think in this case I was just subconsciously deceived by the usual notation F on the LHS of the Fundamental Theorem of Calculus
            $endgroup$
            – Euler_Salter
            Dec 1 '18 at 20:34
















          1












          $begingroup$

          If we have a function defined by $$Q'(x)=f(x)$$
          Then we have that $$F(t)=Q(b+t)-Q(a-t)$$
          Taking $frac{mathrm d}{mathrm dt}(cdot)$ on both sides:
          $$F'(t)=frac{mathrm d}{mathrm dt}Q(b+t)-frac{mathrm d}{mathrm dt}Q(a-t)$$
          $$F'(t)=Q'(b+t)frac{mathrm d}{mathrm dt}(b+t)-Q'(a-t)frac{mathrm d}{mathrm dt}(a-t)$$
          $$F'(t)=Q'(b+t)+Q'(a-t)$$
          $$F'(t)=f(b+t)+f(a-t)$$
          QED






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            This is amazingly simple!
            $endgroup$
            – Euler_Salter
            Dec 1 '18 at 20:29






          • 2




            $begingroup$
            @Euler_Salter Thank you very much. I have often found that many seemingly complex problems in calculus have amazingly simple results.
            $endgroup$
            – clathratus
            Dec 1 '18 at 20:32






          • 1




            $begingroup$
            True! I think in this case I was just subconsciously deceived by the usual notation F on the LHS of the Fundamental Theorem of Calculus
            $endgroup$
            – Euler_Salter
            Dec 1 '18 at 20:34














          1












          1








          1





          $begingroup$

          If we have a function defined by $$Q'(x)=f(x)$$
          Then we have that $$F(t)=Q(b+t)-Q(a-t)$$
          Taking $frac{mathrm d}{mathrm dt}(cdot)$ on both sides:
          $$F'(t)=frac{mathrm d}{mathrm dt}Q(b+t)-frac{mathrm d}{mathrm dt}Q(a-t)$$
          $$F'(t)=Q'(b+t)frac{mathrm d}{mathrm dt}(b+t)-Q'(a-t)frac{mathrm d}{mathrm dt}(a-t)$$
          $$F'(t)=Q'(b+t)+Q'(a-t)$$
          $$F'(t)=f(b+t)+f(a-t)$$
          QED






          share|cite|improve this answer









          $endgroup$



          If we have a function defined by $$Q'(x)=f(x)$$
          Then we have that $$F(t)=Q(b+t)-Q(a-t)$$
          Taking $frac{mathrm d}{mathrm dt}(cdot)$ on both sides:
          $$F'(t)=frac{mathrm d}{mathrm dt}Q(b+t)-frac{mathrm d}{mathrm dt}Q(a-t)$$
          $$F'(t)=Q'(b+t)frac{mathrm d}{mathrm dt}(b+t)-Q'(a-t)frac{mathrm d}{mathrm dt}(a-t)$$
          $$F'(t)=Q'(b+t)+Q'(a-t)$$
          $$F'(t)=f(b+t)+f(a-t)$$
          QED







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 1 '18 at 20:23









          clathratusclathratus

          3,541332




          3,541332








          • 1




            $begingroup$
            This is amazingly simple!
            $endgroup$
            – Euler_Salter
            Dec 1 '18 at 20:29






          • 2




            $begingroup$
            @Euler_Salter Thank you very much. I have often found that many seemingly complex problems in calculus have amazingly simple results.
            $endgroup$
            – clathratus
            Dec 1 '18 at 20:32






          • 1




            $begingroup$
            True! I think in this case I was just subconsciously deceived by the usual notation F on the LHS of the Fundamental Theorem of Calculus
            $endgroup$
            – Euler_Salter
            Dec 1 '18 at 20:34














          • 1




            $begingroup$
            This is amazingly simple!
            $endgroup$
            – Euler_Salter
            Dec 1 '18 at 20:29






          • 2




            $begingroup$
            @Euler_Salter Thank you very much. I have often found that many seemingly complex problems in calculus have amazingly simple results.
            $endgroup$
            – clathratus
            Dec 1 '18 at 20:32






          • 1




            $begingroup$
            True! I think in this case I was just subconsciously deceived by the usual notation F on the LHS of the Fundamental Theorem of Calculus
            $endgroup$
            – Euler_Salter
            Dec 1 '18 at 20:34








          1




          1




          $begingroup$
          This is amazingly simple!
          $endgroup$
          – Euler_Salter
          Dec 1 '18 at 20:29




          $begingroup$
          This is amazingly simple!
          $endgroup$
          – Euler_Salter
          Dec 1 '18 at 20:29




          2




          2




          $begingroup$
          @Euler_Salter Thank you very much. I have often found that many seemingly complex problems in calculus have amazingly simple results.
          $endgroup$
          – clathratus
          Dec 1 '18 at 20:32




          $begingroup$
          @Euler_Salter Thank you very much. I have often found that many seemingly complex problems in calculus have amazingly simple results.
          $endgroup$
          – clathratus
          Dec 1 '18 at 20:32




          1




          1




          $begingroup$
          True! I think in this case I was just subconsciously deceived by the usual notation F on the LHS of the Fundamental Theorem of Calculus
          $endgroup$
          – Euler_Salter
          Dec 1 '18 at 20:34




          $begingroup$
          True! I think in this case I was just subconsciously deceived by the usual notation F on the LHS of the Fundamental Theorem of Calculus
          $endgroup$
          – Euler_Salter
          Dec 1 '18 at 20:34











          0












          $begingroup$

          $$frac{dF}{dt}=frac{d(b+t)}{dt}frac{d}{d(b+t)}int_c^{b+t}f(x)dx-frac{d(a-t)}{dt}frac{d}{d(a-t)}int_c^{a-t}f(x)dx\=1cdot f(b+t)-(-1)cdot f(a-t)=f(b+t)+f(a-t).$$We use the chain rule at the first $=$, then the FTC at the second $=$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            can you please state which results you're using?
            $endgroup$
            – Euler_Salter
            Dec 1 '18 at 20:23






          • 1




            $begingroup$
            @Euler_Salter See edit.
            $endgroup$
            – J.G.
            Dec 1 '18 at 20:29










          • $begingroup$
            Thank you for explaining it more!
            $endgroup$
            – Euler_Salter
            Dec 1 '18 at 20:30
















          0












          $begingroup$

          $$frac{dF}{dt}=frac{d(b+t)}{dt}frac{d}{d(b+t)}int_c^{b+t}f(x)dx-frac{d(a-t)}{dt}frac{d}{d(a-t)}int_c^{a-t}f(x)dx\=1cdot f(b+t)-(-1)cdot f(a-t)=f(b+t)+f(a-t).$$We use the chain rule at the first $=$, then the FTC at the second $=$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            can you please state which results you're using?
            $endgroup$
            – Euler_Salter
            Dec 1 '18 at 20:23






          • 1




            $begingroup$
            @Euler_Salter See edit.
            $endgroup$
            – J.G.
            Dec 1 '18 at 20:29










          • $begingroup$
            Thank you for explaining it more!
            $endgroup$
            – Euler_Salter
            Dec 1 '18 at 20:30














          0












          0








          0





          $begingroup$

          $$frac{dF}{dt}=frac{d(b+t)}{dt}frac{d}{d(b+t)}int_c^{b+t}f(x)dx-frac{d(a-t)}{dt}frac{d}{d(a-t)}int_c^{a-t}f(x)dx\=1cdot f(b+t)-(-1)cdot f(a-t)=f(b+t)+f(a-t).$$We use the chain rule at the first $=$, then the FTC at the second $=$.






          share|cite|improve this answer











          $endgroup$



          $$frac{dF}{dt}=frac{d(b+t)}{dt}frac{d}{d(b+t)}int_c^{b+t}f(x)dx-frac{d(a-t)}{dt}frac{d}{d(a-t)}int_c^{a-t}f(x)dx\=1cdot f(b+t)-(-1)cdot f(a-t)=f(b+t)+f(a-t).$$We use the chain rule at the first $=$, then the FTC at the second $=$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 1 '18 at 20:29

























          answered Dec 1 '18 at 20:16









          J.G.J.G.

          23.9k22539




          23.9k22539












          • $begingroup$
            can you please state which results you're using?
            $endgroup$
            – Euler_Salter
            Dec 1 '18 at 20:23






          • 1




            $begingroup$
            @Euler_Salter See edit.
            $endgroup$
            – J.G.
            Dec 1 '18 at 20:29










          • $begingroup$
            Thank you for explaining it more!
            $endgroup$
            – Euler_Salter
            Dec 1 '18 at 20:30


















          • $begingroup$
            can you please state which results you're using?
            $endgroup$
            – Euler_Salter
            Dec 1 '18 at 20:23






          • 1




            $begingroup$
            @Euler_Salter See edit.
            $endgroup$
            – J.G.
            Dec 1 '18 at 20:29










          • $begingroup$
            Thank you for explaining it more!
            $endgroup$
            – Euler_Salter
            Dec 1 '18 at 20:30
















          $begingroup$
          can you please state which results you're using?
          $endgroup$
          – Euler_Salter
          Dec 1 '18 at 20:23




          $begingroup$
          can you please state which results you're using?
          $endgroup$
          – Euler_Salter
          Dec 1 '18 at 20:23




          1




          1




          $begingroup$
          @Euler_Salter See edit.
          $endgroup$
          – J.G.
          Dec 1 '18 at 20:29




          $begingroup$
          @Euler_Salter See edit.
          $endgroup$
          – J.G.
          Dec 1 '18 at 20:29












          $begingroup$
          Thank you for explaining it more!
          $endgroup$
          – Euler_Salter
          Dec 1 '18 at 20:30




          $begingroup$
          Thank you for explaining it more!
          $endgroup$
          – Euler_Salter
          Dec 1 '18 at 20:30


















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