Proof of an integral property












3












$begingroup$


$$int_0^1 f(x)g'(x)dx=pi$$ If $f(1)g(1)=f(0)g(0)$ then $$int_0^1 f'(x)g(x)dx= -pi$$
So I have to prove this and I have absolutely no idea how to do it. I am guessing I will have to use the fundamental theorem of calculus and it show that the rate of change is $1$ because it didn't change from $f(1)g(1)$ to $f(0)g(0)$






calculus integration definite-integrals







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$endgroup$








  • 5




    $begingroup$
    Do you know integration by parts?
    $endgroup$
    – Minus One-Twelfth
    2 hours ago


















3












$begingroup$


$$int_0^1 f(x)g'(x)dx=pi$$ If $f(1)g(1)=f(0)g(0)$ then $$int_0^1 f'(x)g(x)dx= -pi$$
So I have to prove this and I have absolutely no idea how to do it. I am guessing I will have to use the fundamental theorem of calculus and it show that the rate of change is $1$ because it didn't change from $f(1)g(1)$ to $f(0)g(0)$






calculus integration definite-integrals







share|cite|improve this question









New contributor




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







$endgroup$








  • 5




    $begingroup$
    Do you know integration by parts?
    $endgroup$
    – Minus One-Twelfth
    2 hours ago
















3












3








3





$begingroup$


$$int_0^1 f(x)g'(x)dx=pi$$ If $f(1)g(1)=f(0)g(0)$ then $$int_0^1 f'(x)g(x)dx= -pi$$
So I have to prove this and I have absolutely no idea how to do it. I am guessing I will have to use the fundamental theorem of calculus and it show that the rate of change is $1$ because it didn't change from $f(1)g(1)$ to $f(0)g(0)$






calculus integration definite-integrals







share|cite|improve this question









New contributor




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







$endgroup$




$$int_0^1 f(x)g'(x)dx=pi$$ If $f(1)g(1)=f(0)g(0)$ then $$int_0^1 f'(x)g(x)dx= -pi$$
So I have to prove this and I have absolutely no idea how to do it. I am guessing I will have to use the fundamental theorem of calculus and it show that the rate of change is $1$ because it didn't change from $f(1)g(1)$ to $f(0)g(0)$






calculus integration definite-integrals




calculus integration definite-integrals






share|cite|improve this question









New contributor




adam hany 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









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adam hany 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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edited 2 hours ago









Eevee Trainer

6,39311237




6,39311237






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asked 2 hours ago









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adam hany is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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adam hany is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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  • 5




    $begingroup$
    Do you know integration by parts?
    $endgroup$
    – Minus One-Twelfth
    2 hours ago
















  • 5




    $begingroup$
    Do you know integration by parts?
    $endgroup$
    – Minus One-Twelfth
    2 hours ago










5




5




$begingroup$
Do you know integration by parts?
$endgroup$
– Minus One-Twelfth
2 hours ago






$begingroup$
Do you know integration by parts?
$endgroup$
– Minus One-Twelfth
2 hours ago












2 Answers
2






active

oldest

votes


















4












$begingroup$

Hint:



Utilize integration by parts:



$$int f(x)g'(x)dx = f(x)g(x) - int f'(x)g(x)$$



If we have a definite integral, then this formula becomes



$$int_a^b f(x)g'(x)dx = f(b)g(b) - f(a)g(a) - int_a^b f'(x)g(x)$$






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    thank you so much for the answer ∫baf(x)g′(x)dx=−∫baf′(x)g(x)
    $endgroup$
    – adam hany
    2 hours ago



















1












$begingroup$

Hint



$dfrac{d(f(x)cdot g(x))}{dx}=?$



Integrate both sides wrt $x$ between $[0,1]$






share|cite|improve this answer









$endgroup$













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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4












    $begingroup$

    Hint:



    Utilize integration by parts:



    $$int f(x)g'(x)dx = f(x)g(x) - int f'(x)g(x)$$



    If we have a definite integral, then this formula becomes



    $$int_a^b f(x)g'(x)dx = f(b)g(b) - f(a)g(a) - int_a^b f'(x)g(x)$$






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      thank you so much for the answer ∫baf(x)g′(x)dx=−∫baf′(x)g(x)
      $endgroup$
      – adam hany
      2 hours ago
















    4












    $begingroup$

    Hint:



    Utilize integration by parts:



    $$int f(x)g'(x)dx = f(x)g(x) - int f'(x)g(x)$$



    If we have a definite integral, then this formula becomes



    $$int_a^b f(x)g'(x)dx = f(b)g(b) - f(a)g(a) - int_a^b f'(x)g(x)$$






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      thank you so much for the answer ∫baf(x)g′(x)dx=−∫baf′(x)g(x)
      $endgroup$
      – adam hany
      2 hours ago














    4












    4








    4





    $begingroup$

    Hint:



    Utilize integration by parts:



    $$int f(x)g'(x)dx = f(x)g(x) - int f'(x)g(x)$$



    If we have a definite integral, then this formula becomes



    $$int_a^b f(x)g'(x)dx = f(b)g(b) - f(a)g(a) - int_a^b f'(x)g(x)$$






    share|cite|improve this answer









    $endgroup$



    Hint:



    Utilize integration by parts:



    $$int f(x)g'(x)dx = f(x)g(x) - int f'(x)g(x)$$



    If we have a definite integral, then this formula becomes



    $$int_a^b f(x)g'(x)dx = f(b)g(b) - f(a)g(a) - int_a^b f'(x)g(x)$$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered 2 hours ago









    Eevee TrainerEevee Trainer

    6,39311237




    6,39311237








    • 1




      $begingroup$
      thank you so much for the answer ∫baf(x)g′(x)dx=−∫baf′(x)g(x)
      $endgroup$
      – adam hany
      2 hours ago














    • 1




      $begingroup$
      thank you so much for the answer ∫baf(x)g′(x)dx=−∫baf′(x)g(x)
      $endgroup$
      – adam hany
      2 hours ago








    1




    1




    $begingroup$
    thank you so much for the answer ∫baf(x)g′(x)dx=−∫baf′(x)g(x)
    $endgroup$
    – adam hany
    2 hours ago




    $begingroup$
    thank you so much for the answer ∫baf(x)g′(x)dx=−∫baf′(x)g(x)
    $endgroup$
    – adam hany
    2 hours ago











    1












    $begingroup$

    Hint



    $dfrac{d(f(x)cdot g(x))}{dx}=?$



    Integrate both sides wrt $x$ between $[0,1]$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Hint



      $dfrac{d(f(x)cdot g(x))}{dx}=?$



      Integrate both sides wrt $x$ between $[0,1]$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Hint



        $dfrac{d(f(x)cdot g(x))}{dx}=?$



        Integrate both sides wrt $x$ between $[0,1]$






        share|cite|improve this answer









        $endgroup$



        Hint



        $dfrac{d(f(x)cdot g(x))}{dx}=?$



        Integrate both sides wrt $x$ between $[0,1]$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 2 hours ago









        lab bhattacharjeelab bhattacharjee

        226k15157275




        226k15157275






















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