Definite integrals : how do we approach in solving a problem











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While practicing definite integrals I came across a question and now I am stuck



Question:



let f be a continous satisfying $f(x+y) = f(x) + f(y) + f(x)cdot f(y)$ for all real $x$ and $y$ and $f'(0)= -1$.



Find the value of $int_0^1 f(x) , dx$.



I tried to solve by find the function but no success and i am confused.



Please tell me how to solve these kind of questions.










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  • $g(x):=1+f(x)$ should produce a simpler relation for $g(x+y)$...
    – Raymond Manzoni
    Nov 14 at 13:27












  • I guess you may take $lim yto 0$?
    – Mythomorphic
    Nov 14 at 13:44

















up vote
0
down vote

favorite












While practicing definite integrals I came across a question and now I am stuck



Question:



let f be a continous satisfying $f(x+y) = f(x) + f(y) + f(x)cdot f(y)$ for all real $x$ and $y$ and $f'(0)= -1$.



Find the value of $int_0^1 f(x) , dx$.



I tried to solve by find the function but no success and i am confused.



Please tell me how to solve these kind of questions.










share|cite|improve this question
























  • $g(x):=1+f(x)$ should produce a simpler relation for $g(x+y)$...
    – Raymond Manzoni
    Nov 14 at 13:27












  • I guess you may take $lim yto 0$?
    – Mythomorphic
    Nov 14 at 13:44















up vote
0
down vote

favorite









up vote
0
down vote

favorite











While practicing definite integrals I came across a question and now I am stuck



Question:



let f be a continous satisfying $f(x+y) = f(x) + f(y) + f(x)cdot f(y)$ for all real $x$ and $y$ and $f'(0)= -1$.



Find the value of $int_0^1 f(x) , dx$.



I tried to solve by find the function but no success and i am confused.



Please tell me how to solve these kind of questions.










share|cite|improve this question















While practicing definite integrals I came across a question and now I am stuck



Question:



let f be a continous satisfying $f(x+y) = f(x) + f(y) + f(x)cdot f(y)$ for all real $x$ and $y$ and $f'(0)= -1$.



Find the value of $int_0^1 f(x) , dx$.



I tried to solve by find the function but no success and i am confused.



Please tell me how to solve these kind of questions.







definite-integrals functional-calculus






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edited Nov 14 at 13:31









callculus

17.5k31427




17.5k31427










asked Nov 14 at 13:20









Shantanu Kaushik

223




223












  • $g(x):=1+f(x)$ should produce a simpler relation for $g(x+y)$...
    – Raymond Manzoni
    Nov 14 at 13:27












  • I guess you may take $lim yto 0$?
    – Mythomorphic
    Nov 14 at 13:44




















  • $g(x):=1+f(x)$ should produce a simpler relation for $g(x+y)$...
    – Raymond Manzoni
    Nov 14 at 13:27












  • I guess you may take $lim yto 0$?
    – Mythomorphic
    Nov 14 at 13:44


















$g(x):=1+f(x)$ should produce a simpler relation for $g(x+y)$...
– Raymond Manzoni
Nov 14 at 13:27






$g(x):=1+f(x)$ should produce a simpler relation for $g(x+y)$...
– Raymond Manzoni
Nov 14 at 13:27














I guess you may take $lim yto 0$?
– Mythomorphic
Nov 14 at 13:44






I guess you may take $lim yto 0$?
– Mythomorphic
Nov 14 at 13:44












1 Answer
1






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1
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Rearrange



$$f(x+y)-f(x)=f(y)[1+f(x)]$$



Divide both sides by $y$ and take the limit
$$lim_{yto0}frac{f(x+y)-f(x)}{y}=lim_{yto0}=frac{f(y)[1+f(x)]}{y}$$
$$f'(x)=[1+f(x)]cdotleft[lim_{yto0}frac{f(y)}{y}right]$$



Can you proceed from here?




Notice $f(0)=0$ and given $f'(0)=-1$

$$lim_{yto0}frac{f(y)}{y}=-1$$

So $$f'(x)=-[1+f(x)]$$ $$int^1_0f'(x)dx=int^1_0-1-f(x)dx$$
$$int^1_0f(x)dx=[-x]^1_0-[f(x)]^1_0=-1-f(1)$$







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    1 Answer
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    1 Answer
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    active

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    up vote
    1
    down vote



    accepted










    Rearrange



    $$f(x+y)-f(x)=f(y)[1+f(x)]$$



    Divide both sides by $y$ and take the limit
    $$lim_{yto0}frac{f(x+y)-f(x)}{y}=lim_{yto0}=frac{f(y)[1+f(x)]}{y}$$
    $$f'(x)=[1+f(x)]cdotleft[lim_{yto0}frac{f(y)}{y}right]$$



    Can you proceed from here?




    Notice $f(0)=0$ and given $f'(0)=-1$

    $$lim_{yto0}frac{f(y)}{y}=-1$$

    So $$f'(x)=-[1+f(x)]$$ $$int^1_0f'(x)dx=int^1_0-1-f(x)dx$$
    $$int^1_0f(x)dx=[-x]^1_0-[f(x)]^1_0=-1-f(1)$$







    share|cite|improve this answer

























      up vote
      1
      down vote



      accepted










      Rearrange



      $$f(x+y)-f(x)=f(y)[1+f(x)]$$



      Divide both sides by $y$ and take the limit
      $$lim_{yto0}frac{f(x+y)-f(x)}{y}=lim_{yto0}=frac{f(y)[1+f(x)]}{y}$$
      $$f'(x)=[1+f(x)]cdotleft[lim_{yto0}frac{f(y)}{y}right]$$



      Can you proceed from here?




      Notice $f(0)=0$ and given $f'(0)=-1$

      $$lim_{yto0}frac{f(y)}{y}=-1$$

      So $$f'(x)=-[1+f(x)]$$ $$int^1_0f'(x)dx=int^1_0-1-f(x)dx$$
      $$int^1_0f(x)dx=[-x]^1_0-[f(x)]^1_0=-1-f(1)$$







      share|cite|improve this answer























        up vote
        1
        down vote



        accepted







        up vote
        1
        down vote



        accepted






        Rearrange



        $$f(x+y)-f(x)=f(y)[1+f(x)]$$



        Divide both sides by $y$ and take the limit
        $$lim_{yto0}frac{f(x+y)-f(x)}{y}=lim_{yto0}=frac{f(y)[1+f(x)]}{y}$$
        $$f'(x)=[1+f(x)]cdotleft[lim_{yto0}frac{f(y)}{y}right]$$



        Can you proceed from here?




        Notice $f(0)=0$ and given $f'(0)=-1$

        $$lim_{yto0}frac{f(y)}{y}=-1$$

        So $$f'(x)=-[1+f(x)]$$ $$int^1_0f'(x)dx=int^1_0-1-f(x)dx$$
        $$int^1_0f(x)dx=[-x]^1_0-[f(x)]^1_0=-1-f(1)$$







        share|cite|improve this answer












        Rearrange



        $$f(x+y)-f(x)=f(y)[1+f(x)]$$



        Divide both sides by $y$ and take the limit
        $$lim_{yto0}frac{f(x+y)-f(x)}{y}=lim_{yto0}=frac{f(y)[1+f(x)]}{y}$$
        $$f'(x)=[1+f(x)]cdotleft[lim_{yto0}frac{f(y)}{y}right]$$



        Can you proceed from here?




        Notice $f(0)=0$ and given $f'(0)=-1$

        $$lim_{yto0}frac{f(y)}{y}=-1$$

        So $$f'(x)=-[1+f(x)]$$ $$int^1_0f'(x)dx=int^1_0-1-f(x)dx$$
        $$int^1_0f(x)dx=[-x]^1_0-[f(x)]^1_0=-1-f(1)$$








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        answered Nov 14 at 14:00









        Mythomorphic

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        5,2841733






























             

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