$int_-1^0 sin^2 (fracpi2 f(x+1)) dx +int_0^1 cos^2 (fracpi2 f(x)) dx =?$

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It is well known that we can choose smooth $f:mathbb R to [0,1]$ with the property $f(x)=0$ for $xleq 0$ and $f(x)=1$ for all $xgeq 1.$




Let $0< C<1.$ Can we choose smooth $f:mathbb R to [0,1]$ with the property $f(x)=0$ for $xleq 0$ and $f(x)=1$ for all $xgeq 1$
and it satisfies the following property:
$$int_-1^0 sin^2 (fracpi2 f(x+1)) dx +int_0^1 cos^2 (fracpi2 f(x)) dx =C?$$





calculus real-analysis integration analysis examples-counterexamples




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    Try setting $x+1=y$
    – lab bhattacharjee
    Aug 3 at 1:18














up vote
0
down vote

favorite
1












It is well known that we can choose smooth $f:mathbb R to [0,1]$ with the property $f(x)=0$ for $xleq 0$ and $f(x)=1$ for all $xgeq 1.$




Let $0< C<1.$ Can we choose smooth $f:mathbb R to [0,1]$ with the property $f(x)=0$ for $xleq 0$ and $f(x)=1$ for all $xgeq 1$
and it satisfies the following property:
$$int_-1^0 sin^2 (fracpi2 f(x+1)) dx +int_0^1 cos^2 (fracpi2 f(x)) dx =C?$$





calculus real-analysis integration analysis examples-counterexamples




share|cite|improve this question















  • 3




    Try setting $x+1=y$
    – lab bhattacharjee
    Aug 3 at 1:18












up vote
0
down vote

favorite
1









up vote
0
down vote

favorite
1






1





It is well known that we can choose smooth $f:mathbb R to [0,1]$ with the property $f(x)=0$ for $xleq 0$ and $f(x)=1$ for all $xgeq 1.$




Let $0< C<1.$ Can we choose smooth $f:mathbb R to [0,1]$ with the property $f(x)=0$ for $xleq 0$ and $f(x)=1$ for all $xgeq 1$
and it satisfies the following property:
$$int_-1^0 sin^2 (fracpi2 f(x+1)) dx +int_0^1 cos^2 (fracpi2 f(x)) dx =C?$$





calculus real-analysis integration analysis examples-counterexamples




share|cite|improve this question











It is well known that we can choose smooth $f:mathbb R to [0,1]$ with the property $f(x)=0$ for $xleq 0$ and $f(x)=1$ for all $xgeq 1.$




Let $0< C<1.$ Can we choose smooth $f:mathbb R to [0,1]$ with the property $f(x)=0$ for $xleq 0$ and $f(x)=1$ for all $xgeq 1$
and it satisfies the following property:
$$int_-1^0 sin^2 (fracpi2 f(x+1)) dx +int_0^1 cos^2 (fracpi2 f(x)) dx =C?$$






calculus real-analysis integration analysis examples-counterexamples





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share|cite|improve this question




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asked Aug 3 at 1:12









Math Learner

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  • 3




    Try setting $x+1=y$
    – lab bhattacharjee
    Aug 3 at 1:18












  • 3




    Try setting $x+1=y$
    – lab bhattacharjee
    Aug 3 at 1:18







3




3




Try setting $x+1=y$
– lab bhattacharjee
Aug 3 at 1:18




Try setting $x+1=y$
– lab bhattacharjee
Aug 3 at 1:18










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Hint:



Substituting $t=x+1$ you will see that the integral will in fact be equal to $1$ rather than some constant C.



Note: What I don't understand is that the domain you defined for the function is all real numbers but you did not define the function in $(0,1)$ so do you find the value of function in $(0,1)$ or anything else? If you want to define a function for that interval then I must tell you that there would be infinite functions satisfying the integral if $C=1$ and no function if $Cneq 1$






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    1 Answer
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    1






    active

    oldest

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    active

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    up vote
    2
    down vote



    accepted










    Hint:



    Substituting $t=x+1$ you will see that the integral will in fact be equal to $1$ rather than some constant C.



    Note: What I don't understand is that the domain you defined for the function is all real numbers but you did not define the function in $(0,1)$ so do you find the value of function in $(0,1)$ or anything else? If you want to define a function for that interval then I must tell you that there would be infinite functions satisfying the integral if $C=1$ and no function if $Cneq 1$






    share|cite|improve this answer

























      up vote
      2
      down vote



      accepted










      Hint:



      Substituting $t=x+1$ you will see that the integral will in fact be equal to $1$ rather than some constant C.



      Note: What I don't understand is that the domain you defined for the function is all real numbers but you did not define the function in $(0,1)$ so do you find the value of function in $(0,1)$ or anything else? If you want to define a function for that interval then I must tell you that there would be infinite functions satisfying the integral if $C=1$ and no function if $Cneq 1$






      share|cite|improve this answer























        up vote
        2
        down vote



        accepted







        up vote
        2
        down vote



        accepted






        Hint:



        Substituting $t=x+1$ you will see that the integral will in fact be equal to $1$ rather than some constant C.



        Note: What I don't understand is that the domain you defined for the function is all real numbers but you did not define the function in $(0,1)$ so do you find the value of function in $(0,1)$ or anything else? If you want to define a function for that interval then I must tell you that there would be infinite functions satisfying the integral if $C=1$ and no function if $Cneq 1$






        share|cite|improve this answer













        Hint:



        Substituting $t=x+1$ you will see that the integral will in fact be equal to $1$ rather than some constant C.



        Note: What I don't understand is that the domain you defined for the function is all real numbers but you did not define the function in $(0,1)$ so do you find the value of function in $(0,1)$ or anything else? If you want to define a function for that interval then I must tell you that there would be infinite functions satisfying the integral if $C=1$ and no function if $Cneq 1$







        share|cite|improve this answer













        share|cite|improve this answer



        share|cite|improve this answer











        answered Aug 3 at 5:35









        Manthanein

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