If $ f(x) = x^3+frac{3x}{4}-frac{3x^2}{2}+frac{7}{8}.$ Then $ int^{frac{3}{4}}_{frac{1}{4}}f(f(x))dx$...












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  • Let $f(x) = x^3-frac{3}{2}x^2+x+frac{1}{4}.$ Then the value of $ int^{3/4}_{1/4}f(f(x))mathrm dx$

    3 answers





If $displaystyle f(x) = x^3+frac{3x}{4}-frac{3x^2}{2}+frac{7}{8}.$ Then $displaystyle int^{frac{3}{4}}_{frac{1}{4}}f(f(x))dx$




Try: Given $displaystyle f(x) =x^3+frac{3x}{4}-frac{3x^2}{2}+frac{7}{8}. $



Then $displaystyle f(1-x) = (1-x)^3+frac{3}{4}(1-x)-frac{3}{2}(1-x)^2+frac{7}{8}$



$$f(1-x) = -x^3+frac{3x^2}{2}-frac{3x}{4}+frac{9}{8}$$



$$f(x)+f(1-x) = 2$$



Let $$I = int^{frac{3}{4}}_{frac{1}{4}}f(f(x))dx = int^{frac{3}{4}}_{frac{1}{4}}f(f(1-x))dx$$



I did not understand how to solve from there



could some help me to solve it






definite-integrals







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marked as duplicate by Anurag A, Claude Leibovici, user91500, egreg, Martin Sleziak Jan 12 at 14:49


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


















  • $begingroup$
    I mean, is there much to be said about this? Just compute $f(f(x))$ and then compute the integral.
    $endgroup$
    – MathematicsStudent1122
    Jan 12 at 7:43








  • 3




    $begingroup$
    Notice $$f(x) = left(x-frac12right)^3 + 1implies f(f(x)) = left(left(x-frac12right)^3 + frac12right)^3 + 1$$ Change variable to $u = x - frac12$, the integral becomes $$int_{-frac14}^frac14 left[left(u^3 + frac12right)^3 + 1right] du stackrel{color{red}{text{WHY?}}}{=} 2int_0^frac14 left[frac32 u^6 + frac98right] du = cdots $$
    $endgroup$
    – achille hui
    Jan 12 at 7:53
















2












$begingroup$



This question already has an answer here:




  • Let $f(x) = x^3-frac{3}{2}x^2+x+frac{1}{4}.$ Then the value of $ int^{3/4}_{1/4}f(f(x))mathrm dx$

    3 answers





If $displaystyle f(x) = x^3+frac{3x}{4}-frac{3x^2}{2}+frac{7}{8}.$ Then $displaystyle int^{frac{3}{4}}_{frac{1}{4}}f(f(x))dx$




Try: Given $displaystyle f(x) =x^3+frac{3x}{4}-frac{3x^2}{2}+frac{7}{8}. $



Then $displaystyle f(1-x) = (1-x)^3+frac{3}{4}(1-x)-frac{3}{2}(1-x)^2+frac{7}{8}$



$$f(1-x) = -x^3+frac{3x^2}{2}-frac{3x}{4}+frac{9}{8}$$



$$f(x)+f(1-x) = 2$$



Let $$I = int^{frac{3}{4}}_{frac{1}{4}}f(f(x))dx = int^{frac{3}{4}}_{frac{1}{4}}f(f(1-x))dx$$



I did not understand how to solve from there



could some help me to solve it






definite-integrals







share|cite|improve this question









$endgroup$



marked as duplicate by Anurag A, Claude Leibovici, user91500, egreg, Martin Sleziak Jan 12 at 14:49


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


















  • $begingroup$
    I mean, is there much to be said about this? Just compute $f(f(x))$ and then compute the integral.
    $endgroup$
    – MathematicsStudent1122
    Jan 12 at 7:43








  • 3




    $begingroup$
    Notice $$f(x) = left(x-frac12right)^3 + 1implies f(f(x)) = left(left(x-frac12right)^3 + frac12right)^3 + 1$$ Change variable to $u = x - frac12$, the integral becomes $$int_{-frac14}^frac14 left[left(u^3 + frac12right)^3 + 1right] du stackrel{color{red}{text{WHY?}}}{=} 2int_0^frac14 left[frac32 u^6 + frac98right] du = cdots $$
    $endgroup$
    – achille hui
    Jan 12 at 7:53














2












2








2


1



$begingroup$



This question already has an answer here:




  • Let $f(x) = x^3-frac{3}{2}x^2+x+frac{1}{4}.$ Then the value of $ int^{3/4}_{1/4}f(f(x))mathrm dx$

    3 answers





If $displaystyle f(x) = x^3+frac{3x}{4}-frac{3x^2}{2}+frac{7}{8}.$ Then $displaystyle int^{frac{3}{4}}_{frac{1}{4}}f(f(x))dx$




Try: Given $displaystyle f(x) =x^3+frac{3x}{4}-frac{3x^2}{2}+frac{7}{8}. $



Then $displaystyle f(1-x) = (1-x)^3+frac{3}{4}(1-x)-frac{3}{2}(1-x)^2+frac{7}{8}$



$$f(1-x) = -x^3+frac{3x^2}{2}-frac{3x}{4}+frac{9}{8}$$



$$f(x)+f(1-x) = 2$$



Let $$I = int^{frac{3}{4}}_{frac{1}{4}}f(f(x))dx = int^{frac{3}{4}}_{frac{1}{4}}f(f(1-x))dx$$



I did not understand how to solve from there



could some help me to solve it






definite-integrals







share|cite|improve this question









$endgroup$





This question already has an answer here:




  • Let $f(x) = x^3-frac{3}{2}x^2+x+frac{1}{4}.$ Then the value of $ int^{3/4}_{1/4}f(f(x))mathrm dx$

    3 answers





If $displaystyle f(x) = x^3+frac{3x}{4}-frac{3x^2}{2}+frac{7}{8}.$ Then $displaystyle int^{frac{3}{4}}_{frac{1}{4}}f(f(x))dx$




Try: Given $displaystyle f(x) =x^3+frac{3x}{4}-frac{3x^2}{2}+frac{7}{8}. $



Then $displaystyle f(1-x) = (1-x)^3+frac{3}{4}(1-x)-frac{3}{2}(1-x)^2+frac{7}{8}$



$$f(1-x) = -x^3+frac{3x^2}{2}-frac{3x}{4}+frac{9}{8}$$



$$f(x)+f(1-x) = 2$$



Let $$I = int^{frac{3}{4}}_{frac{1}{4}}f(f(x))dx = int^{frac{3}{4}}_{frac{1}{4}}f(f(1-x))dx$$



I did not understand how to solve from there



could some help me to solve it





This question already has an answer here:




  • Let $f(x) = x^3-frac{3}{2}x^2+x+frac{1}{4}.$ Then the value of $ int^{3/4}_{1/4}f(f(x))mathrm dx$

    3 answers







definite-integrals




definite-integrals






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 12 at 7:28









DXTDXT

5,6892630




5,6892630




marked as duplicate by Anurag A, Claude Leibovici, user91500, egreg, Martin Sleziak Jan 12 at 14:49


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.









marked as duplicate by Anurag A, Claude Leibovici, user91500, egreg, Martin Sleziak Jan 12 at 14:49


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • $begingroup$
    I mean, is there much to be said about this? Just compute $f(f(x))$ and then compute the integral.
    $endgroup$
    – MathematicsStudent1122
    Jan 12 at 7:43








  • 3




    $begingroup$
    Notice $$f(x) = left(x-frac12right)^3 + 1implies f(f(x)) = left(left(x-frac12right)^3 + frac12right)^3 + 1$$ Change variable to $u = x - frac12$, the integral becomes $$int_{-frac14}^frac14 left[left(u^3 + frac12right)^3 + 1right] du stackrel{color{red}{text{WHY?}}}{=} 2int_0^frac14 left[frac32 u^6 + frac98right] du = cdots $$
    $endgroup$
    – achille hui
    Jan 12 at 7:53


















  • $begingroup$
    I mean, is there much to be said about this? Just compute $f(f(x))$ and then compute the integral.
    $endgroup$
    – MathematicsStudent1122
    Jan 12 at 7:43








  • 3




    $begingroup$
    Notice $$f(x) = left(x-frac12right)^3 + 1implies f(f(x)) = left(left(x-frac12right)^3 + frac12right)^3 + 1$$ Change variable to $u = x - frac12$, the integral becomes $$int_{-frac14}^frac14 left[left(u^3 + frac12right)^3 + 1right] du stackrel{color{red}{text{WHY?}}}{=} 2int_0^frac14 left[frac32 u^6 + frac98right] du = cdots $$
    $endgroup$
    – achille hui
    Jan 12 at 7:53
















$begingroup$
I mean, is there much to be said about this? Just compute $f(f(x))$ and then compute the integral.
$endgroup$
– MathematicsStudent1122
Jan 12 at 7:43






$begingroup$
I mean, is there much to be said about this? Just compute $f(f(x))$ and then compute the integral.
$endgroup$
– MathematicsStudent1122
Jan 12 at 7:43






3




3




$begingroup$
Notice $$f(x) = left(x-frac12right)^3 + 1implies f(f(x)) = left(left(x-frac12right)^3 + frac12right)^3 + 1$$ Change variable to $u = x - frac12$, the integral becomes $$int_{-frac14}^frac14 left[left(u^3 + frac12right)^3 + 1right] du stackrel{color{red}{text{WHY?}}}{=} 2int_0^frac14 left[frac32 u^6 + frac98right] du = cdots $$
$endgroup$
– achille hui
Jan 12 at 7:53




$begingroup$
Notice $$f(x) = left(x-frac12right)^3 + 1implies f(f(x)) = left(left(x-frac12right)^3 + frac12right)^3 + 1$$ Change variable to $u = x - frac12$, the integral becomes $$int_{-frac14}^frac14 left[left(u^3 + frac12right)^3 + 1right] du stackrel{color{red}{text{WHY?}}}{=} 2int_0^frac14 left[frac32 u^6 + frac98right] du = cdots $$
$endgroup$
– achille hui
Jan 12 at 7:53










1 Answer
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Hint



Note that $$f(x)={(x-{1over 2})^3}+1$$and $$int^{frac{3}{4}}_{frac{1}{4}}f(f(x))dx=int^{frac{1}{4}}_{-frac{1}{4}}f(f(u+{1over 2}))du=int^{frac{1}{4}}_{-frac{1}{4}}f(u^3+1)du$$






share|cite|improve this answer









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    1 Answer
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    1 Answer
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    active

    oldest

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    active

    oldest

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    active

    oldest

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    3












    $begingroup$

    Hint



    Note that $$f(x)={(x-{1over 2})^3}+1$$and $$int^{frac{3}{4}}_{frac{1}{4}}f(f(x))dx=int^{frac{1}{4}}_{-frac{1}{4}}f(f(u+{1over 2}))du=int^{frac{1}{4}}_{-frac{1}{4}}f(u^3+1)du$$






    share|cite|improve this answer









    $endgroup$


















      3












      $begingroup$

      Hint



      Note that $$f(x)={(x-{1over 2})^3}+1$$and $$int^{frac{3}{4}}_{frac{1}{4}}f(f(x))dx=int^{frac{1}{4}}_{-frac{1}{4}}f(f(u+{1over 2}))du=int^{frac{1}{4}}_{-frac{1}{4}}f(u^3+1)du$$






      share|cite|improve this answer









      $endgroup$
















        3












        3








        3





        $begingroup$

        Hint



        Note that $$f(x)={(x-{1over 2})^3}+1$$and $$int^{frac{3}{4}}_{frac{1}{4}}f(f(x))dx=int^{frac{1}{4}}_{-frac{1}{4}}f(f(u+{1over 2}))du=int^{frac{1}{4}}_{-frac{1}{4}}f(u^3+1)du$$






        share|cite|improve this answer









        $endgroup$



        Hint



        Note that $$f(x)={(x-{1over 2})^3}+1$$and $$int^{frac{3}{4}}_{frac{1}{4}}f(f(x))dx=int^{frac{1}{4}}_{-frac{1}{4}}f(f(u+{1over 2}))du=int^{frac{1}{4}}_{-frac{1}{4}}f(u^3+1)du$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 12 at 9:05









        Mostafa AyazMostafa Ayaz

        15.6k3939




        15.6k3939















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