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$












2












$begingroup$


If $displaystyle f(x) = x^3-frac{3}{2}x^2+x+frac{1}{4}.$ then the value of $displaystyle int^{frac{3}{4}}_{frac{1}{4}}f(f(x))mathrm dx$



$displaystyle int^{frac{3}{4}}_{frac{1}{4}}(f(x))^3-1.5(f(x))^2+f(x)+0.25 mathrm dx$



could some help me with this, thanks






integration definite-integrals







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Is there any reason to not simply expand it out? Or are you looking for something more? If expanding is fine, then wolfram gave an answer of $frac{1}{4}$.
    $endgroup$
    – Nick Guerrero
    Jan 9 '17 at 6:05






  • 2




    $begingroup$
    The substitution $y=x-frac{1}{2}$ simplifies $f$ to a depressed cubic, and the integration interval to a symmetric one around $0,$.
    $endgroup$
    – dxiv
    Jan 9 '17 at 6:12


















2












$begingroup$


If $displaystyle f(x) = x^3-frac{3}{2}x^2+x+frac{1}{4}.$ then the value of $displaystyle int^{frac{3}{4}}_{frac{1}{4}}f(f(x))mathrm dx$



$displaystyle int^{frac{3}{4}}_{frac{1}{4}}(f(x))^3-1.5(f(x))^2+f(x)+0.25 mathrm dx$



could some help me with this, thanks






integration definite-integrals







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Is there any reason to not simply expand it out? Or are you looking for something more? If expanding is fine, then wolfram gave an answer of $frac{1}{4}$.
    $endgroup$
    – Nick Guerrero
    Jan 9 '17 at 6:05






  • 2




    $begingroup$
    The substitution $y=x-frac{1}{2}$ simplifies $f$ to a depressed cubic, and the integration interval to a symmetric one around $0,$.
    $endgroup$
    – dxiv
    Jan 9 '17 at 6:12
















2












2








2


1



$begingroup$


If $displaystyle f(x) = x^3-frac{3}{2}x^2+x+frac{1}{4}.$ then the value of $displaystyle int^{frac{3}{4}}_{frac{1}{4}}f(f(x))mathrm dx$



$displaystyle int^{frac{3}{4}}_{frac{1}{4}}(f(x))^3-1.5(f(x))^2+f(x)+0.25 mathrm dx$



could some help me with this, thanks






integration definite-integrals







share|cite|improve this question











$endgroup$




If $displaystyle f(x) = x^3-frac{3}{2}x^2+x+frac{1}{4}.$ then the value of $displaystyle int^{frac{3}{4}}_{frac{1}{4}}f(f(x))mathrm dx$



$displaystyle int^{frac{3}{4}}_{frac{1}{4}}(f(x))^3-1.5(f(x))^2+f(x)+0.25 mathrm dx$



could some help me with this, thanks






integration definite-integrals




integration definite-integrals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 12 at 14:45









Martin Sleziak

44.8k10118272




44.8k10118272










asked Jan 9 '17 at 6:00









DXTDXT

5,6892630




5,6892630








  • 1




    $begingroup$
    Is there any reason to not simply expand it out? Or are you looking for something more? If expanding is fine, then wolfram gave an answer of $frac{1}{4}$.
    $endgroup$
    – Nick Guerrero
    Jan 9 '17 at 6:05






  • 2




    $begingroup$
    The substitution $y=x-frac{1}{2}$ simplifies $f$ to a depressed cubic, and the integration interval to a symmetric one around $0,$.
    $endgroup$
    – dxiv
    Jan 9 '17 at 6:12
















  • 1




    $begingroup$
    Is there any reason to not simply expand it out? Or are you looking for something more? If expanding is fine, then wolfram gave an answer of $frac{1}{4}$.
    $endgroup$
    – Nick Guerrero
    Jan 9 '17 at 6:05






  • 2




    $begingroup$
    The substitution $y=x-frac{1}{2}$ simplifies $f$ to a depressed cubic, and the integration interval to a symmetric one around $0,$.
    $endgroup$
    – dxiv
    Jan 9 '17 at 6:12










1




1




$begingroup$
Is there any reason to not simply expand it out? Or are you looking for something more? If expanding is fine, then wolfram gave an answer of $frac{1}{4}$.
$endgroup$
– Nick Guerrero
Jan 9 '17 at 6:05




$begingroup$
Is there any reason to not simply expand it out? Or are you looking for something more? If expanding is fine, then wolfram gave an answer of $frac{1}{4}$.
$endgroup$
– Nick Guerrero
Jan 9 '17 at 6:05




2




2




$begingroup$
The substitution $y=x-frac{1}{2}$ simplifies $f$ to a depressed cubic, and the integration interval to a symmetric one around $0,$.
$endgroup$
– dxiv
Jan 9 '17 at 6:12






$begingroup$
The substitution $y=x-frac{1}{2}$ simplifies $f$ to a depressed cubic, and the integration interval to a symmetric one around $0,$.
$endgroup$
– dxiv
Jan 9 '17 at 6:12












3 Answers
3






active

oldest

votes


















7












$begingroup$

As @dxiv says in the comments, we can write $$f (x) = (x-frac {1}{2})^3 +frac {1}{4}x +frac {3}{8} $$ Thus, substituting $u=x- frac {1}{2}$ gives us $f (u)=u^3+frac {1}{4}(u+frac {1}{2}) +frac {3}{8} = u^3+frac {1}{4}u +frac {1}{2} $. Thus, $$f (f (u)) =(u^3+frac {1}{4}u +frac {1}{2})^3 +frac {1}{4}[u^3 +frac {1}{4}u +frac {1}{2}] +frac {1}{2}$$ As $x $ goes from $frac {1}{4} $ to $frac {3}{4} $,. $u $ goes from $-frac {1}{4} $ to $frac {1}{4} $.



Hope you can take it from here.






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    Apply kings rule, I=int(f(f(1-x))) =integral( -x3 +1.5x2 -x +0.75)

    Add the two 2I = int(1)

    I= 0.25.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Sorry for the unformatted answer, on mobile, in journey
      $endgroup$
      – Red Floyd
      Jan 9 '17 at 7:24










    • $begingroup$
      How could $int f(f(1-x))dx=int -x^3+1.5x^2-x+0.75 dx$ I mean $int f(1-x)=int -x^3+1.5x^2-x+0.75 dx$ and not the one you stated.Thanks.
      $endgroup$
      – Navin
      Apr 7 '17 at 20:13



















    1












    $begingroup$

    You can use the property given here:
    $$color{blue}{int_a^b f(x)dx=int_a^{(a+b)/2}[f(x)+f(a+b-x)]dx Rightarrow \ int_a^b f(f(x))dx=int_a^{(a+b)/2}[f(f(x))+f(f(a+b-x))]dx};\
    int_{1/4}^{3/4} f(f(x))dx=int_{1/4}^{1/2}[f(f(x))+f(f(1-x))]dx=\
    int_{1/4}^{1/2}[f(f(x))+f(f(1-x))]dx=\
    int_{1/4}^{1/2}[1]dx=frac14,$$
    because:
    $$f(f(x))=x^3-frac32x^2+x+frac14;\
    f(f(1-x))=(1-x)^3-frac32(1-x)^2+(1-x)+frac14=
    -x^3+frac32x^2-x+frac34.$$
    Also, the property mentioned by RedFloyd is given in the above source:
    $$color{blue}{int_a^b f(x)dx=int_a^b f(a+b-x)dx Rightarrow \
    int_a^b f(f(x))dx=int_a^b f(f(a+b-x))dx}.$$






    share|cite|improve this answer









    $endgroup$













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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      7












      $begingroup$

      As @dxiv says in the comments, we can write $$f (x) = (x-frac {1}{2})^3 +frac {1}{4}x +frac {3}{8} $$ Thus, substituting $u=x- frac {1}{2}$ gives us $f (u)=u^3+frac {1}{4}(u+frac {1}{2}) +frac {3}{8} = u^3+frac {1}{4}u +frac {1}{2} $. Thus, $$f (f (u)) =(u^3+frac {1}{4}u +frac {1}{2})^3 +frac {1}{4}[u^3 +frac {1}{4}u +frac {1}{2}] +frac {1}{2}$$ As $x $ goes from $frac {1}{4} $ to $frac {3}{4} $,. $u $ goes from $-frac {1}{4} $ to $frac {1}{4} $.



      Hope you can take it from here.






      share|cite|improve this answer









      $endgroup$


















        7












        $begingroup$

        As @dxiv says in the comments, we can write $$f (x) = (x-frac {1}{2})^3 +frac {1}{4}x +frac {3}{8} $$ Thus, substituting $u=x- frac {1}{2}$ gives us $f (u)=u^3+frac {1}{4}(u+frac {1}{2}) +frac {3}{8} = u^3+frac {1}{4}u +frac {1}{2} $. Thus, $$f (f (u)) =(u^3+frac {1}{4}u +frac {1}{2})^3 +frac {1}{4}[u^3 +frac {1}{4}u +frac {1}{2}] +frac {1}{2}$$ As $x $ goes from $frac {1}{4} $ to $frac {3}{4} $,. $u $ goes from $-frac {1}{4} $ to $frac {1}{4} $.



        Hope you can take it from here.






        share|cite|improve this answer









        $endgroup$
















          7












          7








          7





          $begingroup$

          As @dxiv says in the comments, we can write $$f (x) = (x-frac {1}{2})^3 +frac {1}{4}x +frac {3}{8} $$ Thus, substituting $u=x- frac {1}{2}$ gives us $f (u)=u^3+frac {1}{4}(u+frac {1}{2}) +frac {3}{8} = u^3+frac {1}{4}u +frac {1}{2} $. Thus, $$f (f (u)) =(u^3+frac {1}{4}u +frac {1}{2})^3 +frac {1}{4}[u^3 +frac {1}{4}u +frac {1}{2}] +frac {1}{2}$$ As $x $ goes from $frac {1}{4} $ to $frac {3}{4} $,. $u $ goes from $-frac {1}{4} $ to $frac {1}{4} $.



          Hope you can take it from here.






          share|cite|improve this answer









          $endgroup$



          As @dxiv says in the comments, we can write $$f (x) = (x-frac {1}{2})^3 +frac {1}{4}x +frac {3}{8} $$ Thus, substituting $u=x- frac {1}{2}$ gives us $f (u)=u^3+frac {1}{4}(u+frac {1}{2}) +frac {3}{8} = u^3+frac {1}{4}u +frac {1}{2} $. Thus, $$f (f (u)) =(u^3+frac {1}{4}u +frac {1}{2})^3 +frac {1}{4}[u^3 +frac {1}{4}u +frac {1}{2}] +frac {1}{2}$$ As $x $ goes from $frac {1}{4} $ to $frac {3}{4} $,. $u $ goes from $-frac {1}{4} $ to $frac {1}{4} $.



          Hope you can take it from here.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 9 '17 at 6:23









          RohanRohan

          27.8k42444




          27.8k42444























              2












              $begingroup$

              Apply kings rule, I=int(f(f(1-x))) =integral( -x3 +1.5x2 -x +0.75)

              Add the two 2I = int(1)

              I= 0.25.






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                Sorry for the unformatted answer, on mobile, in journey
                $endgroup$
                – Red Floyd
                Jan 9 '17 at 7:24










              • $begingroup$
                How could $int f(f(1-x))dx=int -x^3+1.5x^2-x+0.75 dx$ I mean $int f(1-x)=int -x^3+1.5x^2-x+0.75 dx$ and not the one you stated.Thanks.
                $endgroup$
                – Navin
                Apr 7 '17 at 20:13
















              2












              $begingroup$

              Apply kings rule, I=int(f(f(1-x))) =integral( -x3 +1.5x2 -x +0.75)

              Add the two 2I = int(1)

              I= 0.25.






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                Sorry for the unformatted answer, on mobile, in journey
                $endgroup$
                – Red Floyd
                Jan 9 '17 at 7:24










              • $begingroup$
                How could $int f(f(1-x))dx=int -x^3+1.5x^2-x+0.75 dx$ I mean $int f(1-x)=int -x^3+1.5x^2-x+0.75 dx$ and not the one you stated.Thanks.
                $endgroup$
                – Navin
                Apr 7 '17 at 20:13














              2












              2








              2





              $begingroup$

              Apply kings rule, I=int(f(f(1-x))) =integral( -x3 +1.5x2 -x +0.75)

              Add the two 2I = int(1)

              I= 0.25.






              share|cite|improve this answer











              $endgroup$



              Apply kings rule, I=int(f(f(1-x))) =integral( -x3 +1.5x2 -x +0.75)

              Add the two 2I = int(1)

              I= 0.25.







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Feb 5 '17 at 18:46

























              answered Jan 9 '17 at 7:10









              Red FloydRed Floyd

              273314




              273314












              • $begingroup$
                Sorry for the unformatted answer, on mobile, in journey
                $endgroup$
                – Red Floyd
                Jan 9 '17 at 7:24










              • $begingroup$
                How could $int f(f(1-x))dx=int -x^3+1.5x^2-x+0.75 dx$ I mean $int f(1-x)=int -x^3+1.5x^2-x+0.75 dx$ and not the one you stated.Thanks.
                $endgroup$
                – Navin
                Apr 7 '17 at 20:13


















              • $begingroup$
                Sorry for the unformatted answer, on mobile, in journey
                $endgroup$
                – Red Floyd
                Jan 9 '17 at 7:24










              • $begingroup$
                How could $int f(f(1-x))dx=int -x^3+1.5x^2-x+0.75 dx$ I mean $int f(1-x)=int -x^3+1.5x^2-x+0.75 dx$ and not the one you stated.Thanks.
                $endgroup$
                – Navin
                Apr 7 '17 at 20:13
















              $begingroup$
              Sorry for the unformatted answer, on mobile, in journey
              $endgroup$
              – Red Floyd
              Jan 9 '17 at 7:24




              $begingroup$
              Sorry for the unformatted answer, on mobile, in journey
              $endgroup$
              – Red Floyd
              Jan 9 '17 at 7:24












              $begingroup$
              How could $int f(f(1-x))dx=int -x^3+1.5x^2-x+0.75 dx$ I mean $int f(1-x)=int -x^3+1.5x^2-x+0.75 dx$ and not the one you stated.Thanks.
              $endgroup$
              – Navin
              Apr 7 '17 at 20:13




              $begingroup$
              How could $int f(f(1-x))dx=int -x^3+1.5x^2-x+0.75 dx$ I mean $int f(1-x)=int -x^3+1.5x^2-x+0.75 dx$ and not the one you stated.Thanks.
              $endgroup$
              – Navin
              Apr 7 '17 at 20:13











              1












              $begingroup$

              You can use the property given here:
              $$color{blue}{int_a^b f(x)dx=int_a^{(a+b)/2}[f(x)+f(a+b-x)]dx Rightarrow \ int_a^b f(f(x))dx=int_a^{(a+b)/2}[f(f(x))+f(f(a+b-x))]dx};\
              int_{1/4}^{3/4} f(f(x))dx=int_{1/4}^{1/2}[f(f(x))+f(f(1-x))]dx=\
              int_{1/4}^{1/2}[f(f(x))+f(f(1-x))]dx=\
              int_{1/4}^{1/2}[1]dx=frac14,$$
              because:
              $$f(f(x))=x^3-frac32x^2+x+frac14;\
              f(f(1-x))=(1-x)^3-frac32(1-x)^2+(1-x)+frac14=
              -x^3+frac32x^2-x+frac34.$$
              Also, the property mentioned by RedFloyd is given in the above source:
              $$color{blue}{int_a^b f(x)dx=int_a^b f(a+b-x)dx Rightarrow \
              int_a^b f(f(x))dx=int_a^b f(f(a+b-x))dx}.$$






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                You can use the property given here:
                $$color{blue}{int_a^b f(x)dx=int_a^{(a+b)/2}[f(x)+f(a+b-x)]dx Rightarrow \ int_a^b f(f(x))dx=int_a^{(a+b)/2}[f(f(x))+f(f(a+b-x))]dx};\
                int_{1/4}^{3/4} f(f(x))dx=int_{1/4}^{1/2}[f(f(x))+f(f(1-x))]dx=\
                int_{1/4}^{1/2}[f(f(x))+f(f(1-x))]dx=\
                int_{1/4}^{1/2}[1]dx=frac14,$$
                because:
                $$f(f(x))=x^3-frac32x^2+x+frac14;\
                f(f(1-x))=(1-x)^3-frac32(1-x)^2+(1-x)+frac14=
                -x^3+frac32x^2-x+frac34.$$
                Also, the property mentioned by RedFloyd is given in the above source:
                $$color{blue}{int_a^b f(x)dx=int_a^b f(a+b-x)dx Rightarrow \
                int_a^b f(f(x))dx=int_a^b f(f(a+b-x))dx}.$$






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  You can use the property given here:
                  $$color{blue}{int_a^b f(x)dx=int_a^{(a+b)/2}[f(x)+f(a+b-x)]dx Rightarrow \ int_a^b f(f(x))dx=int_a^{(a+b)/2}[f(f(x))+f(f(a+b-x))]dx};\
                  int_{1/4}^{3/4} f(f(x))dx=int_{1/4}^{1/2}[f(f(x))+f(f(1-x))]dx=\
                  int_{1/4}^{1/2}[f(f(x))+f(f(1-x))]dx=\
                  int_{1/4}^{1/2}[1]dx=frac14,$$
                  because:
                  $$f(f(x))=x^3-frac32x^2+x+frac14;\
                  f(f(1-x))=(1-x)^3-frac32(1-x)^2+(1-x)+frac14=
                  -x^3+frac32x^2-x+frac34.$$
                  Also, the property mentioned by RedFloyd is given in the above source:
                  $$color{blue}{int_a^b f(x)dx=int_a^b f(a+b-x)dx Rightarrow \
                  int_a^b f(f(x))dx=int_a^b f(f(a+b-x))dx}.$$






                  share|cite|improve this answer









                  $endgroup$



                  You can use the property given here:
                  $$color{blue}{int_a^b f(x)dx=int_a^{(a+b)/2}[f(x)+f(a+b-x)]dx Rightarrow \ int_a^b f(f(x))dx=int_a^{(a+b)/2}[f(f(x))+f(f(a+b-x))]dx};\
                  int_{1/4}^{3/4} f(f(x))dx=int_{1/4}^{1/2}[f(f(x))+f(f(1-x))]dx=\
                  int_{1/4}^{1/2}[f(f(x))+f(f(1-x))]dx=\
                  int_{1/4}^{1/2}[1]dx=frac14,$$
                  because:
                  $$f(f(x))=x^3-frac32x^2+x+frac14;\
                  f(f(1-x))=(1-x)^3-frac32(1-x)^2+(1-x)+frac14=
                  -x^3+frac32x^2-x+frac34.$$
                  Also, the property mentioned by RedFloyd is given in the above source:
                  $$color{blue}{int_a^b f(x)dx=int_a^b f(a+b-x)dx Rightarrow \
                  int_a^b f(f(x))dx=int_a^b f(f(a+b-x))dx}.$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 12 at 16:24









                  farruhotafarruhota

                  20.2k2738




                  20.2k2738






























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                      MongoDB - Not Authorized To Execute Command

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                      .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 0 I have successfully enabled authorization on MongoDB and I have created an account on the admin database and then I created an account for my database called test. The following connection string to connect to my test database works successfully: mongo --host 192.168.17.52 --port 27017 -u user1 -p password --authenticationDatabase test Only problem I have now is, I cannot execute commands such as: show dbs. I get the following error when I try to do so: "errmsg" : "not authorized on admin to execute command { listDatabases: 1.0, lsid: { id: UUID("a1d5bc0d-bc58-485e-b232-270758a89455") }, $db: "admin...
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                      How to fix TextFormField cause rebuild widget in Flutter

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                      .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 0 I've got the same issue at #9280. I tried all solutions from that topic with/without GlobalKey for Scaffold and Form but the problem wasn't fixed. I tried to build a authentication form with login and signup. But the keyboard doesn't show when I tap the TextFormField at the first time. Then I try to input some letter from physical keyboard, the widget is rebuilt. Like this: login screen Nothing to be log when I run flutter run --verbose, so I logged manually. Every tapping on field, the widget called dispose -> init -> build. [√] Flutter (Channel beta, v1.0.0, on Microsoft Windows [Version 10.0.17763.195], ...
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                      in spring boot 2.1 many test slices are not allowed anymore due to multiple @BootstrapWith

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                      1 I tried to upgrade a yummy sandwich made of two test slices (@JsonTest and @JdbcTest in my case, crunchy test code in between) adding spring boot 2.1 flavour to it. But it seems it was not much of a success. I cannot annotate my tests with many @...Test since they are now each bringing their own XxxTestContextBootstrapper. It used to work when they all used same SpringBootTestContextBootstrapper. @RunWith(SpringRunner.class) @JdbcTest @JsonTest public class Test { @Test public void test() { System.out.printn("Hello, World !"); } } The error I get from BootstrapUtils is illegalStateException : Configuration error: found multiple declarations of @BootstrapWith for test class I understand I might be doing something wrong here but is there an easy way I could load both Json and Jdbc contex...
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