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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This question already has an answer here:

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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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asked Jan 12 at 7:28

DXTDXT

5,6892630

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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.

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  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
  
  
  

  
  
  
  


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  $begingroup$
  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
  

  
  
  
  

add a comment | 

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

asked Jan 12 at 7:28

DXTDXT

5,6892630

$endgroup$

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
  

  
  
  
  

add a comment | 

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

asked Jan 12 at 7:28

DXTDXT

5,6892630

$endgroup$

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

asked Jan 12 at 7:28

DXTDXT

5,6892630

share|cite|improve this question

asked Jan 12 at 7:28

DXTDXT

5,6892630

share|cite|improve this question

share|cite|improve this question

asked Jan 12 at 7:28

DXTDXT

5,6892630

asked Jan 12 at 7:28

DXTDXT

5,6892630

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
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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

add a comment | 

1 Answer
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Hint

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

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answered Jan 12 at 9:05

Mostafa AyazMostafa Ayaz

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

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

answered Jan 12 at 9:05

Mostafa AyazMostafa Ayaz

15.6k3939

$endgroup$

add a comment | 

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

answered Jan 12 at 9:05

Mostafa AyazMostafa Ayaz

15.6k3939

$endgroup$

add a comment | 

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

answered Jan 12 at 9:05

Mostafa AyazMostafa Ayaz

15.6k3939

$endgroup$

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

answered Jan 12 at 9:05

Mostafa AyazMostafa Ayaz

15.6k3939

share|cite|improve this answer

share|cite|improve this answer

answered Jan 12 at 9:05

Mostafa AyazMostafa Ayaz

15.6k3939

answered Jan 12 at 9:05

Mostafa AyazMostafa Ayaz

15.6k3939

answered Jan 12 at 9:05

Mostafa AyazMostafa Ayaz

15.6k3939

15.6k3939

add a comment | 

add a comment | 

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