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:
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
asked Jan 12 at 7:28
DXTDXT
5,8742733
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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.
add a comment |
$begingroup$
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
asked Jan 12 at 7:28
DXTDXT
5,8742733
$endgroup$
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.
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I mean, is there much to be said about this? Just compute $f(f(x))$ and then compute the integral.
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– 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 $$int_{-frac14}^frac14 left[left(u^3 + frac12right)^3 + 1right] du stackrel{color{red}{text{WHY?}}}{=} 2int_0^frac14 left[frac32 u^6 + frac98right] du = cdots $$
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– achille hui
Jan 12 at 7:53
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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
asked Jan 12 at 7:28
DXTDXT
5,8742733
$endgroup$
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
asked Jan 12 at 7:28
DXTDXT
5,8742733
asked Jan 12 at 7:28
DXTDXT
5,8742733
asked Jan 12 at 7:28
DXTDXT
5,8742733
asked Jan 12 at 7:28
DXTDXT
5,8742733
asked Jan 12 at 7:28
DXTDXT
5,8742733
5,8742733
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
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1 Answer
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Hint
Note that $$f(x)={(x-{1over 2})^3}+1$$and $$int^{frac{3}{4}}_{frac{1}{4}}f(f(x))dx=int^{frac{1}{4}}_{-frac{1}{4}}f(f(u+{1over 2}))du=int^{frac{1}{4}}_{-frac{1}{4}}f(u^3+1)du$$
answered Jan 12 at 9:05
Mostafa AyazMostafa Ayaz
18.1k31040
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add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint
Note that $$f(x)={(x-{1over 2})^3}+1$$and $$int^{frac{3}{4}}_{frac{1}{4}}f(f(x))dx=int^{frac{1}{4}}_{-frac{1}{4}}f(f(u+{1over 2}))du=int^{frac{1}{4}}_{-frac{1}{4}}f(u^3+1)du$$
answered Jan 12 at 9:05
Mostafa AyazMostafa Ayaz
18.1k31040
$endgroup$
add a comment |
$begingroup$
Hint
Note that $$f(x)={(x-{1over 2})^3}+1$$and $$int^{frac{3}{4}}_{frac{1}{4}}f(f(x))dx=int^{frac{1}{4}}_{-frac{1}{4}}f(f(u+{1over 2}))du=int^{frac{1}{4}}_{-frac{1}{4}}f(u^3+1)du$$
answered Jan 12 at 9:05
Mostafa AyazMostafa Ayaz
18.1k31040
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add a comment |
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Hint
Note that $$f(x)={(x-{1over 2})^3}+1$$and $$int^{frac{3}{4}}_{frac{1}{4}}f(f(x))dx=int^{frac{1}{4}}_{-frac{1}{4}}f(f(u+{1over 2}))du=int^{frac{1}{4}}_{-frac{1}{4}}f(u^3+1)du$$
answered Jan 12 at 9:05
Mostafa AyazMostafa Ayaz
18.1k31040
$endgroup$
Hint
Note that $$f(x)={(x-{1over 2})^3}+1$$and $$int^{frac{3}{4}}_{frac{1}{4}}f(f(x))dx=int^{frac{1}{4}}_{-frac{1}{4}}f(f(u+{1over 2}))du=int^{frac{1}{4}}_{-frac{1}{4}}f(u^3+1)du$$
answered Jan 12 at 9:05
Mostafa AyazMostafa Ayaz
18.1k31040
answered Jan 12 at 9:05
Mostafa AyazMostafa Ayaz
18.1k31040
answered Jan 12 at 9:05
Mostafa AyazMostafa Ayaz
18.1k31040
answered Jan 12 at 9:05
Mostafa AyazMostafa Ayaz
18.1k31040
18.1k31040
add a comment |
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