If $2int_0^1 xf(x)dxgeqint_0^1(f(x))^2dx$, prove than $int_0^1(f(x))^2dxgeqfrac43$Prove that $int_0^1|f(x)f'(x)|dxleqfrac12int_0^1|f'(x)|^2dx$Integral inequality $int_0^1h(x)^2dxint_0^1x^2h(x)^2dxge cleft(int_0^1h(x)dxright)^4$Prove that $int_0^1f(x)^2dxgeq 4$$int_0^1f(x)cdot g(x)dx=int_0^1f(x)dxcdot int_0^1g(x)dx$Prove that:$f(f(x)) = x^2 implies int_0^1(f(x))^2dx geq frac313$how can we explain that $int_0^L1 + frac12(u'(x))^2dx - L = int_0^Lfrac12(u'(x))^2dx$?For a continuous function with $int_0^1 f(x)dx=1$, and $M=maxf(x)$, show that $1-frac12Mgeqint_0^1xf(x)dxgeqfrac12M$Let $f:mathbbRrightarrowmathbbR$ be a continuous function such that $int_x^1f(t)dtgeq(1-x)^2$. Prove that $f(1)=0$Prove that $int_0^1f(x)arctan x dx=fracπ8int_0^1f(x)dx$Prove that $f(x)=0$ for any $xin[0,1]$.
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If $2int_0^1 xf(x)dxgeqint_0^1(f(x))^2dx$, prove than $int_0^1(f(x))^2dxgeqfrac43$
Prove that $int_0^1|f(x)f'(x)|dxleqfrac12int_0^1|f'(x)|^2dx$Integral inequality $int_0^1h(x)^2dxint_0^1x^2h(x)^2dxge cleft(int_0^1h(x)dxright)^4$Prove that $int_0^1f(x)^2dxgeq 4$$int_0^1f(x)cdot g(x)dx=int_0^1f(x)dxcdot int_0^1g(x)dx$Prove that:$f(f(x)) = x^2 implies int_0^1(f(x))^2dx geq frac313$how can we explain that $int_0^L1 + frac12(u'(x))^2dx - L = int_0^Lfrac12(u'(x))^2dx$?For a continuous function with $int_0^1 f(x)dx=1$, and $M=maxf(x)$, show that $1-frac12Mgeqint_0^1xf(x)dxgeqfrac12M$Let $f:mathbbRrightarrowmathbbR$ be a continuous function such that $int_x^1f(t)dtgeq(1-x)^2$. Prove that $f(1)=0$Prove that $int_0^1f(x)arctan x dx=fracπ8int_0^1f(x)dx$Prove that $f(x)=0$ for any $xin[0,1]$.
$begingroup$
If $f:[0,1]rightarrowmathbbR$ is a continuous function such that $2int_0^1 xf(x)dxgeqint_0^1(f(x))^2dx$, prove that $int_0^1(f(x))^2dxgeqfrac43$.
However, using CBS I proved that $int_0^1(f(x))^2dxint_0^1x^2dxgeq(int_0^1(f(x))xdx)^2geqfrac14(int_0^1(f(x))^2dx)^2$ and so I obtain exactly the reverse of what I am asked to:$frac43geqint_0^1(f(x))^2dx$. Is there a mistake in the task or in my proof?
integration analysis inequality definite-integrals continuity
edited 2 days ago
Michael Rozenberg
107k1895200
asked Mar 7 at 8:42
Septimiu CristianSeptimiu Cristian
1
$endgroup$
add a comment |
$begingroup$
If $f:[0,1]rightarrowmathbbR$ is a continuous function such that $2int_0^1 xf(x)dxgeqint_0^1(f(x))^2dx$, prove that $int_0^1(f(x))^2dxgeqfrac43$.
However, using CBS I proved that $int_0^1(f(x))^2dxint_0^1x^2dxgeq(int_0^1(f(x))xdx)^2geqfrac14(int_0^1(f(x))^2dx)^2$ and so I obtain exactly the reverse of what I am asked to:$frac43geqint_0^1(f(x))^2dx$. Is there a mistake in the task or in my proof?
integration analysis inequality definite-integrals continuity
edited 2 days ago
Michael Rozenberg
107k1895200
asked Mar 7 at 8:42
Septimiu CristianSeptimiu Cristian
1
$endgroup$
add a comment |
$begingroup$
If $f:[0,1]rightarrowmathbbR$ is a continuous function such that $2int_0^1 xf(x)dxgeqint_0^1(f(x))^2dx$, prove that $int_0^1(f(x))^2dxgeqfrac43$.
However, using CBS I proved that $int_0^1(f(x))^2dxint_0^1x^2dxgeq(int_0^1(f(x))xdx)^2geqfrac14(int_0^1(f(x))^2dx)^2$ and so I obtain exactly the reverse of what I am asked to:$frac43geqint_0^1(f(x))^2dx$. Is there a mistake in the task or in my proof?
integration analysis inequality definite-integrals continuity
edited 2 days ago
Michael Rozenberg
107k1895200
asked Mar 7 at 8:42
Septimiu CristianSeptimiu Cristian
1
$endgroup$
If $f:[0,1]rightarrowmathbbR$ is a continuous function such that $2int_0^1 xf(x)dxgeqint_0^1(f(x))^2dx$, prove that $int_0^1(f(x))^2dxgeqfrac43$.
However, using CBS I proved that $int_0^1(f(x))^2dxint_0^1x^2dxgeq(int_0^1(f(x))xdx)^2geqfrac14(int_0^1(f(x))^2dx)^2$ and so I obtain exactly the reverse of what I am asked to:$frac43geqint_0^1(f(x))^2dx$. Is there a mistake in the task or in my proof?
integration analysis inequality definite-integrals continuity
integration analysis inequality definite-integrals continuity
edited 2 days ago
Michael Rozenberg
107k1895200
asked Mar 7 at 8:42
Septimiu CristianSeptimiu Cristian
1
edited 2 days ago
Michael Rozenberg
107k1895200
asked Mar 7 at 8:42
Septimiu CristianSeptimiu Cristian
1
edited 2 days ago
Michael Rozenberg
107k1895200
edited 2 days ago
Michael Rozenberg
107k1895200
edited 2 days ago
Michael Rozenberg
107k1895200
107k1895200
asked Mar 7 at 8:42
Septimiu CristianSeptimiu Cristian
1
asked Mar 7 at 8:42
Septimiu CristianSeptimiu Cristian
1
asked Mar 7 at 8:42
Septimiu CristianSeptimiu Cristian
1
1
add a comment |
add a comment |
2 Answers
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oldest
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$begingroup$
Yes, the inequality $int_0^1(f(x))^2dxgeqfrac43$ should be reversed. Note that $f=0$ satisfies
$$2int_0^1 xf(x)dxgeqint_0^1(f(x))^2dx$$
but $int_0^1(f(x))^2dxgeqfrac43$ does not hold.
On the other hand, by Cauchy-Schwarz (note that $int_0^1 2xf(x)dxgeq 0$)
$$left(int_0^1(f(x))^2dxright)^2leqleft(int_0^1 2xf(x)dxright)^2leq int_0^14x^2dxint_0^1(f(x))^2dx=frac43int_0^1(f(x))^2dx$$
which implies
$$int_0^1(f(x))^2dxleq frac43.$$
answered Mar 7 at 8:47
Robert ZRobert Z
100k1069140
$endgroup$
add a comment |
$begingroup$
You need to write that the condition gives $intlimits_0^1xf(x)dxgeq0$.
Otherwise, your solution is not full.
Now, by C-S $$intlimits_0^1f(x)^2dxintlimits_0^1x^2dxgeqleft(intlimits_0^1xf(x)dxright)^2geqfrac14left(intlimits_0^1f(x)^2dxright)^2$$
and since $intlimits_0^1x^2dx=frac13$, we obtain:
$$intlimits_0^1f(x)^2dxleqfrac43.$$
answered Mar 7 at 9:01
Michael RozenbergMichael Rozenberg
107k1895200
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Yes, the inequality $int_0^1(f(x))^2dxgeqfrac43$ should be reversed. Note that $f=0$ satisfies
$$2int_0^1 xf(x)dxgeqint_0^1(f(x))^2dx$$
but $int_0^1(f(x))^2dxgeqfrac43$ does not hold.
On the other hand, by Cauchy-Schwarz (note that $int_0^1 2xf(x)dxgeq 0$)
$$left(int_0^1(f(x))^2dxright)^2leqleft(int_0^1 2xf(x)dxright)^2leq int_0^14x^2dxint_0^1(f(x))^2dx=frac43int_0^1(f(x))^2dx$$
which implies
$$int_0^1(f(x))^2dxleq frac43.$$
answered Mar 7 at 8:47
Robert ZRobert Z
100k1069140
$endgroup$
add a comment |
$begingroup$
Yes, the inequality $int_0^1(f(x))^2dxgeqfrac43$ should be reversed. Note that $f=0$ satisfies
$$2int_0^1 xf(x)dxgeqint_0^1(f(x))^2dx$$
but $int_0^1(f(x))^2dxgeqfrac43$ does not hold.
On the other hand, by Cauchy-Schwarz (note that $int_0^1 2xf(x)dxgeq 0$)
$$left(int_0^1(f(x))^2dxright)^2leqleft(int_0^1 2xf(x)dxright)^2leq int_0^14x^2dxint_0^1(f(x))^2dx=frac43int_0^1(f(x))^2dx$$
which implies
$$int_0^1(f(x))^2dxleq frac43.$$
answered Mar 7 at 8:47
Robert ZRobert Z
100k1069140
$endgroup$
add a comment |
$begingroup$
Yes, the inequality $int_0^1(f(x))^2dxgeqfrac43$ should be reversed. Note that $f=0$ satisfies
$$2int_0^1 xf(x)dxgeqint_0^1(f(x))^2dx$$
but $int_0^1(f(x))^2dxgeqfrac43$ does not hold.
On the other hand, by Cauchy-Schwarz (note that $int_0^1 2xf(x)dxgeq 0$)
$$left(int_0^1(f(x))^2dxright)^2leqleft(int_0^1 2xf(x)dxright)^2leq int_0^14x^2dxint_0^1(f(x))^2dx=frac43int_0^1(f(x))^2dx$$
which implies
$$int_0^1(f(x))^2dxleq frac43.$$
answered Mar 7 at 8:47
Robert ZRobert Z
100k1069140
$endgroup$
Yes, the inequality $int_0^1(f(x))^2dxgeqfrac43$ should be reversed. Note that $f=0$ satisfies
$$2int_0^1 xf(x)dxgeqint_0^1(f(x))^2dx$$
but $int_0^1(f(x))^2dxgeqfrac43$ does not hold.
On the other hand, by Cauchy-Schwarz (note that $int_0^1 2xf(x)dxgeq 0$)
$$left(int_0^1(f(x))^2dxright)^2leqleft(int_0^1 2xf(x)dxright)^2leq int_0^14x^2dxint_0^1(f(x))^2dx=frac43int_0^1(f(x))^2dx$$
which implies
$$int_0^1(f(x))^2dxleq frac43.$$
answered Mar 7 at 8:47
Robert ZRobert Z
100k1069140
edited Mar 7 at 9:02
answered Mar 7 at 8:47
Robert ZRobert Z
100k1069140
answered Mar 7 at 8:47
Robert ZRobert Z
100k1069140
answered Mar 7 at 8:47
Robert ZRobert Z
100k1069140
100k1069140
add a comment |
add a comment |
$begingroup$
You need to write that the condition gives $intlimits_0^1xf(x)dxgeq0$.
Otherwise, your solution is not full.
Now, by C-S $$intlimits_0^1f(x)^2dxintlimits_0^1x^2dxgeqleft(intlimits_0^1xf(x)dxright)^2geqfrac14left(intlimits_0^1f(x)^2dxright)^2$$
and since $intlimits_0^1x^2dx=frac13$, we obtain:
$$intlimits_0^1f(x)^2dxleqfrac43.$$
answered Mar 7 at 9:01
Michael RozenbergMichael Rozenberg
107k1895200
$endgroup$
add a comment |
$begingroup$
You need to write that the condition gives $intlimits_0^1xf(x)dxgeq0$.
Otherwise, your solution is not full.
Now, by C-S $$intlimits_0^1f(x)^2dxintlimits_0^1x^2dxgeqleft(intlimits_0^1xf(x)dxright)^2geqfrac14left(intlimits_0^1f(x)^2dxright)^2$$
and since $intlimits_0^1x^2dx=frac13$, we obtain:
$$intlimits_0^1f(x)^2dxleqfrac43.$$
answered Mar 7 at 9:01
Michael RozenbergMichael Rozenberg
107k1895200
$endgroup$
add a comment |
$begingroup$
You need to write that the condition gives $intlimits_0^1xf(x)dxgeq0$.
Otherwise, your solution is not full.
Now, by C-S $$intlimits_0^1f(x)^2dxintlimits_0^1x^2dxgeqleft(intlimits_0^1xf(x)dxright)^2geqfrac14left(intlimits_0^1f(x)^2dxright)^2$$
and since $intlimits_0^1x^2dx=frac13$, we obtain:
$$intlimits_0^1f(x)^2dxleqfrac43.$$
answered Mar 7 at 9:01
Michael RozenbergMichael Rozenberg
107k1895200
$endgroup$
You need to write that the condition gives $intlimits_0^1xf(x)dxgeq0$.
Otherwise, your solution is not full.
Now, by C-S $$intlimits_0^1f(x)^2dxintlimits_0^1x^2dxgeqleft(intlimits_0^1xf(x)dxright)^2geqfrac14left(intlimits_0^1f(x)^2dxright)^2$$
and since $intlimits_0^1x^2dx=frac13$, we obtain:
$$intlimits_0^1f(x)^2dxleqfrac43.$$
answered Mar 7 at 9:01
Michael RozenbergMichael Rozenberg
107k1895200
edited 2 days ago
answered Mar 7 at 9:01
Michael RozenbergMichael Rozenberg
107k1895200
answered Mar 7 at 9:01
Michael RozenbergMichael Rozenberg
107k1895200
answered Mar 7 at 9:01
Michael RozenbergMichael Rozenberg
107k1895200
107k1895200
add a comment |
add a comment |
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