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Proving definite integrals fit in numerical constraints
Two Multiple IntegralsDifference of definite integrals inequalityDefinite integration problem (trig).Evaluate the definite integral.How to find a bound for these (simple) integralsConcentration of sequence of integralsLimits of a definite integrals (Demidovich)Proving an inequality with $arcsin$Help with this Advanced Definite Integralproving definite integral value by induction
$begingroup$
The question I have been given is
Prove that $$frac12<int_0^frac12 frac1sqrt1-x^2n dxleq 0.52359$$
for any integer $ngeq 1$, and that $$frac12<int_0^1 frac1sqrt4-x+x^3 dxleq 0.52359$$
I can show that it works for $n=1,2$ but I don't know how to approach proving this for all $ngeq 1$. I also can't see how the two equations relate. I have asked my professor and he did not know what to do.
If there is any direction you could point me in or any part of the proof you could show I would really appreciate it.
Thank you!
calculus definite-integrals approximation
asked Mar 12 at 11:50
tjsptjsp
82
New contributor
tjsp is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
The question I have been given is
Prove that $$frac12<int_0^frac12 frac1sqrt1-x^2n dxleq 0.52359$$
for any integer $ngeq 1$, and that $$frac12<int_0^1 frac1sqrt4-x+x^3 dxleq 0.52359$$
I can show that it works for $n=1,2$ but I don't know how to approach proving this for all $ngeq 1$. I also can't see how the two equations relate. I have asked my professor and he did not know what to do.
If there is any direction you could point me in or any part of the proof you could show I would really appreciate it.
Thank you!
calculus definite-integrals approximation
asked Mar 12 at 11:50
tjsptjsp
82
New contributor
tjsp is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
2
$begingroup$
The sequence $a_n=int_0^1/2 frac1sqrt1-x^2ndx$ is decreasing in $n$, so you only need to check it for $n=1$.
$endgroup$
– Song
Mar 12 at 11:59
add a comment |
$begingroup$
The question I have been given is
Prove that $$frac12<int_0^frac12 frac1sqrt1-x^2n dxleq 0.52359$$
for any integer $ngeq 1$, and that $$frac12<int_0^1 frac1sqrt4-x+x^3 dxleq 0.52359$$
I can show that it works for $n=1,2$ but I don't know how to approach proving this for all $ngeq 1$. I also can't see how the two equations relate. I have asked my professor and he did not know what to do.
If there is any direction you could point me in or any part of the proof you could show I would really appreciate it.
Thank you!
calculus definite-integrals approximation
asked Mar 12 at 11:50
tjsptjsp
82
New contributor
tjsp is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
The question I have been given is
Prove that $$frac12<int_0^frac12 frac1sqrt1-x^2n dxleq 0.52359$$
for any integer $ngeq 1$, and that $$frac12<int_0^1 frac1sqrt4-x+x^3 dxleq 0.52359$$
I can show that it works for $n=1,2$ but I don't know how to approach proving this for all $ngeq 1$. I also can't see how the two equations relate. I have asked my professor and he did not know what to do.
If there is any direction you could point me in or any part of the proof you could show I would really appreciate it.
Thank you!
calculus definite-integrals approximation
calculus definite-integrals approximation
asked Mar 12 at 11:50
tjsptjsp
82
New contributor
tjsp is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 12 at 11:50
tjsptjsp
82
New contributor
tjsp is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 12 at 11:50
tjsptjsp
82
New contributor
tjsp is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 12 at 11:50
tjsptjsp
82
asked Mar 12 at 11:50
tjsptjsp
82
82
New contributor
tjsp is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
tjsp is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
tjsp is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
2
$begingroup$
The sequence $a_n=int_0^1/2 frac1sqrt1-x^2ndx$ is decreasing in $n$, so you only need to check it for $n=1$.
$endgroup$
– Song
Mar 12 at 11:59
add a comment |
2
$begingroup$
The sequence $a_n=int_0^1/2 frac1sqrt1-x^2ndx$ is decreasing in $n$, so you only need to check it for $n=1$.
$endgroup$
– Song
Mar 12 at 11:59
2
2
$begingroup$
The sequence $a_n=int_0^1/2 frac1sqrt1-x^2ndx$ is decreasing in $n$, so you only need to check it for $n=1$.
$endgroup$
– Song
Mar 12 at 11:59
$begingroup$
The sequence $a_n=int_0^1/2 frac1sqrt1-x^2ndx$ is decreasing in $n$, so you only need to check it for $n=1$.
$endgroup$
– Song
Mar 12 at 11:59
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
For the first integral:
That $0.52359$ constant is supposed to be $frac pi6$. Thats the integral when $n=1$.
The integrand is decreasing on $(0,frac 12)$, and it's graph forms something very close to a rectangle with the lines $x=0$ and $x=1$. This can be seen since $frac11-(0)^2n=1$ regardless the value of $n$, and $frac11-(frac 12)^2n$ decreases from $frac 43$ to just over $1$ as $n$ increases, and the function is continuous in the interim.
answered Mar 12 at 12:12
Rhys HughesRhys Hughes
7,0301630
$endgroup$
$begingroup$
Thank you so much, I've proved it now!
$endgroup$
– tjsp
Mar 12 at 13:20
add a comment |
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1 Answer
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1 Answer
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$begingroup$
For the first integral:
That $0.52359$ constant is supposed to be $frac pi6$. Thats the integral when $n=1$.
The integrand is decreasing on $(0,frac 12)$, and it's graph forms something very close to a rectangle with the lines $x=0$ and $x=1$. This can be seen since $frac11-(0)^2n=1$ regardless the value of $n$, and $frac11-(frac 12)^2n$ decreases from $frac 43$ to just over $1$ as $n$ increases, and the function is continuous in the interim.
answered Mar 12 at 12:12
Rhys HughesRhys Hughes
7,0301630
$endgroup$
$begingroup$
Thank you so much, I've proved it now!
$endgroup$
– tjsp
Mar 12 at 13:20
add a comment |
$begingroup$
For the first integral:
That $0.52359$ constant is supposed to be $frac pi6$. Thats the integral when $n=1$.
The integrand is decreasing on $(0,frac 12)$, and it's graph forms something very close to a rectangle with the lines $x=0$ and $x=1$. This can be seen since $frac11-(0)^2n=1$ regardless the value of $n$, and $frac11-(frac 12)^2n$ decreases from $frac 43$ to just over $1$ as $n$ increases, and the function is continuous in the interim.
answered Mar 12 at 12:12
Rhys HughesRhys Hughes
7,0301630
$endgroup$
$begingroup$
Thank you so much, I've proved it now!
$endgroup$
– tjsp
Mar 12 at 13:20
add a comment |
$begingroup$
For the first integral:
That $0.52359$ constant is supposed to be $frac pi6$. Thats the integral when $n=1$.
The integrand is decreasing on $(0,frac 12)$, and it's graph forms something very close to a rectangle with the lines $x=0$ and $x=1$. This can be seen since $frac11-(0)^2n=1$ regardless the value of $n$, and $frac11-(frac 12)^2n$ decreases from $frac 43$ to just over $1$ as $n$ increases, and the function is continuous in the interim.
answered Mar 12 at 12:12
Rhys HughesRhys Hughes
7,0301630
$endgroup$
For the first integral:
That $0.52359$ constant is supposed to be $frac pi6$. Thats the integral when $n=1$.
The integrand is decreasing on $(0,frac 12)$, and it's graph forms something very close to a rectangle with the lines $x=0$ and $x=1$. This can be seen since $frac11-(0)^2n=1$ regardless the value of $n$, and $frac11-(frac 12)^2n$ decreases from $frac 43$ to just over $1$ as $n$ increases, and the function is continuous in the interim.
answered Mar 12 at 12:12
Rhys HughesRhys Hughes
7,0301630
answered Mar 12 at 12:12
Rhys HughesRhys Hughes
7,0301630
answered Mar 12 at 12:12
Rhys HughesRhys Hughes
7,0301630
answered Mar 12 at 12:12
Rhys HughesRhys Hughes
7,0301630
7,0301630
$begingroup$
Thank you so much, I've proved it now!
$endgroup$
– tjsp
Mar 12 at 13:20
add a comment |
$begingroup$
Thank you so much, I've proved it now!
$endgroup$
– tjsp
Mar 12 at 13:20
$begingroup$
Thank you so much, I've proved it now!
$endgroup$
– tjsp
Mar 12 at 13:20
$begingroup$
Thank you so much, I've proved it now!
$endgroup$
– tjsp
Mar 12 at 13:20
add a comment |
tjsp is a new contributor. Be nice, and check out our Code of Conduct.
tjsp is a new contributor. Be nice, and check out our Code of Conduct.
tjsp is a new contributor. Be nice, and check out our Code of Conduct.
tjsp is a new contributor. Be nice, and check out our Code of Conduct.
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2
$begingroup$
The sequence $a_n=int_0^1/2 frac1sqrt1-x^2ndx$ is decreasing in $n$, so you only need to check it for $n=1$.
$endgroup$
– Song
Mar 12 at 11:59