Mixing
Prove the following inequality: $frac{1}{20sqrt{2}}<int_0^1 frac{x^{19}}{sqrt{1+x^2}}dx<frac1{20}$
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I'm having difficulty with proving this inequality: $$frac{1}{20sqrt{2}}<int_0^1 frac{x^{19}}{sqrt{1+x^2}}dx<frac1{20}$$
I thought this task should be fairly easy to prove with the following theorem:
If $f(x)$ and $g(x)$ are Riemann-integrable and $f(x) le g(x)$ for each $x$ in $[a,b]$ then $int_a^b f(x)dx le int_a^bg(x)dx$
Especially with the specific version of the above lemma:
If $f(x)$ is Riemann-integrable then it is necessarily bounded on this interval. Thus there are real numbers m and M such that $m le f(x) le M$ for all $x$ in $[a,b]$. Since the lower and upper sums of $f$ over $[a,b]$ are therfore bounded by $m(b-a)$ and $M(b-a)$ we have $m(b-a)le int_a^b f(x)dx le M(b-a)$
However, this bound seems to be too weak. ($frac{x^{19}}{sqrt{1+x^2}}<frac1{20}$ doesn't hold for all $x$ in $[0,1]$) Any help is appreciated. Thanks.
integration inequality definite-integrals
asked Feb 2 at 19:15
Rafał SzypulskiRafał Szypulski
35015
$endgroup$
add a comment |
$begingroup$
I'm having difficulty with proving this inequality: $$frac{1}{20sqrt{2}}<int_0^1 frac{x^{19}}{sqrt{1+x^2}}dx<frac1{20}$$
I thought this task should be fairly easy to prove with the following theorem:
If $f(x)$ and $g(x)$ are Riemann-integrable and $f(x) le g(x)$ for each $x$ in $[a,b]$ then $int_a^b f(x)dx le int_a^bg(x)dx$
Especially with the specific version of the above lemma:
If $f(x)$ is Riemann-integrable then it is necessarily bounded on this interval. Thus there are real numbers m and M such that $m le f(x) le M$ for all $x$ in $[a,b]$. Since the lower and upper sums of $f$ over $[a,b]$ are therfore bounded by $m(b-a)$ and $M(b-a)$ we have $m(b-a)le int_a^b f(x)dx le M(b-a)$
However, this bound seems to be too weak. ($frac{x^{19}}{sqrt{1+x^2}}<frac1{20}$ doesn't hold for all $x$ in $[0,1]$) Any help is appreciated. Thanks.
integration inequality definite-integrals
asked Feb 2 at 19:15
Rafał SzypulskiRafał Szypulski
35015
$endgroup$
add a comment |
$begingroup$
I'm having difficulty with proving this inequality: $$frac{1}{20sqrt{2}}<int_0^1 frac{x^{19}}{sqrt{1+x^2}}dx<frac1{20}$$
I thought this task should be fairly easy to prove with the following theorem:
If $f(x)$ and $g(x)$ are Riemann-integrable and $f(x) le g(x)$ for each $x$ in $[a,b]$ then $int_a^b f(x)dx le int_a^bg(x)dx$
Especially with the specific version of the above lemma:
If $f(x)$ is Riemann-integrable then it is necessarily bounded on this interval. Thus there are real numbers m and M such that $m le f(x) le M$ for all $x$ in $[a,b]$. Since the lower and upper sums of $f$ over $[a,b]$ are therfore bounded by $m(b-a)$ and $M(b-a)$ we have $m(b-a)le int_a^b f(x)dx le M(b-a)$
However, this bound seems to be too weak. ($frac{x^{19}}{sqrt{1+x^2}}<frac1{20}$ doesn't hold for all $x$ in $[0,1]$) Any help is appreciated. Thanks.
integration inequality definite-integrals
asked Feb 2 at 19:15
Rafał SzypulskiRafał Szypulski
35015
$endgroup$
I'm having difficulty with proving this inequality: $$frac{1}{20sqrt{2}}<int_0^1 frac{x^{19}}{sqrt{1+x^2}}dx<frac1{20}$$
I thought this task should be fairly easy to prove with the following theorem:
If $f(x)$ and $g(x)$ are Riemann-integrable and $f(x) le g(x)$ for each $x$ in $[a,b]$ then $int_a^b f(x)dx le int_a^bg(x)dx$
Especially with the specific version of the above lemma:
If $f(x)$ is Riemann-integrable then it is necessarily bounded on this interval. Thus there are real numbers m and M such that $m le f(x) le M$ for all $x$ in $[a,b]$. Since the lower and upper sums of $f$ over $[a,b]$ are therfore bounded by $m(b-a)$ and $M(b-a)$ we have $m(b-a)le int_a^b f(x)dx le M(b-a)$
However, this bound seems to be too weak. ($frac{x^{19}}{sqrt{1+x^2}}<frac1{20}$ doesn't hold for all $x$ in $[0,1]$) Any help is appreciated. Thanks.
integration inequality definite-integrals
integration inequality definite-integrals
asked Feb 2 at 19:15
Rafał SzypulskiRafał Szypulski
35015
asked Feb 2 at 19:15
Rafał SzypulskiRafał Szypulski
35015
asked Feb 2 at 19:15
Rafał SzypulskiRafał Szypulski
35015
asked Feb 2 at 19:15
Rafał SzypulskiRafał Szypulski
35015
asked Feb 2 at 19:15
Rafał SzypulskiRafał Szypulski
35015
35015
add a comment |
add a comment |
1 Answer
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$begingroup$
HINT:
If $xin[0,1],$ then $$frac{x^{19}}{sqrt{2}}leqfrac{x^{19}}{sqrt{1+x^2}}leq x^{19}$$
answered Feb 2 at 19:19
user376343user376343
3,9834829
$endgroup$
1
$begingroup$
Oh my, how come have I not looked at it this way. Thank you so much!
$endgroup$
– Rafał Szypulski
Feb 2 at 19:22
add a comment |
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1 Answer
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1 Answer
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active
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$begingroup$
HINT:
If $xin[0,1],$ then $$frac{x^{19}}{sqrt{2}}leqfrac{x^{19}}{sqrt{1+x^2}}leq x^{19}$$
answered Feb 2 at 19:19
user376343user376343
3,9834829
$endgroup$
1
$begingroup$
Oh my, how come have I not looked at it this way. Thank you so much!
$endgroup$
– Rafał Szypulski
Feb 2 at 19:22
add a comment |
$begingroup$
HINT:
If $xin[0,1],$ then $$frac{x^{19}}{sqrt{2}}leqfrac{x^{19}}{sqrt{1+x^2}}leq x^{19}$$
answered Feb 2 at 19:19
user376343user376343
3,9834829
$endgroup$
1
$begingroup$
Oh my, how come have I not looked at it this way. Thank you so much!
$endgroup$
– Rafał Szypulski
Feb 2 at 19:22
add a comment |
$begingroup$
HINT:
If $xin[0,1],$ then $$frac{x^{19}}{sqrt{2}}leqfrac{x^{19}}{sqrt{1+x^2}}leq x^{19}$$
answered Feb 2 at 19:19
user376343user376343
3,9834829
$endgroup$
HINT:
If $xin[0,1],$ then $$frac{x^{19}}{sqrt{2}}leqfrac{x^{19}}{sqrt{1+x^2}}leq x^{19}$$
answered Feb 2 at 19:19
user376343user376343
3,9834829
answered Feb 2 at 19:19
user376343user376343
3,9834829
answered Feb 2 at 19:19
user376343user376343
3,9834829
answered Feb 2 at 19:19
user376343user376343
3,9834829
3,9834829
1
$begingroup$
Oh my, how come have I not looked at it this way. Thank you so much!
$endgroup$
– Rafał Szypulski
Feb 2 at 19:22
add a comment |
1
$begingroup$
Oh my, how come have I not looked at it this way. Thank you so much!
$endgroup$
– Rafał Szypulski
Feb 2 at 19:22
1
1
$begingroup$
Oh my, how come have I not looked at it this way. Thank you so much!
$endgroup$
– Rafał Szypulski
Feb 2 at 19:22
$begingroup$
Oh my, how come have I not looked at it this way. Thank you so much!
$endgroup$
– Rafał Szypulski
Feb 2 at 19:22
add a comment |
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