Mixing
Definite integral with interval depends on $n$!
$begingroup$
The following that
$$int_0^1 f(x) ,mathrm dx=lim_{nrightarrowinfty}sum_{k=1}^{n-1}f(frac{k}{n})frac{1}{n} , $$
is well-known fact! But if $$lim_{nrightarrowinfty}frac{r_n}{n}=alpha,$$
then is it true that $$int_0^{alpha} f(x) ,mathrm dx=lim_{nrightarrowinfty}sum_{k=1}^{r_n-1}f(frac{k}{n})frac{1}{n} , $$?
If this is true, why? Help me with big mercy!!
calculus definite-integrals
asked Jan 29 at 14:06
Hs PHs P
275
$endgroup$
add a comment |
$begingroup$
The following that
$$int_0^1 f(x) ,mathrm dx=lim_{nrightarrowinfty}sum_{k=1}^{n-1}f(frac{k}{n})frac{1}{n} , $$
is well-known fact! But if $$lim_{nrightarrowinfty}frac{r_n}{n}=alpha,$$
then is it true that $$int_0^{alpha} f(x) ,mathrm dx=lim_{nrightarrowinfty}sum_{k=1}^{r_n-1}f(frac{k}{n})frac{1}{n} , $$?
If this is true, why? Help me with big mercy!!
calculus definite-integrals
asked Jan 29 at 14:06
Hs PHs P
275
$endgroup$
$begingroup$
Substitute $y=x/alpha$, then define $g(y):=f(yalpha)$.
$endgroup$
– J.G.
Jan 29 at 20:33
add a comment |
$begingroup$
The following that
$$int_0^1 f(x) ,mathrm dx=lim_{nrightarrowinfty}sum_{k=1}^{n-1}f(frac{k}{n})frac{1}{n} , $$
is well-known fact! But if $$lim_{nrightarrowinfty}frac{r_n}{n}=alpha,$$
then is it true that $$int_0^{alpha} f(x) ,mathrm dx=lim_{nrightarrowinfty}sum_{k=1}^{r_n-1}f(frac{k}{n})frac{1}{n} , $$?
If this is true, why? Help me with big mercy!!
calculus definite-integrals
asked Jan 29 at 14:06
Hs PHs P
275
$endgroup$
The following that
$$int_0^1 f(x) ,mathrm dx=lim_{nrightarrowinfty}sum_{k=1}^{n-1}f(frac{k}{n})frac{1}{n} , $$
is well-known fact! But if $$lim_{nrightarrowinfty}frac{r_n}{n}=alpha,$$
then is it true that $$int_0^{alpha} f(x) ,mathrm dx=lim_{nrightarrowinfty}sum_{k=1}^{r_n-1}f(frac{k}{n})frac{1}{n} , $$?
If this is true, why? Help me with big mercy!!
calculus definite-integrals
calculus definite-integrals
asked Jan 29 at 14:06
Hs PHs P
275
asked Jan 29 at 14:06
Hs PHs P
275
asked Jan 29 at 14:06
Hs PHs P
275
asked Jan 29 at 14:06
Hs PHs P
275
asked Jan 29 at 14:06
Hs PHs P
275
275
$begingroup$
Substitute $y=x/alpha$, then define $g(y):=f(yalpha)$.
$endgroup$
– J.G.
Jan 29 at 20:33
add a comment |
$begingroup$
Substitute $y=x/alpha$, then define $g(y):=f(yalpha)$.
$endgroup$
– J.G.
Jan 29 at 20:33
$begingroup$
Substitute $y=x/alpha$, then define $g(y):=f(yalpha)$.
$endgroup$
– J.G.
Jan 29 at 20:33
$begingroup$
Substitute $y=x/alpha$, then define $g(y):=f(yalpha)$.
$endgroup$
– J.G.
Jan 29 at 20:33
add a comment |
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$begingroup$
Substitute $y=x/alpha$, then define $g(y):=f(yalpha)$.
$endgroup$
– J.G.
Jan 29 at 20:33