Change of variables in integral: the limits of $y = int_{sigma_0}^{R_M} frac{dR}{R} $
$begingroup$
I have this equation:
$$y = int_{sigma_0}^{R_M} frac{dR}{R} $$
where $x = R/sigma_0$
If I want to do a change of variables, I would have:
$$y = int frac{sigma_0}{xsigma_0} $$
$$y = int frac{1}{x} $$
My question is: What do the limits become?
integration change-of-variable
asked Jan 9 at 20:34
Jackson HartJackson Hart
5052626
$endgroup$
add a comment |
$begingroup$
I have this equation:
$$y = int_{sigma_0}^{R_M} frac{dR}{R} $$
where $x = R/sigma_0$
If I want to do a change of variables, I would have:
$$y = int frac{sigma_0}{xsigma_0} $$
$$y = int frac{1}{x} $$
My question is: What do the limits become?
integration change-of-variable
asked Jan 9 at 20:34
Jackson HartJackson Hart
5052626
$endgroup$
add a comment |
$begingroup$
I have this equation:
$$y = int_{sigma_0}^{R_M} frac{dR}{R} $$
where $x = R/sigma_0$
If I want to do a change of variables, I would have:
$$y = int frac{sigma_0}{xsigma_0} $$
$$y = int frac{1}{x} $$
My question is: What do the limits become?
integration change-of-variable
asked Jan 9 at 20:34
Jackson HartJackson Hart
5052626
$endgroup$
I have this equation:
$$y = int_{sigma_0}^{R_M} frac{dR}{R} $$
where $x = R/sigma_0$
If I want to do a change of variables, I would have:
$$y = int frac{sigma_0}{xsigma_0} $$
$$y = int frac{1}{x} $$
My question is: What do the limits become?
integration change-of-variable
integration change-of-variable
asked Jan 9 at 20:34
Jackson HartJackson Hart
5052626
asked Jan 9 at 20:34
Jackson HartJackson Hart
5052626
edited Jan 9 at 21:02
Jackson Hart
asked Jan 9 at 20:34
Jackson HartJackson Hart
5052626
asked Jan 9 at 20:34
Jackson HartJackson Hart
5052626
asked Jan 9 at 20:34
Jackson HartJackson Hart
5052626
5052626
add a comment |
add a comment |
2 Answers
2
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oldest
votes
$begingroup$
$R; $ becomes $; xsigma_0$.
$dR; $ becomes $; sigma_0dx$.
$sigma_0; $ changes to $; frac{sigma_0}{sigma_0}$.
and
$R_M$ will become $frac{R_M}{sigma_0}$.
the integral is then
$$int_1^{frac{R_M}{sigma_0}}frac{sigma_0 dx}{xsigma_0}=$$
$$int_1^{frac{R_M}{sigma_0}}frac{dx}{x}.$$
answered Jan 9 at 20:53
hamam_Abdallahhamam_Abdallah
38k21634
$endgroup$
$begingroup$
Sorry I made a mistake. $R=xsigma_0$. Could you fix the answer?
$endgroup$
– Jackson Hart
Jan 9 at 21:02
add a comment |
$begingroup$
$R=sigma _0$ implies that $ x= frac {sigma_0}{sigma}$
$R=R_M $implies that $x=frac {R_M}{sigma}$
Thus the new bounds are the old ones divided by $sigma$
answered Jan 9 at 21:03
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.5k42061
$endgroup$
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
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active
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active
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votes
$begingroup$
$R; $ becomes $; xsigma_0$.
$dR; $ becomes $; sigma_0dx$.
$sigma_0; $ changes to $; frac{sigma_0}{sigma_0}$.
and
$R_M$ will become $frac{R_M}{sigma_0}$.
the integral is then
$$int_1^{frac{R_M}{sigma_0}}frac{sigma_0 dx}{xsigma_0}=$$
$$int_1^{frac{R_M}{sigma_0}}frac{dx}{x}.$$
answered Jan 9 at 20:53
hamam_Abdallahhamam_Abdallah
38k21634
$endgroup$
$begingroup$
Sorry I made a mistake. $R=xsigma_0$. Could you fix the answer?
$endgroup$
– Jackson Hart
Jan 9 at 21:02
add a comment |
$begingroup$
$R; $ becomes $; xsigma_0$.
$dR; $ becomes $; sigma_0dx$.
$sigma_0; $ changes to $; frac{sigma_0}{sigma_0}$.
and
$R_M$ will become $frac{R_M}{sigma_0}$.
the integral is then
$$int_1^{frac{R_M}{sigma_0}}frac{sigma_0 dx}{xsigma_0}=$$
$$int_1^{frac{R_M}{sigma_0}}frac{dx}{x}.$$
answered Jan 9 at 20:53
hamam_Abdallahhamam_Abdallah
38k21634
$endgroup$
$begingroup$
Sorry I made a mistake. $R=xsigma_0$. Could you fix the answer?
$endgroup$
– Jackson Hart
Jan 9 at 21:02
add a comment |
$begingroup$
$R; $ becomes $; xsigma_0$.
$dR; $ becomes $; sigma_0dx$.
$sigma_0; $ changes to $; frac{sigma_0}{sigma_0}$.
and
$R_M$ will become $frac{R_M}{sigma_0}$.
the integral is then
$$int_1^{frac{R_M}{sigma_0}}frac{sigma_0 dx}{xsigma_0}=$$
$$int_1^{frac{R_M}{sigma_0}}frac{dx}{x}.$$
answered Jan 9 at 20:53
hamam_Abdallahhamam_Abdallah
38k21634
$endgroup$
$R; $ becomes $; xsigma_0$.
$dR; $ becomes $; sigma_0dx$.
$sigma_0; $ changes to $; frac{sigma_0}{sigma_0}$.
and
$R_M$ will become $frac{R_M}{sigma_0}$.
the integral is then
$$int_1^{frac{R_M}{sigma_0}}frac{sigma_0 dx}{xsigma_0}=$$
$$int_1^{frac{R_M}{sigma_0}}frac{dx}{x}.$$
answered Jan 9 at 20:53
hamam_Abdallahhamam_Abdallah
38k21634
edited Jan 9 at 21:07
answered Jan 9 at 20:53
hamam_Abdallahhamam_Abdallah
38k21634
answered Jan 9 at 20:53
hamam_Abdallahhamam_Abdallah
38k21634
answered Jan 9 at 20:53
hamam_Abdallahhamam_Abdallah
38k21634
38k21634
$begingroup$
Sorry I made a mistake. $R=xsigma_0$. Could you fix the answer?
$endgroup$
– Jackson Hart
Jan 9 at 21:02
add a comment |
$begingroup$
Sorry I made a mistake. $R=xsigma_0$. Could you fix the answer?
$endgroup$
– Jackson Hart
Jan 9 at 21:02
$begingroup$
Sorry I made a mistake. $R=xsigma_0$. Could you fix the answer?
$endgroup$
– Jackson Hart
Jan 9 at 21:02
$begingroup$
Sorry I made a mistake. $R=xsigma_0$. Could you fix the answer?
$endgroup$
– Jackson Hart
Jan 9 at 21:02
add a comment |
$begingroup$
$R=sigma _0$ implies that $ x= frac {sigma_0}{sigma}$
$R=R_M $implies that $x=frac {R_M}{sigma}$
Thus the new bounds are the old ones divided by $sigma$
answered Jan 9 at 21:03
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.5k42061
$endgroup$
add a comment |
$begingroup$
$R=sigma _0$ implies that $ x= frac {sigma_0}{sigma}$
$R=R_M $implies that $x=frac {R_M}{sigma}$
Thus the new bounds are the old ones divided by $sigma$
answered Jan 9 at 21:03
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.5k42061
$endgroup$
add a comment |
$begingroup$
$R=sigma _0$ implies that $ x= frac {sigma_0}{sigma}$
$R=R_M $implies that $x=frac {R_M}{sigma}$
Thus the new bounds are the old ones divided by $sigma$
answered Jan 9 at 21:03
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.5k42061
$endgroup$
$R=sigma _0$ implies that $ x= frac {sigma_0}{sigma}$
$R=R_M $implies that $x=frac {R_M}{sigma}$
Thus the new bounds are the old ones divided by $sigma$
answered Jan 9 at 21:03
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.5k42061
answered Jan 9 at 21:03
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.5k42061
answered Jan 9 at 21:03
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.5k42061
answered Jan 9 at 21:03
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.5k42061
41.5k42061
add a comment |
add a comment |
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