Mixing
Integrating derivatives over functions problem
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I had a question from a student which I'm unable to answer.
We were practicing the rule $int fracf'(x)f(x) , dx=ln(f(x))$.
A student noticed that if applied naively it gives the following two results $$int frac22x , dx = ln(2x) +c$$
and $$int frac22x , dx = int frac1x , dx = ln(x) +c$$
which appear contradictory.
The whole maths department here is stumped, so I'd be terribly grateful if anyone could point out our error, and why we can't apply this integration method in this way.
secondary-education calculus
edited 16 mins ago
Xander Henderson
2,315626
asked 1 hour ago
Paul Dean
1062
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Paul Dean is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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up vote
1
down vote
favorite
I had a question from a student which I'm unable to answer.
We were practicing the rule $int fracf'(x)f(x) , dx=ln(f(x))$.
A student noticed that if applied naively it gives the following two results $$int frac22x , dx = ln(2x) +c$$
and $$int frac22x , dx = int frac1x , dx = ln(x) +c$$
which appear contradictory.
The whole maths department here is stumped, so I'd be terribly grateful if anyone could point out our error, and why we can't apply this integration method in this way.
secondary-education calculus
edited 16 mins ago
Xander Henderson
2,315626
asked 1 hour ago
Paul Dean
1062
New contributor
Paul Dean is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
They're equivalent. Remember your basic rules of logarithms. ln(2x) = ln(2) + ln (x). Since you have an arbitrary constant of integration, it's no big deal to have an extra ln(2) hanging around. IOW, C' = ln(2) + C. Capisce?
– guest
1 hour ago
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I had a question from a student which I'm unable to answer.
We were practicing the rule $int fracf'(x)f(x) , dx=ln(f(x))$.
A student noticed that if applied naively it gives the following two results $$int frac22x , dx = ln(2x) +c$$
and $$int frac22x , dx = int frac1x , dx = ln(x) +c$$
which appear contradictory.
The whole maths department here is stumped, so I'd be terribly grateful if anyone could point out our error, and why we can't apply this integration method in this way.
secondary-education calculus
edited 16 mins ago
Xander Henderson
2,315626
asked 1 hour ago
Paul Dean
1062
New contributor
Paul Dean is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I had a question from a student which I'm unable to answer.
We were practicing the rule $int fracf'(x)f(x) , dx=ln(f(x))$.
A student noticed that if applied naively it gives the following two results $$int frac22x , dx = ln(2x) +c$$
and $$int frac22x , dx = int frac1x , dx = ln(x) +c$$
which appear contradictory.
The whole maths department here is stumped, so I'd be terribly grateful if anyone could point out our error, and why we can't apply this integration method in this way.
secondary-education calculus
secondary-education calculus
edited 16 mins ago
Xander Henderson
2,315626
asked 1 hour ago
Paul Dean
1062
New contributor
Paul Dean is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 16 mins ago
Xander Henderson
2,315626
asked 1 hour ago
Paul Dean
1062
New contributor
Paul Dean is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 16 mins ago
Xander Henderson
2,315626
edited 16 mins ago
Xander Henderson
2,315626
edited 16 mins ago
Xander Henderson
2,315626
2,315626
asked 1 hour ago
Paul Dean
1062
New contributor
Paul Dean is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
Paul Dean
1062
asked 1 hour ago
Paul Dean
1062
1062
New contributor
Paul Dean is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Paul Dean is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Paul Dean is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
They're equivalent. Remember your basic rules of logarithms. ln(2x) = ln(2) + ln (x). Since you have an arbitrary constant of integration, it's no big deal to have an extra ln(2) hanging around. IOW, C' = ln(2) + C. Capisce?
– guest
1 hour ago
add a comment |Â
1
They're equivalent. Remember your basic rules of logarithms. ln(2x) = ln(2) + ln (x). Since you have an arbitrary constant of integration, it's no big deal to have an extra ln(2) hanging around. IOW, C' = ln(2) + C. Capisce?
– guest
1 hour ago
1
1
They're equivalent. Remember your basic rules of logarithms. ln(2x) = ln(2) + ln (x). Since you have an arbitrary constant of integration, it's no big deal to have an extra ln(2) hanging around. IOW, C' = ln(2) + C. Capisce?
– guest
1 hour ago
They're equivalent. Remember your basic rules of logarithms. ln(2x) = ln(2) + ln (x). Since you have an arbitrary constant of integration, it's no big deal to have an extra ln(2) hanging around. IOW, C' = ln(2) + C. Capisce?
– guest
1 hour ago
add a comment |Â
1 Answer
1
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up vote
4
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As guest mentioned in a comment, the two expressions are equivalent. Suppose $x > 0$ so the lograrithm is defined without any problems.
For logarithms, $log(2x) = log (2) + log (x)$.
Hence, the first integral equals $log (x) + log (2) + C_1$, where $C_1$ is the constant of integration.
The second integral equals $log (x) + C_2$, where $C_2$ is the constant of integration.
To see that these are the same thing, choose $C_1 = C_2 - log(2)$.
In particular, as $C_1$ goes through all the real numbers, and as $C_2$ does so, you get precisely the same collections of functions from both integrals.
This would be a nice moment to discuss the meaning of the constant of integration, and of the indefinite integral in general. You might want to see this questions and answers for motivation: Should we avoid indefinite integrals?
This might also be useful: Explaining the symbols in definite and indefinite integrals
answered 1 hour ago
Tommi Brander
1,2211726
This is also a nice moment to plot $y=ln(x)$ and $y=ln(2x)$ to visually verify that the two functions differ by a constant. A good window would be $x=1..10$, $y=0..5$
– user52817
10 mins ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
As guest mentioned in a comment, the two expressions are equivalent. Suppose $x > 0$ so the lograrithm is defined without any problems.
For logarithms, $log(2x) = log (2) + log (x)$.
Hence, the first integral equals $log (x) + log (2) + C_1$, where $C_1$ is the constant of integration.
The second integral equals $log (x) + C_2$, where $C_2$ is the constant of integration.
To see that these are the same thing, choose $C_1 = C_2 - log(2)$.
In particular, as $C_1$ goes through all the real numbers, and as $C_2$ does so, you get precisely the same collections of functions from both integrals.
This would be a nice moment to discuss the meaning of the constant of integration, and of the indefinite integral in general. You might want to see this questions and answers for motivation: Should we avoid indefinite integrals?
This might also be useful: Explaining the symbols in definite and indefinite integrals
answered 1 hour ago
Tommi Brander
1,2211726
This is also a nice moment to plot $y=ln(x)$ and $y=ln(2x)$ to visually verify that the two functions differ by a constant. A good window would be $x=1..10$, $y=0..5$
– user52817
10 mins ago
add a comment |Â
up vote
4
down vote
As guest mentioned in a comment, the two expressions are equivalent. Suppose $x > 0$ so the lograrithm is defined without any problems.
For logarithms, $log(2x) = log (2) + log (x)$.
Hence, the first integral equals $log (x) + log (2) + C_1$, where $C_1$ is the constant of integration.
The second integral equals $log (x) + C_2$, where $C_2$ is the constant of integration.
To see that these are the same thing, choose $C_1 = C_2 - log(2)$.
In particular, as $C_1$ goes through all the real numbers, and as $C_2$ does so, you get precisely the same collections of functions from both integrals.
This would be a nice moment to discuss the meaning of the constant of integration, and of the indefinite integral in general. You might want to see this questions and answers for motivation: Should we avoid indefinite integrals?
This might also be useful: Explaining the symbols in definite and indefinite integrals
answered 1 hour ago
Tommi Brander
1,2211726
This is also a nice moment to plot $y=ln(x)$ and $y=ln(2x)$ to visually verify that the two functions differ by a constant. A good window would be $x=1..10$, $y=0..5$
– user52817
10 mins ago
add a comment |Â
up vote
4
down vote
up vote
4
down vote
As guest mentioned in a comment, the two expressions are equivalent. Suppose $x > 0$ so the lograrithm is defined without any problems.
For logarithms, $log(2x) = log (2) + log (x)$.
Hence, the first integral equals $log (x) + log (2) + C_1$, where $C_1$ is the constant of integration.
The second integral equals $log (x) + C_2$, where $C_2$ is the constant of integration.
To see that these are the same thing, choose $C_1 = C_2 - log(2)$.
In particular, as $C_1$ goes through all the real numbers, and as $C_2$ does so, you get precisely the same collections of functions from both integrals.
This would be a nice moment to discuss the meaning of the constant of integration, and of the indefinite integral in general. You might want to see this questions and answers for motivation: Should we avoid indefinite integrals?
This might also be useful: Explaining the symbols in definite and indefinite integrals
answered 1 hour ago
Tommi Brander
1,2211726
As guest mentioned in a comment, the two expressions are equivalent. Suppose $x > 0$ so the lograrithm is defined without any problems.
For logarithms, $log(2x) = log (2) + log (x)$.
Hence, the first integral equals $log (x) + log (2) + C_1$, where $C_1$ is the constant of integration.
The second integral equals $log (x) + C_2$, where $C_2$ is the constant of integration.
To see that these are the same thing, choose $C_1 = C_2 - log(2)$.
In particular, as $C_1$ goes through all the real numbers, and as $C_2$ does so, you get precisely the same collections of functions from both integrals.
This would be a nice moment to discuss the meaning of the constant of integration, and of the indefinite integral in general. You might want to see this questions and answers for motivation: Should we avoid indefinite integrals?
This might also be useful: Explaining the symbols in definite and indefinite integrals
answered 1 hour ago
Tommi Brander
1,2211726
answered 1 hour ago
Tommi Brander
1,2211726
answered 1 hour ago
Tommi Brander
1,2211726
answered 1 hour ago
Tommi Brander
1,2211726
1,2211726
This is also a nice moment to plot $y=ln(x)$ and $y=ln(2x)$ to visually verify that the two functions differ by a constant. A good window would be $x=1..10$, $y=0..5$
– user52817
10 mins ago
add a comment |Â
This is also a nice moment to plot $y=ln(x)$ and $y=ln(2x)$ to visually verify that the two functions differ by a constant. A good window would be $x=1..10$, $y=0..5$
– user52817
10 mins ago
This is also a nice moment to plot $y=ln(x)$ and $y=ln(2x)$ to visually verify that the two functions differ by a constant. A good window would be $x=1..10$, $y=0..5$
– user52817
10 mins ago
This is also a nice moment to plot $y=ln(x)$ and $y=ln(2x)$ to visually verify that the two functions differ by a constant. A good window would be $x=1..10$, $y=0..5$
– user52817
10 mins ago
add a comment |Â
Paul Dean is a new contributor. Be nice, and check out our Code of Conduct.
Paul Dean is a new contributor. Be nice, and check out our Code of Conduct.
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1
They're equivalent. Remember your basic rules of logarithms. ln(2x) = ln(2) + ln (x). Since you have an arbitrary constant of integration, it's no big deal to have an extra ln(2) hanging around. IOW, C' = ln(2) + C. Capisce?
– guest
1 hour ago